text:
(* ========================================================================= *)
(* Mizar Light II                                                            *)
(*                                                                           *)
(*                   Freek Wiedijk, University of Nijmegen                   *)
(* ========================================================================= *)

type mterm = string;;

let parse_context_term s env =
  let ptm,l = (parse_preterm o lex o explode) s in
  if l = [] then
   (term_of_preterm o retypecheck
     (map ((fun (s,ty) -> s,pretype_of_type ty) o dest_var) env)) ptm
  else failwith "Unexpected junk after term";;

let goal_frees (asl,w as g) =
  frees (itlist (curry mk_imp) (map (concl o snd) asl) w);;

let (parse_mterm: mterm -> goal -> term) =
  let ttm = mk_var("thesis",bool_ty) in
  let atm = mk_var("antecedent",bool_ty) in
  let otm = mk_var("opposite",bool_ty) in
  fun s (asl,w as g) ->
    let ant = try fst (dest_imp w) with Failure _ -> atm in
    let opp = try dest_neg w with Failure _ -> mk_neg w in
    let t =
     (subst [w,ttm; ant,atm; opp,otm]
       (parse_context_term s ((goal_frees g) @ [ttm; atm; otm]))) in
    try
      let lhs = lhand (concl (snd (hd asl))) in
      let itm = mk_var("...",type_of lhs) in
      subst [lhs,itm] t
    with Failure _ -> t;;

type stepinfo =
  (goal -> term) option * int option *
  (goal -> thm list) * (thm list -> tactic);;

type step = (stepinfo -> tactic) * stepinfo;;

let TRY' tac thl = TRY (tac thl);;

let (then'_) = fun tac1 tac2 thl -> tac1 thl THEN tac2 thl;;

let standard_prover = TRY' (REWRITE_TAC THEN' MESON_TAC);;
let sketch_prover = K CHEAT_TAC;;
let current_prover = ref standard_prover;;

let (default_stepinfo: (goal -> term) option -> stepinfo) =
  fun t -> t,None,
    (map snd o filter ((=) "=" o fst) o fst),
    (fun thl -> !current_prover thl);;

let ((st'): step -> (goal -> term)  -> step) =
  fun (tac,(t,l,thl,just)) t' -> (tac,(Some t',l,thl,just));;

let (st) = fun stp -> (st') stp o parse_mterm;;

let (((at)): step -> int -> step) =
  fun (tac,(t,l,thl,just)) l' -> (tac,(t,Some l',thl,just));;

let (((from)): step -> int list -> step) =
  fun (tac,(t,l,thl,just)) ll -> (tac,(t,l,
    (fun (asl,_ as g) -> thl g @
      let n = length asl in
      map
        (fun l ->
          if l < 0 then snd (el ((n - l - 1) mod n) asl)
            else assoc (string_of_int l) asl)
        ll),
    just));;

let so x = fun y -> x y from [-1];;

let (((by)): step -> thm list -> step) =
  fun (tac,(t,l,thl,just)) thl' -> (tac,(t,l,(fun g -> thl g @ thl'),just));;

let (((using)): step -> (thm list -> tactic) -> step) =
  fun (tac,(t,l,thl,just)) just' -> (tac,(t,l,thl,just' THEN' just));;

let (step: step -> tactic) = fun (f,x) -> f x;;

let (steps: step list -> tactic) =
  fun stpl ->
    itlist (fun tac1 tac2 -> tac1 THENL [tac2]) (map step stpl) ALL_TAC;;

let (tactics: tactic list -> step) =
  fun tacl -> ((K (itlist ((THEN)) tacl ALL_TAC)),
    default_stepinfo None);;

let (theorem': (goal -> term) -> step list -> thm) =
  let g = ([],`T`) in
  fun t stpl -> prove(t g,steps stpl);;

let (((proof)): step -> step list -> step) =
  fun (tac,(t,l,thl,just)) prf -> (tac,(t,l,thl,K (steps prf)));;

let (N_ASSUME_TAC: int option -> thm_tactic) =
  fun l th (asl,_ as g) ->
    match l with
      None -> ASSUME_TAC th g
    | Some n ->
        warn (n >= 0 && length asl <> n) "*** out of sequence label ***";
        LABEL_TAC (if n < 0 then "=" else string_of_int n) th g;;

let (per: step -> step list list -> step) =
  let F = `F` in
  fun (_,(_,_,thl,just)) cases ->
    (fun (_,_,thl',just') g ->
       let tx (t',_,_,_) =
         match t' with None -> failwith "per" | Some t -> t g in
       let dj = itlist (curry mk_disj)
         (map (tx o snd o hd) cases) F in
       (SUBGOAL_THEN dj
          (EVERY_TCL (map (fun case -> let _,l,_,_ = snd (hd case) in
            (DISJ_CASES_THEN2 (N_ASSUME_TAC l))) cases) CONTR_TAC) THENL
        ([(just' THEN' just) ((thl' g) @ (thl g))] @
           map (steps o tl) cases)) g),
    (None,None,(K []),(K ALL_TAC));;

let (cases: step) =
  (fun _ -> failwith "cases"),default_stepinfo None;;

let (suppose': (goal -> term) -> step) =
  fun t -> (fun _ -> failwith "suppose"),default_stepinfo (Some t);;

let (consider': (goal -> term) list -> step) =
  let T = `T` in
  fun tl' ->
  (fun (t',l,thl,just) (asl,w as g) ->
    let tl = map (fun t' -> t' g) tl' in
    let g' = ((asl @ (map (fun t -> ("",REFL t)) tl)),w) in
    let ex = itlist (curry mk_exists) tl
      (match t' with
         None -> failwith "consider"
       | Some t -> t g') in
    (SUBGOAL_THEN ex
       ((EVERY_TCL (map X_CHOOSE_THEN tl)) (N_ASSUME_TAC l)) THENL
     [just (thl g); ALL_TAC]) g),
  default_stepinfo (Some
    (fun g -> end_itlist (curry mk_conj)
      (map (fun t' -> let t = t' g in mk_eq(t,t)) tl')));;

let (take': (goal -> term) list -> step) =
  fun tl ->
    (fun _ g -> (MAP_EVERY EXISTS_TAC o map (fun t -> t g)) tl g),
    default_stepinfo None;;

let (assume': (goal -> term) -> step) =
  fun t ->
    (fun (t',l,thl,just) g ->
       match t' with
         None -> failwith "assume"
       | Some t ->
           (DISJ_CASES_THEN2
              (fun th -> REWRITE_TAC[th] THEN
                 N_ASSUME_TAC l th)
              (fun th -> just ((REWRITE_RULE[] th)::(thl g)))
            (SPEC (t g) EXCLUDED_MIDDLE)) g),
    default_stepinfo (Some t);;

let (have': (goal -> term) -> step) =
  fun t ->
    (fun (t',l,thl,just) g ->
       match t' with
         None -> failwith "have"
       | Some t ->
           (SUBGOAL_THEN (t g) (N_ASSUME_TAC l) THENL
            [just (thl g); ALL_TAC]) g),
    default_stepinfo (Some t);;

let (thus': (goal -> term) -> step) =
  fun t ->
    (fun (t',l,thl,just) g ->
       match t' with
         None -> failwith "thus"
       | Some t ->
           (SUBGOAL_THEN (t g) ASSUME_TAC THENL
            [just (thl g);
             POP_ASSUM (fun th ->
               N_ASSUME_TAC l th THEN
               EVERY (map (fun th -> REWRITE_TAC[EQT_INTRO th])
                 (CONJUNCTS th)))])
           g),
    default_stepinfo (Some t);;

let (fix': (goal -> term) list -> step) =
  fun tl ->
    (fun _ g -> (MAP_EVERY X_GEN_TAC o (map (fun t -> t g))) tl g),
    default_stepinfo None;;

let (set': (goal -> term) -> step) =
  fun t ->
    let stp =
      (fun (t',l,_,_) g ->
       match t' with
         None -> failwith "set"
       | Some t ->
           let eq = t g in
           let lhs,rhs = dest_eq eq in
           let lv,largs = strip_comb lhs in
           let rtm = list_mk_abs(largs,rhs) in
           let ex = mk_exists(lv,mk_eq(lv,rtm)) in
           (SUBGOAL_THEN ex (X_CHOOSE_THEN lv
              (fun th -> (N_ASSUME_TAC l) (prove(eq,REWRITE_TAC[th])))) THENL
            [REWRITE_TAC[EXISTS_REFL];
             ALL_TAC]) g),
      default_stepinfo (Some t) in
    stp at -1;;

let theorem = theorem' o parse_mterm;;
let suppose = suppose' o parse_mterm;;
let consider = consider' o map parse_mterm;;
let take = take' o map parse_mterm;;
let assume = assume' o parse_mterm;;
let have = have' o parse_mterm;;
let thus = thus' o parse_mterm;;
let fix = fix' o map parse_mterm;;
let set = set' o parse_mterm;;

let iff prfs = tactics [EQ_TAC THENL map steps prfs];;

let (otherwise: ('a -> step) -> ('a -> step)) =
  fun stp x ->
    let tac,si = stp x in
    ((fun (t,l,thl,just) g ->
       REFUTE_THEN (fun th ->
         tac (t,l,K ([REWRITE_RULE[] th] @ thl g),just)) g),
     si);;

let (thesis:mterm) = "thesis";;
let (antecedent:mterm) = "antecedent";;
let (opposite:mterm) = "opposite";;
let (contradiction:mterm) = "F";;

let hence = so thus;;
let qed = hence thesis;;

let h = g o parse_term;;
let f = e o step;;
let ff = e o steps;;
let ee = e o EVERY;;
let fp = f o (fun x -> x proof []);;

let GOAL_HERE = tactics [GOAL_TAC];;

subset_name: curated


TITLE: Isomorphism of preduals implies isomorphism of the $W^*$-algebras or not?
QUESTION [0 upvotes]: Let $M$ and $N$ are two von Neumann algebras such that their preduals $M_{∗}$ and $N_{∗}$ are isomorphic in the sense of Banach spaces, does it imply M and N are $∗$-isomorphic or not??

REPLY [7 votes]: No. Take $M={\bf C}^{\oplus 4}$ (i.e. $\ell_\infty$ on a 4-element set) and take $N=M_2({\bf C})$.
Things should get more interesting if you require stricter bounds on the norm of the Banach space isomorphism between the preduals. For instance, isometric isomorphism of the preduals as Banach spaces immediately implies the two von Neumann algebras are isometrically isomorphic, and then based on Kadison's theorem (1951, Annals of Math.) this places various restrictions on how much the two von Neumann algebras can differ.


TITLE: find $\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$
QUESTION [3 upvotes]: If the equation of the tangent to the graph of the function $y=f(x)$ at $x=2$ is $4x-y-3=0$ and at this point tangent cuts the graph also,then find $\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$

$$\lim_{x\to 2}\frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2}$$
As the denominator is tending to zero,so numerator must br tending to zero.
So I applied L Hospital rule two times to get
$$\lim_{x\to 2}\frac{f'(x^2-2).2x-f'(f(x)-3).f'(x)}{2(x-2)}$$ for the first time and $$\lim_{x\to 2}\frac{4x^2f''(x^2-2)+2f'(x^2-2)-f'(f(x)-3).f''(x)-f''(f(x)-3).(f'(x))^2}{2}$$
As the equation of tangent at the point $x=2$ is $y=4x-3$
So slope of tangent at $x=2$ is $4$ i.e. $f'(2)=4$ and as the tangent cuts the graph at $x=2$,so put $x=2$ in $y=4x-3$ to get $y=5$, so $f(2)=5$ but
I don't know how to find $f''(2)$.

I am stuck here, please help me. Thanks.

REPLY [0 votes]: In general, the equation of a tangent is $y-f(x_0)=f'(x_0)(x-x_0)$.
Rearrange $4x-y-3=0$, we have $y-5=4(x-2)$,
i.e. $f(2)=5$ and $f'(2)=4$.
As $t\rightarrow 2$, $f(x^2-2)-f(f(x)-3) \rightarrow f(4-2)-f(5-3)=0$.
Now
$\lim_{x\rightarrow 2} \frac{f(x^2-2)-f(f(x)-3)}
                            {(x-2)^2}
=\lim_{x\rightarrow 2} \frac{f'(x^2-2)\cdot 2x-f'(f(x)-3)\cdot f'(x)}
                            {2(x-2)} $
As $t\rightarrow 2$, $f'(x^2-2)\cdot 2x-f'(f(x)-3)\cdot f'(x) \rightarrow f'(2)\times 4-f'(5-3)\times 4=0$.
Then applying  L'Hospital rule once more,
$\begin{eqnarray*}
   &&
   \lim_{x\rightarrow 2}
   \frac{f(x^2-2)-f(f(x)-3)}{(x-2)^2} \\
   &=&
   \lim_{x\rightarrow 2}
   \frac{f''(x^2-2) \cdot (2x)^2+f'(x^2-2)\cdot 2-
         f''(f(x)-3)\cdot [f'(x)]^2-f'(f(x)-3)\cdot f''(x)}{2} \\
   &=&
   \frac{f''(2)(4)^2+f'(2)\times 2-f''(2)[f'(2)]^2-f'(2)\times f''(2)}{2} \\
   &=& 4-2f''(2)
\end{eqnarray*}$
Simply leaves $f''(2)$ in your answer.


TITLE: Height of algebraic numbers
QUESTION [9 upvotes]: I would like to find effective upper bound for the height of $a+b$ and $a/b$ and $ab$ knowing the heights of $a$ and $b$. Thanks.

REPLY [2 votes]: I too was looking for the answer to the same question. It seems necessary to define "height" since there are of course several variants in use. The height $h(a)$ I am interested in is the maximum of the absolute values of the coefficients of the minimal polynomial of the algebraic number $a$. Note that the first answer refers to a different notion of height, since, for example,
$9 = h(9) = h(3 \cdot 3) \not \leq h(3) + h(3) = 3 + 3 = 6.$
I imagine there must be upper bounds of the form
$h(ab) \leq f(d) h(a)^{g(d)} h(b)^{g(d)}$
for some simple functions $f$ and $g$, where $d$ is the degree of a field extension of $\mathbb{Q}$ containing both $a$ and $b$, for example.
Are there any such results in the literature, and similarly for $h(a+b)$ and $h(a/b)$?


TITLE: Gödel's incompleteness theorem and real closed fields
QUESTION [21 upvotes]: I am familiar with the result of Gödel's incompleteness theorem. I find it hard though, to convince myself that when we replace normal number arithmetic with real closed fields, that there is an axiomatic system that is complete. After all, $\mathbb{N} \subset \mathbb{R}$, so why can't we use the more 'general' system for arithmetic? I know other people also struggle with this. Is there a more intuitive explantion as to why this result is possible and not contradictory to the incompleteness theorem?

REPLY [11 votes]: As has been mentioned, one cannot apply the theory of real-closed fields to solve Diophantine problems since $\rm\:\mathbb Z\:$ is not definable in the elementary language of real-closed fields. However, unless you have studied model theory, such an answer will probably lend little intuition on the matter. Towards such, consider the following remarks.
The primary reason that the first-order theory of the reals proves to be much simpler than Diophantine (integer) analogs is that the associated geometry is much simpler. Namely, the sets that are definable by real-polynomial inequations - so-called semi-algebraic sets - can be decomposed into a finite number of cells (e.g. cylinders) where, on each cell, every defining polynomial  has constant sign. Therefore testing the truth of a first-order statement reduces to a finite number of tests on constant-sign cells, which is trivially decided by simply evaluating the polynomials at any point in the cell. This yields an algorithm for deciding the truth of first order statements about the reals - the so-called cylindrical algebraic decomposition (CAD) algorithm  (an effective realization of Tarski's famous method of quantifier elimination for the reals).
For example, in the one-dimensional case the sets definable in the first-order language of $\rm\:\mathbb R\:$ are simply finite unions of intervals. So, e.g. to test if $\rm\ f(x) > 0\ \Rightarrow g(x) > 0\ $ we can employ Sturm's Theorem to partition $\:\mathbb R\:$ by the finite number of roots of $\rm\:f,\:g\:,\:$ and then pick a point $\rm\:r_0\:$ in each interval and test if $\rm\ f(r_0)>0\ \Rightarrow\ g(r_0)>0\:.\:$  The CAD algorithm works analogously in higher dimensions, by projecting higher-dimensional cells down to lower dimensions. The key property is a result of Tarski and Seidenberg that semi-algebraic sets are preserved by projections. The essential innate structure has been generalized in model-theoretic study of o-minimal structures.
Contrast this to the study of (first-order) Diophantine equations. Anyone who has studied number theory quickly appreciates the immense complexity of the structure of solution sets of integer polynomial equations, e.g. Pell equations, elliptic curves, etc. Undecidability results imply that there can be no simple uniform recursive decision procedure like that for the reals. Each leap into a higher dimension may require completely novel ideas. But this shouldn't be viewed as a detriment. Rather, it yields an unending source of rich problems that will constantly challenge our ingenuity.


TITLE: Units of average-Is it units of $y$ or units of $y/x$
QUESTION [1 upvotes]: Imagine we have a collection of 10 heights of individuals in feet.
We calculate the average value as:
$$
\mu=\frac{1}{10}\sum_{i=1}^{10}h_{i}
$$
where $h_{i}$ is the height of an individual $i.$ In this case,
is the average $\mu$ in units of feet, or units of feet/individual?
From economics, the average cost of producing $q$ units is given
as:
$$
AC\left(q\right)=\frac{TC\left(q\right)}{q}
$$
where $TC$ is total cost for producing $q$ units, $AC$ is average
cost, and $q$ is total quantity. Here, the average cost is in units
of say \$/unit of output. My question: in which circumstances are
the units of height in units of the variable being measured- say $y,$
and in which are they two dimensional, say $y$ per $x$ (like cost
per unit, or average speed per hour)? Does it matter?

REPLY [3 votes]: Compare:
   ✔    “The average height of the students is $178.3\textrm{ cm}.$”
   ✗    “The average height of the students is $178.3\textrm{ cm}\text{/student}.$”
   ✔    “The average production cost of the units is $\$459.$”
   ✗    “TThe average production cost of the units is $\$459\text{/unit}.$”
   ✔    “The average speed of the car for journey A is $75.1\textrm{ km h}^{-1}.$”
   ✗    “The average speed of the car for journey A is $75.1\textrm{ km h}^{-1}\textrm{/h}.$”
   ✔    “The average number of goals among the teams is $13.4.$”
   ✗    “The average number of goals among the teams is $13.4\text{ goals/team}.$”
Average height is after all still a height, and makes sense to have exactly the same dimension as individual height.  That we are referring to student heights (as opposed to tree heights) belongs to the context, rather than to the quantity's specification per se.
The third example makes this even clearer: surely the average speed doesn't have units $75.1\textrm{ km h}^{-2},$ which is that of acceleration!
In the same vein:
   ✔    “The number of goals scored by the team is $7.$”
   ✗    “The number of goals scored by the team is $7\text{ goals}.$”

The gradient of a displacement-time graph is instantaneous velocity—not average displacement nor average velocity —so has units $$\textrm{km h}^{-1}$$ instead of merely $$\textrm{km}.$$

In the graph of Total Production Cost $(TC)$ against Quantity $(q),$ the gradient is Marginal Cost $(MC),$ not Average Production Cost $(AC),$ which is just $\displaystyle\frac{TC(q)}q.$
$TC, MC,$ and $AC$ all have unit $$\$.$$


TITLE: how to rearrange for y for: $\ln{|y|}= \frac{x^3}{3}$?
QUESTION [0 upvotes]: Is it $y= \pm e^{\frac{x^3}{3}}$?
I am not sure about whether the plus minus sign is correct.
Just here to look for confirmation or correction! Thanks!

REPLY [1 votes]: Hint:
$$\log|y|=\log \sqrt{y^2}=\frac{1}{2}\log y^2$$
so the equation becomes
$$\frac{1}{2}\log y^2=\frac{x^3}{3}$$
$$ y^2=e^{\frac{2}{3}x^3}$$
now we see the plus minus sign is necessary
$$ y=\pm e^{\frac{1}{3}x^3}$$


TITLE: Magnetic field of uniformly charged rotating hollow sphere
QUESTION [2 upvotes]: I want to compute the magnetic field due to a homogeneously charged, rotating sphere with radius $R$, angular velocity $\vec{\omega} = \omega\hat{z}$ and total charge $Q$. I want to use Biot-Savart law, so I first computed the current density $\vec{j}(r,\theta,\phi) = \frac{\omega Q}{4 \pi R} \delta(R-r) \cdot (-\sin\theta \sin\phi,\sin\theta\cos\phi,0)^T$. After taking out all the constants with respect to integration and removing the integral over the radial component with the $\delta$ function, what stays in the Biot-Savart integral is
$$\vec{B}(\vec{r}) \propto \iint_{\theta,\phi}^{\pi,2\pi}\frac{\vec{j}(\vec{y}) \times (\vec{r} - \vec{y})}{|\vec{r}-\vec{y}|^3}\sin{\theta}\;\text{d}\phi\;\text{d}\theta$$
where $\vec{r} = |\vec{r}|\cdot(\sin\theta_r\cos\phi_r, \sin\theta_r\sin\phi_r,\cos\theta_r)^T$ is the point where the $B$-field is evaluated, $\vec{y} = R\cdot(\sin\theta\cos\phi, \sin\theta\sin\phi,\cos\theta)$ is a point on the sphere. and the $\vec{j}$ is now missing the $\delta$ part, and some constants.
The problem I have is, that the vector inside the integral is
$$\begin{pmatrix} \sin\theta\cos\phi(|\vec{r}|\cos{\theta_r} - R\cos\theta )\\\sin\theta\sin\phi(|\vec{r}|\cos\theta_r-R\cos\theta)\\-|\vec{r}|\sin\theta\sin\phi\sin\theta_r\sin\phi_r-|\vec{r}|\sin\theta\cos\phi\sin\theta_r\cos\phi_r+R\sin^2\theta\end{pmatrix}.$$
Since all terms involving $\sin\phi$ or $\cos\phi$ will integrate to zero, this implies that I will get a $B$-field that is along the $z$ axis, no matter where it's evaluated. This can't be, because in the next part I'll show that far away the field looks like that of a dipole.
I have been sitting over this for hours, can someone help me clear this up? Where am I going wrong?

REPLY [1 votes]: The problem is with your current density $\vec{j}$, let's start with it's basic definition as current over area, $\vec{j} = \frac{Q\vec{v}}{4\pi R^2}$, now knowing that the angular velocity can be used to get the linear velocity of the charge on the spherical surface using $\vec{v}=\vec{\omega}\times R\hat{r} = \omega R \hat{z} \times \hat{r} = \omega R (\cos\theta\hat{r}-\sin\theta\hat{\theta})\times\hat{r} = \omega R\sin\theta \hat{\phi}$
So that your current density is:
$$\vec{j} = \frac{Q\omega}{4\pi R} \sin\theta \delta(r-R)\hat{\phi}$$
Notice that this makes sense since the charges should only be moving in the $\phi$ direction around the z-axis, you should never have had any $r$ or $\theta$ components before. Also, at the poles of the sphere the current density goes to zero as expected and with the maximal current density at the equator.
With this current density your integrand will have a $\sin ^2\theta$ term which will clear up your previous issues.


TITLE: Let $V$ an inner product space and $T\in\mathcal L(V)$ normal. Show that the minimal polynomial of $T$ have no repeated zeros
QUESTION [1 upvotes]: This is the exercise 13 in page 268 of Linear algebra done right, third edition of Axler.

Let $V$ an inner product space and $T\in\mathcal L(V)$ normal. Show that the minimal polynomial of $T$ have no repeated zeros.

Here it is assumed that $V$ is finite-dimensional.
The result, in the context of the book, is immediate if $V$ is a complex finite-dimensional vector space. However if $V$ is a real vector space it is not so clear if it is probable in the context where the exercise is asked.
Every theorem of the chapter, except perhaps some general definitions as the minimal polynomial, are done in the context of $V$ as a complex finite-dimensional vector space.
This question was asked previously here without an answer (the comment on the question seems not correct because an operator in $\Bbb C^n$ doesnt necessarily can be seen as an operator in $\Bbb R^n$, by example $T(x,y)=(ix,x-iy)$ cannot be real.)
Questions:

Can someone clarify if the exercise can be solved for real $V$ in the context of the book?
Can someone confirm if the statement is true or false for real $V$?

REPLY [3 votes]: Masacroso: I did indeed intend for that exercise to apply to both real and complex finite-dimensional inner product spaces. Thus you asked a good question.
Itay4: You wrote that the solution you posted here is one that you "came up with". However, it is word-for-word and symbol-for-symbol identical to the solution that I wrote for the Instructor's Solutions Manual many years ago, except that you have a few typos and you omitted a parenthetical remark. I do not distribute the Instructor's Solutions Manual or parts of it, but versions from earlier editions of the book have leaked out to the web in a few places. I have attached below a screen shot of my solution to this exercise; this was written 20 years ago. As you can see, this 20-year-old solution is virtually identical to what you submitted without proper attribution.


TITLE: Compute the least upper bound of the set of all real numbers of the form $Re(iz^3+1)$ such that |z|<2.
QUESTION [0 upvotes]: Can someone please tell me how to begin? I found another question similar to this but the final result I get is 9. Is this correct?
EDIT: $Re(iz^3+1)≤|iz^3+1|≤|iz^3|+|1|=|i|.|z^3|+1=|i|.|z|^3+1=|z|^3+1<2^3+1=9$
So the supreme is less than 9?

REPLY [0 votes]: The first step to consider here is the fact that is that Re (z) is always less than or equals to modulus of z. Your answer I think can be found here.
Compute the least upper bound of the set of all real numbers of the form $Re(iz^3 +1)$ such that $|z| \leq 1$.


text:
\section{Preliminaries from algebra}\label{preliminaries}
\subsection{Conventions and definitions}\label{divisionalg}
\begin{convention}
Throughout the paper, let $F$ be a field of characteristic unequal to 2. We will denote elements of $F$ (scalars) by lower case Greek letters, vector spaces over $F$ will be denoted by capital roman letters and elements of such vector spaces will be denoted by lower case roman letters.
\end{convention}

\begin{definition}\label{algebra}
An \emph{algebra} over $F$ is a vector space $A$ over $F$ equiped with an bilinear multiplication map $A \times A \to A$. An algebra $A$ is called \emph{unital} if there exists an element $1 \in F$ such that $1a = a1 = a$ for all $a \in A$. It is called associative when $(ab)c = a(bc)$ for all $a, b, c$ in $A$.
\end{definition}

\begin{convention}
Throughout this paper, by an \emph{algebra} we will denote a finite-dimensional, unital, but not necessarily associative algebra over $F$.
\end{convention}

Let $A$ be an algebra. The unit element of $A$ will be denoted $1$ and the one-dimensional subalgebra of scalar multiples of 1 will be denoted $F1$. By bilinearity of the product on $A$ we have that

all elements of $F1$ commute and associate with all elements of $A$.

\begin{notation}
Let $A$ be an algebra. For any subset $B$ of $A$ we denote by $\langle B \rangle$ the subalgebra of $A$ generated by the elements of $B$. That is: the intersection of all subalgebras of $A$ containing $B$. When $B$ is finite, e.g. $B = \{u, v\}$, we will write $\langle u, v \rangle$ for $\langle \{u, v\}\rangle$.
\end{notation}
\begin{convention} Since we required all our algebras to be unital, we demand the same of their subalgebras. In particular we have that $F1 \subseteq \langle B \rangle$ for any subset $B$ of $A$.
\end{convention}
The following elementary observation is stated explicitly, not because I think the reader is stupid but because I woke up one night in panick, believing that there were a hole in the proof amounting to the following fact being false. However it is true and promoted to a lemma so that I can refer back to it later.
\begin{lemma}\label{generated}
Let $A$ be an algebra, let $U \subseteq A$ be a linear subspace and let $u_1, \ldots, u_k$ be a basis of $U$. Then $\langle u_1, \ldots, u_k \rangle = \langle U \rangle$.
\end{lemma}
\begin{proof}
The inclusion $\langle u_1, \ldots, u_k \rangle \subseteq \langle U \rangle$ is obvious from the inclusion $\{ u_1, \ldots, u_k \} \subseteq  U$. For the converse inclusion we note that every algebra containing $\{u_1, \ldots, u_k\}$ contains $U$ and hence so does the intersection $\langle u_1, \ldots, u_k \rangle$ of all such algebras. But this means that $\langle u_1, \ldots, u_k \rangle$ is an algebra containing $U$ and hence it certainly contains the intersection of all such algebras, that is $\langle u_1, \ldots, u_k \rangle \supseteq \langle U \rangle$.
\end{proof}

\begin{definition}
An algebra is \emph{alternative} if every subalgebra generated by two elements is associative.
\end{definition}

Alternative algebras need not be associative, but many results that are familiar for associative algebras do hold for more general alternative algebras as well. We will see an example below in Lemma \ref{division}, many more examples can be found in the book by Schafer \cite{Schafer}.

\begin{definition}
We say that an algebra $A$ is a \emph{division algebra} if for every non-zero $a \in A$ the linear maps $L_a \colon x \mapsto ax$ and $R_a \colon x \mapsto xa$ are invertible in $\End(A)$.
\end{definition}
\begin{definition}
A non-zero element $x$ in an algebra $A$ is called a \emph{zero-divisor} if there exist non-zero $y \in A$ such that $xy = 0$ or $yx = 0$.
\end{definition}
\begin{lemma}\label{division}
Let $A$ be a finite-dimensional alternative algebra. Then the following are equivalent:
\begin{enumerate}
\item $A$ is a division algebra
\item $A$ has no zero-divisors
\item For every non-zero $a \in A$ there exist an element $a^{-1} \in A$ such that $aa^{-1} = a^{-1}a = 1$.
\end{enumerate}
\end{lemma}
The proof is left as an exercise to the reader. A counterexample to the implications $2 \implies 1$ and $2 \implies 3$ when we drop the condition that $A$ is finite-dimensional is given by the algebra $F[x]$ of polynomials over $F$. A counterexample to the implication $3 \implies 2$ when we drop the condition that $A$ is alternative is given in Section \ref{CayleyDickson}.

\begin{definition}\label{quadraticdef}
An algebra is \emph{quadratic} if the subalgebra generated by the single element $a$ is two-dimensional for all $a \nin F1$.
\end{definition}

\begin{definition}
By an \emph{involution} $*$ of an algebra $A$ we denote a linear anti-automorphism of order dividing 2. In other words and involution is a linear map $a \mapsto a^*$ satisfying

\begin{equation}\label{star}
(ab)^* = b^*a^* \qquad (a^*)^* = a
\end{equation}
for all $a, b \in A$.
\end{definition}

By linearity $\alpha^* = \alpha$ for all $\alpha \in F1$.

\begin{definition}\label{stronginv}
We say that an involution $*$ is a \emph{strong involution} if the converse to the above holds, i.e. if $a^* = a$ if and only if $a \in F1$.
\end{definition}

The most famous example of a strong involution is complex conjugation in the algebra $A = \mathbb{C}$ over $F = \mathbb{R}$. The quaternion conjugation (\ref{quaternionconj}) on the $\mathbb{R}$-algebra $\HH$ discussed in the Introduction is another example of a strong involution. Moreover $\CC$ and $\HH$ are both examples of quadratic algebras over $\mathbb{R}$. We will see in Section \ref{nice} that this is no coincidence.

\subsection{A theorem of Artin}
We defined an alternative algebra as an algebra in which the subalgebra generated by any two elements is associative. Alternativity can be viewed as a `controlled' form of non-associativity, where we bear in mind that all associative algebras are automatically alternative. Clearly for any $a, b$ in an alternative algebra we have that
\begin{eqnarray}
(aa)b = a(ab) \label{leftalt}\\
(ab)a = a(ba) \label{flexible} \\
(ba)a = b(aa) \label{rightalt}
\end{eqnarray}
It is an interesting result of Artin that the converse holds:
\begin{theorem}[\cite{Schafer}, Thm III.3.1]\label{associator}
Let $A$ be an algebra. The following are equivalent
\begin{enumerate}
\item $A$ is alternative
\item At least two of the three equations (\ref{leftalt}, \ref{flexible}, \ref{rightalt}) hold for all $a, b \in A$
\item The \emph{associator}, the trilinear map $(. , ., .) \colon A \times A \times A \to A$ defined by $(a, b, c) = (ab)c - a(bc)$ \emph{alternates}, that is: changes sign under odd permutations of its entries.
\end{enumerate}
\end{theorem}
It is the third item that explains the name alternative. Note that $(a, b, c)  = 0$ for all $a, b, c$ when $A$ is associative, thus providing a second way of seeing that all associative algebras are alternative.

\subsection{A particularly nice class of algebras}\label{nice}
Let $A$ be a quadratic algebra and $a \in A, a \nin F1$. By definition of quadratic algebra there are unique scalars $\tau(a), n(a) \in F$ such that
\begin{equation}\label{taun}
a^2 = 2\tau(a)a - n(a)
\end{equation}
This defines maps $\tau, n$ from $A \backslash F1 \to F$ and we extend these maps to all of $A$ by setting $\tau(\alpha) = \alpha, n(\alpha) = \alpha^2$ for $\alpha \in F1$. Note that with this convention equation (\ref{taun}) holds for all $a \in A$.

It is an easy but important observation that the map $\tau \colon A \to F$ is linear. Moreover, and this is the reason for the appearance of the number 2 in (\ref{taun}), we have that $\tau$ is a linear projection operator onto the space $F1$, that is, we have that $\tau \circ \tau = \Id$. When $F = \mathbb{R}$ we think of $\tau$ as `taking the real part of an element'.

We define $\im(A) = \ker \tau$. Since $\tau$ is a projection operator we obtain a decompostion (of linear spaces, not of algebras):
\begin{equation}\label{ReIm}
A = \tau(A) \oplus \im(A).
\end{equation}

The elements of the co-dimension-1 linear space $\im(A)$ are called \emph{imaginary}. The reason for this terminology stems from the special case $F = \mathbb{R}, A = \mathbb{C}$ and becomes apparent from the following alternative characterization of $\im(A)$ (the equivalence of the two characterizations dates back to Frobenius and follows easily from (\ref{taun})):
\begin{equation}\label{im}
\im(A) = \{a \in A \colon a^2 \in F1 \textrm{ but } a \nin F1\} \union \{0\}.
\end{equation}
We will be interested in quadratic algebras satisfying the additional condition
\begin{equation}\label{tauab}
\tau(ab) = \tau(ba) \textnormal{ for all } a, b \in A.
\end{equation}
We have:
\begin{theorem}\label{quadratic}
Let $A$ be an algebra. The following are equivalent:
\begin{enumerate}
\item $A$ is a quadratic algebra satisfying (\ref{tauab}),
\item $A$ possesses a strong involution.
\end{enumerate}
\end{theorem}
\begin{proof}
`$1 \implies 2$': Given the quadratic algebra structure we write $\tau$ for the map $a \mapsto \tau(a)1$ (in other words, we reinterpret $\tau$ as mapping into the subalgebra $F1$ of $A$ rather than into the `external' field $F$.) We write $\im = \Id - \tau$ for the linear projection onto the space $\im(A)$ along the decompostion (\ref{ReIm}).

The involution $*$ is then given by
\begin{equation}\label{starfromtau}
a^* = \tau(a) - \im(a)
\end{equation}
Let $u, v \in \im(A)$. From (\ref{im}) we have that $(u + v)^2$, $u^2$, and $v^2$ are all in $F1$ so that from expanding $(u + v)^2$ we find that $uv + vu \in F1$ as well. It follows that $\tau(uv + vu) = uv + vu$ and hence $\im(uv + vu) = 0$ so that $\im(uv) = -\im(vu)$.

Combining this with (\ref{tauab}) and (\ref{starfromtau}) yields $(uv)^* = \tau(uv) - \im(uv) = \tau(vu) + \im(vu) = vu$. Now since $u, v \in \im(A)$, $v^* = -v$ and $u^* = -u$ by (\ref{starfromtau}) and hence the equation $(uv)^* = vu$ extends to $(uv)^* = v^*u^*$, showing that for $u, v \in \im(A)$ the first relation in (\ref{star}) holds. Writing $a = \alpha + u$, $b = \beta + v$ with $\alpha, \beta \in F1, u, v \in \im(A)$ we can easily extend this to general $a, b \in A$ exploiting the linearity of the $*$-operator.

The second equation in (\ref{star}) is obvious from (\ref{starfromtau}). This shows that ${}^*$ is an involution; the fact that it is strong follows directly from (\ref{starfromtau}).

`$2 \implies 1$': Given a strong involution $*$ we define the maps $\tau$ and $n$ by
\begin{equation}\label{taustar}
\tau(a) = \frac{1}{2}(a + a^*)
\end{equation}
 and
\begin{equation}\label{n}
n(a) = a^*a
\end{equation}
Note that the strongness of the involution guarantees that $\tau$ and $n$ take values inside $F1$, allowing us to reinterpret them as taking values in $F$. It is then straightforward to verify from (\ref{star}) that (\ref{taun}) and (\ref{tauab}) hold. (\ref{taun}) in turn implies that $A$ is quadratic.
\end{proof}

Theorem \ref{quadratic} provides a third characterization of $\im(A)$ for algebras $A$ with a strong involution:
\begin{equation}\label{im2}
\im(A) = \{a \in A \colon a^* = -a\}
\end{equation}
In other words: $\im(A)$ is the $(-1)$-eigenspace of the strong involution. Since by definition of `strongness' the $1$-eigenspace is one-dimensional and since linear transformations of order $\leq 2$ cannot have proper Jordan blocks, we obtain a second proof of a fact that we will now give its own number:

\begin{lemma}\label{codim1}
The co-dimension of $\im(A)$ in an algebra $A$ equipped with a strong involution equals 1.
\end{lemma}

A very useful property of imaginary elements is
\begin{equation}\label{anticommutative}
uv = -vu \qquad \textrm{ for all } u, v \in \im(A)
\end{equation}
Which follows from combining (\ref{im2}) with (\ref{star}).

Finally we note that from (\ref{n}) and (\ref{star}) if follows that in alternative algebras $A$ with a strong involution the function $n$ is multiplicative:
\begin{equation}\label{nmult}
n(ab) = n(a)n(b) \qquad \textnormal{for all } a, b \in A.
\end{equation}
For the $(F = \mathbb{R})$-algebras discussed in the introduction we have that the function $n$ equals the square of the Euclidean norm on $A$ and the proof of (\ref{nmult}) is spelled out in (\ref{normmult}).

\subsection{The Dickson Double}
\begin{definition}\label{DicksonDouble}
Let $A$ be an $F$-algebra with a strong involution $*$ and let $\gamma \in F \backslash \{0\}$ be a non-zero element of $F$. The \emph{Dickson double} $D_\gamma(A)$ of $A$ is an algebra with dimension twice the dimension of $A$ formed by attaching to $A$ an element $i$ satisfying
\begin{equation}\label{i2}
i^2 = \gamma.
\end{equation}
In more detail we have that as a vector space $D_\gamma(A)$ equals $A \oplus iA$ where elements of the subspace $iA$ are interpreted as the product of the `new' element $i = i1$ with elements of $A$. Multiplication is defined by the formula
\begin{equation}\label{eyesleft}
(a + ib)(c + id) = (ac + \gamma d b^*) + i(a^*d + cb)
\end{equation}
The involution $*$ on $A$ is extended to $D_\gamma(A)$ as the unique involution agreeing with the original $*$ on the subalgebra $A$ and satisfying $i^* = -i$.
\end{definition}

\begin{lemma}\label{multiplication}
The multiplication (\ref{eyesleft}) can be equivalently be defined by the three equations
\begin{eqnarray}
p(iq) = i(p^*q)  \label{pq1}\\
(pi)q = (pq^*)i \label{pq2} \\
(ip)(qi) = \gamma(pq)^* \label{pq3}
\end{eqnarray}
for all $p, q \in A$ or by the single equation
\begin{equation}
(a + bi)(c + di) = (ac + \gamma d^*b) + (da + bc^*)i.
\end{equation}
\end{lemma}
Particularly useful in computations is also the corollary
\begin{equation}\label{xi}
xi = ix^* \textnormal{ for all } x \in A
\end{equation}
from which in turn it follow that
\begin{equation}\label{xistar}
(xi)^* = -(xi) \textnormal{ for all } x \in A
\end{equation}

From (\ref{xistar}) we note:
\begin{corollary}\label{strong}
The involution $*$ on $D_\gamma(A)$ is strong.
\end{corollary}

We'll use the last corollary to try and avoid doing computations using (\ref{eyesleft} - \ref{xistar}) in this paper and instead rely on the properties of algebras with a strong involution derived in Section \ref{nice}.

\begin{remark}\label{allfromxi}
One of the things apparent from equations (\ref{pq1}, \ref{pq2}, \ref{pq3}) is that we cannot assume the Dickson double $D$ of an algebra $A$ to be associative, even if $A$ is. On the other hand it is easy to see that if we happen to know that $A$ is a subalgebra of an associative algebra $D$ and $i$ is an element of $D$ not contained in $A$ then (\ref{xi}), if true in $D$, implies equations (\ref{pq1}, \ref{pq2}, \ref{pq3}) to hold in $D$ as well.
\end{remark}

The following theorem, due to Albert, (which can be derived directly from equations (\ref{pq1}, \ref{pq2}, \ref{pq3}) in conjunction with Artin's theorem above) states however that for $A$ of dimension 1, 2 or 4 (and $D$ subsequently of dimension 2, 4 or 8) the non-associativity is controlled.

\begin{theorem}\label{Albert}
Let $A$ be a an $F$-algebra with a strong involution and let $D$ be its Dickson double. Then:

$D$ is commutative if and only if $A = F$.

$D$ is associative if and only if $A$ is commutative.

$D$ is alternative if and only if $A$ is associative.
\end{theorem}

Now that alternativity is on the table it is interesting to know that Remark \ref{allfromxi} in fact generalizes to the alternative case though with a much more complicated proof:

\begin{lemma}\label{allfromxi2}
Let $D$ be an alternative algebra with strong involution $*$, $A$ a subalgebra of $D$ that is closed under application of $*$, and let $i \in \im(D)$ be an element \emph{not} contained in $A$ satisfying (\ref{xi}). Define $\gamma \in F$ by $i^2 = \gamma1$. Then $(\ref{pq1}, \ref{pq2}, \ref{pq3})$ hold in $D$ for all $p, q \in A$. It follows that if $\gamma \neq 0$ then $D$ contains a copy of the Dickson double $D_\gamma(A)$ of $A$.
\end{lemma}
\begin{proof}
We'll start by writing out the proof of (\ref{pq2}).
$$(pi)q = p(iq) + (p, i, q) = p(q^*i) + (p, i, q) = (pq^*)i - (p, q^*, i) + (p, i, q)$$
where the first and third equality apply the definition of the associator $(.,.,.)$ given in Theorem \ref{associator} and the second equality applies (\ref{xi}). Hence equation (\ref{pq2}) follows as soon as we show that $(p, i, q) - (p, q^*, i) = 0$. But by alternativity and trilineairty of the associator (Thm \ref{associator}) we find that
$$(p, i, q) - (p, q^*, i) = (p, i, q) + (p, i, q^*) = (p, i, q + q^*) = (p, i, 2\tau(q)1)$$
where the last equality uses (\ref{taustar}). And since elements of $F1$ associate with all elements of $D$ we find that $(p, i, 2\tau(q)1) = 0$ as  desired.

Equality (\ref{pq1} can be derived from (\ref{pq2}) and (\ref{xi}) as follows. By (\ref{xi}) and (\ref{star}) we see that the right hand side of (\ref{pq1}) equals $(q^*p)i$. Applying (\ref{pq2}) with $q^*$ in the role of $p$ and $p^*$ in the role of $q$ we obtain that this equals $(q^*i)p^*$. We have thus obtained the equality $(pq^*)i = (q^*i)p^*$.

Applying the $*$-operator to both sides of this last equality we find on te left hand side $-(pq^*)i$ by (\ref{xistar}) and on the right hand side $p(i^*q)$ by applying (\ref{star}) both inside and outside the brackets. Since $i^* = -i$ the right hand side simplifies to $-p(iq)$ and multiplying both sides with $-1$ we finally obtain $(pq^*)i = p(iq)$, which is (\ref{pq1}).

Finally to see that (\ref{pq3}) holds as well we first establish that (\ref{xi}) also holds with $qi$ in the role of $i$.
Concretely, for any $x \in A$ we have
\begin{equation}\label{xqi}
x(qi) = \tau(x)(qi) + \im(x)(qi) = (qi)\tau(x) - (qi)\im(x) = (qi)x^*.
\end{equation}
Here the second equality uses (\ref{anticommutative}) and the fact that elements of $F1$ commute with everything and the third equality uses (\ref{taustar}).

Now we use this to attack (\ref{pq3}). Using the definition of the associator for the first and third equality and (\ref{xqi}) for the second we obtain:
\begin{equation}\label{ipqi1}
(ip)(qi) = i(p(qi)) + (i, p, qi) = i((qi)p^*) + (i, p, qi) = (i(qi))p^* - (i, qi, p^*) + (i, p, qi).
\end{equation}
As above we find that $-(i, qi, p^*) + (i, p, qi) = (i, 2\tau(p), qi) = 0$ so that (\ref{ipqi1}) reduces to
\begin{equation}\label{ipqi2}
(ip)(qi) = (i(qi))p^*.
\end{equation}
The left hand side of (\ref{ipqi2}) equals the left hand side of (\ref{pq1}). We look at the term $i(qi)$ within the outermost brackets of the right hand side of (\ref{ipqi2}). From (\ref{xi}) we see that it equals $i(iq^*)$ and since this term lives in the associative subalgebra generated by the elements $i$ and $q$ we can shift the brackets and conclude that it equals $i^2q^* = \gamma q^*$. The right hand side of (\ref{ipqi2}) then becomes $\gamma q^*p^*$ which needs no brackets as $\gamma \in F$. By (\ref{star}) this equals the right hand side of (\ref{pq3}).
\end{proof}

\subsection{The Cayley-Dickson algebras}\label{CayleyDickson}
We inductively define the class of Cayley-Dickson algebras. Each such algebra is a finite-dimensional $F$-algebra with a strong involution.
\begin{definition}
The (unique) one-dimensional $F$-algebra $F$ is a Cayley-Dickson algebra where the strong involution is the identity.
An $n$-dimensional $F$-algebra $A$ for $n > 1$ is a Cayley-Dickson algebra if it is the Dickson Double of a Cayley-Dickson algebra $B$ of lower dimension.
\end{definition}

It follows that every Cayley-Dickson algebra has dimension equal to a power of 2.

It also follows from Theorem \ref{quadratic} that all Cayley-Dickson algebras $A$ are quadratic algebras with a co-dimension 1 subspace $\im(A)$ of imaginary elements $u$ satisfying $u^2 \in F1$, $u^* = -u$. We note the similarity to equations (\ref{i2}) and (\ref{xistar}) implying that the `special' element $i$ used in the construction of $A$ from the smaller algebra $B$ is always contained in the space $\im(A)$.

\begin{remark}\label{special}
In fact the element $i$ and subalgebra $B$ are not \emph{that} special. Once we are presented with a Cayley-Dickson algebra $A$ of dimension $2^n \leq 8$, it is impossible to tell which of its many $2^{n-1}$-dimensional Cayley-Dickson subalgebras $B$ was used to construct $A$, in the sense that $A$ can be realized as a Dickson double of each of them. Various choices of the subalgebra $B$ will allow various elements $u \in \im(A)$ with $u^2 \neq 0$ to play  the role of the `special' element $i$. We will split the proof this remarkable fact into two parts. The proof in case that $A$ is a division algebra is presented here (Prop. \ref{doubling} below); the remaining cases will be dealt with in Theorem \ref{doubling2}. A different proof of the division algebra case in the special case that $\dim A = 4$ appears as part of the proof of Proposition \ref{main}.
\end{remark}

\begin{proposition}\label{doubling}
Let $A$ be a Cayley-Dickson division algebra of dimension $2^{n}$ for $n \in \{1, 2, 3\}$ and let $B$ be a $2^{n-1}$-dimensional subalgebra of $A$ closed under the action of the strong involution. Then there exists a $\gamma \in F^\times$ and $i \in \im(A)$, not contained in $B$ such that $A = B \oplus iB$ as a vector space and such that $(\ref{pq1}, \ref{pq2}, \ref{pq3})$ hold in $A$ for all $p, q \in B$. Consequently, $A$ is isomorphic to the Dickson double $D_\gamma(B)$.
\end{proposition}
\begin{proof}
Let $\{b_1, \ldots, b_{2^{n-1}}\}$ be a basis of $B$. Let the linear maps $R_{b_j}$ be defined as in Lemma \ref{division}. Since $\ker \tau$ has co-dimension 1 (Lemma \ref{codim1}), for each $j \in \{1, \ldots, 2^{n-1}\}$ the space $\ker (\tau \circ R_{b_j})$ has co-dimension at most 1. (In fact these co-dimensions are exactly 1 by Lemma \ref{division}.) It follows that the space $\bigcap_{j = 1}^{\dim B} \ker (\tau \circ R_{b_j})$ of elements $i$ such that $ib \in \im(A)$ for every $b \in B$ has co-dimension at most $2^{n-1}$ and hence in particular is non-zero. Pick any $i \neq 0$ from this space. We note that $i \in \im(A)$ and hence $i^2 = \gamma1$ for some $\gamma \in F$ where moreover $\gamma \neq 0$ since otherwise $i$ would be a divisor of zero.

Now since $A$ is a division algebra, $L_i$ is invertible (Lemma \ref{division}) and hence the space $iB$ is $2^{n-1}$-dimensional. We also have that $iB \cap B = \{0\}$ since if $ia = b$ for some $a, b \in B$ with $a \neq 0$ it follows that $i = ba^{-1} \in B$, and hence $i^{-1} = \frac{1}{\gamma} i \in B$. But this would imply that $1 = ii^{-1} \in \im(A)$ by our choice of $i$ -- a clear contradiction. We conclude that $B + iB$ is $2^n$-dimensional and hence equal to all of $A$.

It remains to verify $i$ and $B$ satisfy (\ref{pq1}, \ref{pq2}, \ref{pq3}). For this we first note that $i$ satisfies (\ref{xistar}) for all $x \in B$ since, by our construction of $i$, $ix \in \im(A)$ for each $x \in B$. From (\ref{xistar}) we then conclude that (\ref{xi}) holds for all $x \in B$ as well and then deduce (\ref{pq1}, \ref{pq2}, \ref{pq3}) from (\ref{xi}) as in Lemma \ref{allfromxi2}. In this last step we exploit that, as $\dim(A) \leq 8$ we have that $A$ is alternative by Theorem \ref{Albert}.
\end{proof}

\begin{definition}
A Cayley-Dickson algebra of dimension 8 is called a \emph{Cayley algebra} or an \emph{octonion algebra}. A Cayley-Dickson algebra of dimension 4 is called a \emph{quaternion algebra}.
\end{definition}

By Theorem \ref{Albert}, all Cayley algebras are alternative and all quaternion algebras are associative. In the literature on associative algebras a different, yet equivalent, definition of the term quaternion algebra is used, we'll come back to that issue in Section \ref{nilp}.

By the same theorem, all two-dimensional Cayley-Dickson algebras are commutative. In particular, when they are division algebras, they are fields. And, as they are formed by attaching to $F$ a new squareroot (the element $i$ from the definition) of some $\alpha \in F$, they are quadratic field extensions of $F$. In particular:

\begin{lemma}\label{2dim}
Let $\alpha \in F^\times$. The two-dimensional Cayley-Dickson algebra $D_\alpha(F)$ is a quadratic field extension of $F$ if and only if $\alpha$ is a non-square in $F$ and contains divisors of zero if and only if $\alpha$ is a square in $F$.
\end{lemma}
\begin{proof}
We only need to prove the `if' directions. In both cases we tacitly exploit that $D_\alpha(F)$ is commutative by Theorem \ref{Albert}. If $\alpha$ is a non-square the polynomial $x^2 - \alpha$ is irreducible over $F$ and attaching the root $i$ of this polynomial to $F$ yields a quadratic field extension by the standard argument. On the other hand when $\alpha = \beta^2$ for some $\beta \in F$ we find that $(\beta - i)(\beta + i)  = 0$ by the `strange product' formula.
\end{proof}

The following surprising fact forms the basis underlying all our considerations from Section \ref{split} onwards.

\begin{theorem}[Uniqueness theorem, \cite{Schafer}, p. 24-27]\label{splitunique}
In each of the dimensions 2, 4, 8 there is up to isomorphism exactly one Cayley-Dickson algebra containing divisors of zero, called the \emph{split} Cayley Dickson algebra in that dimension.
\end{theorem}

The number of alternative Cayley-Dickson algebras \emph{not} containing divisors of zero (hence division algebras by Lemma \ref{division}) depends on the field $F$. Cayley-Dickson algebras of dimension 16 and greater (equivalently: non-alternative Cayley-Dickson algebras) will be of no concern of us, although we note that the 16-dimensional $\mathbb{R}$-algebra $D_{-1}(D_{-1}(D_{-1}(D_{-1}(\mathbb{R}))))$ (the sedenions) are interesting as an example of an algebra containing zero-divisors while at the same time every non-zero element has a multiplicative inverse (cf Lemma \ref{division}).

We collect some consequences of the uniqueness theorem.

\begin{corollary}\label{sqrt1}
An alternative Cayley-Dickson algebra is split if and only if it contains an imaginary square root of 1
\end{corollary}
\begin{proof}
For the `if' direction, let $e$ be a imaginary square-root of 1. Then $(1 + e) \neq 0, (1 - e) \neq 0, (1+e)(1-e) = 0$, showing the existence of zero-divisors. Conversely: $D_1(F)$ and hence $D_\beta(D_1(F))$ and $D_\alpha(D_\beta(D_1(F)))$ contain an imaginary square-root of 1 by construction and hence the `only if' direction follows from the uniqueness theorem.
\end{proof}

We can however be a bit more explicit.

\begin{proposition}\label{CD2}
The (by Theorem \ref{splitunique}) unique two-dimensional split Cayley-Dickson algebra is isomorphic to the algebra $F \oplus F$ of pairs of elements in $F$ with pointwise addition and multiplication. Here the $*$ operation is given by $(\alpha, \beta)^* = (\beta, \alpha)$
\end{proposition}
\begin{proof}
Taking uniqueness for granted we only need to show that this algebra is isomorphic to $D_1(F)$. Let $i$ be the element of that algebra used in the definition of the doubling process (so in this case $i^2 = 1$). Then $\{1, i\}$ is a basis of $D_1(F)$ and the isomorphism is given by $1 \mapsto (1, 1)$, $i \mapsto (1, -1)$.
\end{proof}
\begin{proposition}\label{Mat2F}
With $\tau, n, *$ as in Section \ref{nice}, the (by Theorem \ref{splitunique} unique) four-dimensional Cayley-Dickson algebra over $F$ containing divisors of zero is isomorphic to the matrix algebra $\Mat(2, F)$ with $1 = I$, $\tau(X) = \frac{1}{2} \Tr(X)$, $n(x) = \det(X)$ and $\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}^* = \begin{pmatrix}
d & -b \\ -c & a
\end{pmatrix}$.
\end{proposition}
This relation to the matrix trace is also the reason that the operator $\tau$ is called $\tau$.
\begin{proof} Taking uniqueness for granted, we only need to show an isomorphism between the matrix algebra and $D_1(D_{-1}(F))$. It is easy to verify that the subalgebra $\{\begin{pmatrix}
a & - b \\ b & a
\end{pmatrix} \colon a, b \in F\}$ of $\Mat(2, F)$ is isomorphic to $D_{-1}(F)$ with the involution being the restriction to this subalgebra of the the involution described in the proposition. Denoting this subalgebra by $B$ for the moment, it takes only a little more effort to verify that the algebra $\Mat(2, F)$ is isomorphic to the Dickson Double $D_1(B)$ of $B$ where the role of the element $i$ (the `new square-root of 1') is played by the matrix $i = \begin{pmatrix}
1 & 0 \\
0 & -1
\end{pmatrix}$.
\end{proof}

In a sense the split Cayley and quaternion algebras are the richest Cayley and quaternion algebras:

\begin{corollary}\label{rich}
Every isomorphism class of Cayley-Dickson algebras of dimension $2^{n-1}$ $(n = 1, 2, 3)$ is represented among the subalgebras of the split Cayley-Dickson algebra of dimension $2^n$.
\end{corollary}
\begin{proof}
Given a $2^{n-1}$-dimensional Cayley-Dickson algebra $A$ we can realize the split Cayley-Dickson algebra of dimension $2^n$ as $D_1(A)$.
\end{proof}

However, over certain important fields this richness is not too impressive, as the following two corollaries show.

\begin{corollary}
Every alternative Cayley-Dickson algebra over an algebraically closed field $F$ is split.
\end{corollary}
\begin{proof}
Let $u$ be a non-zero imaginary element in such an algebra. By (\ref{im}), $u^2 \in F$ and by algebraic closedness there exists an $\alpha \in F$ such that $\alpha^2 = u^2$. If $u^2  = 0$ we have that $u$ is a zero-divisor and we are done. Otherwise $(u/\alpha)$ is an imaginary square-root of 1 and we can apply Corollary \ref{sqrt1}.
\end{proof}

\begin{corollary}\label{splitfinite}
Over a finite field $F$ there is up to isomorphism only one quaternion algebra and only one Cayley algebra.
\end{corollary}
\begin{proof}
By Wedderburn's little theorem, every division algebra over $F$ is commutative. By Theorem \ref{Albert} this means that no 4-dimensional Cayley-Dickson algebra can be a division algebra and hence the last proposition implies that all 4-dimensional Cayley-Dickson algebras over $F$ are isomorphic to $\Mat(2, F)$. Moreover, since every Cayley algebra $C$ over $F$ is obtained by doubling a quaternion algebra $Q$ over $F$, which it contains as a subalgebra, any zero-divisor present in the quaternion algebra used to obtain $C$ will also be an element of $C$. As we just saw that the presence of zero-divisors in $Q$ is inevitable, the uniqueness theorem tells us that $C$ is unique up to isomorphism.
\end{proof}

To end this subsection on a happy note we recall that we already saw in the introduction an example of a Cayley algebra that \emph{is} a division algebra, the algebra $\OO \coloneqq D_{-1}(D_{-1}(D_{-1}(\mathbb{R})))$ over $F = \mathbb{R}$. The proof presented there that $\OO$ is a division algebra relied on the fact that the Euclidean norm $\|.\|$ on $\RR^8$ respects the octonion multiplication: $\|x\|^2\|y\|^2 = \|xy\|^2$. In the context of more general Cayley-Dickson algebras this is just equation (\ref{nmult}) combined with the fact that in the special case of $\OO$ the function $n$ equals the square of the Euclidean norm. This suggests a path forward to finding other fields over which a Cayley division algebra might exist.

\begin{lemma}\label{splitn0}
Let $A$ be a Cayley-Dickson algebra. Then $A$ contains a zero divisor if and only it contains a non-zero element $x$ such that $n(x) = 0$.
\end{lemma}
\begin{proof}
For the if-direction it is enough to recall that $xx^* = n(x)$ (equation (\ref{n})). For the only-if-direction let $a, b \in A$ such that $ab = 0$. We note that by (\ref{taun}) we have that $n(0) = 0$ so that the multiplicativity of $n$ (equation (\ref{nmult})) implies that $n(a) = 0$ or $n(b) = 0$.
\end{proof}

Apparently the key to understanding the (non-)existence of zero divisors is understanding the quadratic form $n \colon A \to F$. We'll return to this in Section \ref{Serre}.

\subsection{Multiplication table and relation to the Fano plane}\label{OtoFano}
In Section \ref{introduction} we gave a multiplication table (Table \ref{multtabclassic}) of $\OO = D_{-1}(D_{-1}(D_{-1}(\RR)))$ with respect to the `standard' basis. We also indicated how, when viewing the seven imaginary elements among these standard basis elements as the points in a classical Fano plane, the lines in that Fano plane can be used to describe the multiplication. In this section we will provide the proof of those statements by providing a similar table, Table \ref{multtabgeneral}, for the general Cayley algebra $\mathcal{O} \coloneqq D_\gamma(D_\beta(D_\alpha(F)))$, where $F$ is any field and $\alpha, \beta, \gamma$ are elements of $F^\times$. A special case of interest (see also Section \ref{Serre}) occurs when $\alpha, \beta, \gamma$ are algebraically independent, transcedental elements w. r. t. a subfield $K$ of $F$, but this requirement is not necessary for the correctness of the table. In particular we see that when choosing $\alpha = \beta = \gamma = -1$ in Table \ref{multtabgeneral} we recover Table \ref{multtabclassic}, showing that Table \ref{multtabclassic} is indeed the correct multiplication table, not only of $\OO = D_{-1}(D_{-1}(D_{-1}(\RR)))$ but for the Cayley algebras $D_{-1}(D_{-1}(D_{-1}(F)))$ over any $F$.

Although Table \ref{multtabgeneral} is a bit less transparant than Table \ref{multtabclassic}, the underlying Fano plane structure is still visible when igorning the scalars. In particular: for $a, b, c \in \{e_0, \ldots, e_6\}$ we have that $abc \in F1$ (for both placements of the brackets) if and only if $\{a, b, c\}$ is a line in the standard Fano plane structure on the $e_i$. Since moreover for \emph{any} imaginary elements $a, b \in \im(\mathcal{O})$ with $a^2 \neq 0$ we have that $ab \in F1$ if and only if $b \in Fa$, we find that the product of two imaginary basis elements $e_i, e_j$ is always a scalar multiple of the third basis element $e_k$ lying on the line spanned by $e_i, e_j$ in the Fano plane structure on $\{e_0, \ldots, e_6\}$.

The goal of this subsection is to prove these statements and compute the mentioned `scalar multiples', i.e. to verify the correctness of Table \ref{multtabgeneral}. In other words, our goal is to recover the points and lines of the Fano plane, together with the additional information in Table \ref{multtabgeneral}, from the Cayley algebra $\mathcal{O}$. The opposite direction, obtaining the Cayley algebra (and subsequent $q$-covering design) from the Fano plane together with some additonal information as in Theorem \ref{mainintro}, will be the subject of Section \ref{combinatorics}.

\begin{notation}
For the duration of this subsection let $\mathcal{O} = D_\gamma(D_\beta(D_\alpha(F)))$ and let $e_0, e_1, e_2 \in \im(\mathcal{O})$ denote the elements $i$ used in the first, second and third Dickson-doubling step respectively. Hence we have $e_0^2 = \alpha1$, $e_1^2 = \beta 1$ and $e_2^2 = \gamma 1$.
\end{notation}
Altough the doubling process puts a clear `hierarchy' on the elements $e_0, e_1, e_2$, here we adopt a different perspective where the three elements $e_0, e_1, e_2$ are treated on equal footing, as \emph{generators} of $\mathcal{O}$; the remaining basis elements are then derived from them as products: $e_3 \coloneqq e_0e_1$, $e_4 \coloneqq e_1e_2$, $e_5 = e_0(e_1e_2)$ and $e_6 = e_0e_2$ respectively.

What `allows' us to treat $e_0, e_1, e_2$ as of equal importance is the fact that both these elements and their product are imaginary: taking $x = e_0, i = e_2$ in (\ref{xi}) reads $e_0e_2 = -e_2e_0$ while taking $x = e_2, i = e_0$ reads $e_2e_0 = -e_0e_2$, which is clearly equivalent. The same holds for the pairs $(e_0, e_1)$ and $(e_1, e_2)$ and hence it follows that when presented with $\mathcal{O}$ there is no meaningful way to tell whether if it was created as $D_\gamma(D_\beta(D_\alpha(F)))$, as $D_\gamma(D_\alpha(D_\beta(F)))$, as $D_\beta(D_\gamma(D_\alpha(F)))$, as $D_\beta(D_\alpha(D_\gamma(F)))$, as $D_\alpha(D_\gamma(D_\beta(F)))$ or as $D_\alpha(D_\beta(D_\gamma(F)))$, even if the three elements $e_0, e_1, e_2$ used as $i$ in each doubling step are fixed and known.

\begin{lemma}\label{imbasis}
Let $e_3 = e_0e_1$, $e_4 = e_1e_2$, $e_6 = e_0e_2$ and $e_5 = -e_3e_2$ then each of $e_0 \ldots, e_6$ is imaginary and their squares are as given as in Table \ref{multtabgeneral}.
\end{lemma}
\begin{proof}
For the proof we return to the viewpoint of $\mathcal{O}$ as $D_\gamma(D_\beta(D_\alpha(F)))$. $e_0, e_1, e_2$ are imaginary with squares $\alpha, \beta, \gamma$ respectively by definition. $e_3 = e_0e_1$ appears in the four-dimensional subalgebra $D_\beta(D_\alpha(F))$. Viewing $D_\alpha(F)$ as $A$ in equation (\ref{xistar}) and $e_0$ as $x \in A$, $e_1$ as $i$, we find that $e_3^* = -e_3$ which by (\ref{im2}) means that $e_3 \in \im(\mathcal{O})$. For the remaining three products we use the same reasoning but with $A = D_\beta(D_\alpha(F))$ and $i = e_2$ to obtain imaginarity of $e_4, e_5, e_6$.

To prove the claim about squares we note that since we established that for $i < j \in \{0, 1, 2, 3\}$ we have that both $e_i$, $e_j$ and $e_ie_j$ are imaginary we obtain by alternativity and (\ref{anticommutative}) that $(e_ie_j)^2 = e_i(e_je_i)e_j =e_i(-e_ie_j)e_j = -e_i^2e_j^2$. Note that the right hand side is just an ordinary product of scalars in $F$.
\end{proof}
From the lemma and (\ref{anticommutative}) it also follows that $e_ie_j = - e_je_i$ whenever $i \neq j$. The usefulness of Table \ref{multtabgeneral} hinges on the following:
\begin{lemma}
The set $\{1, e_0, \ldots, e_6\}$ is a basis of $\mathcal{O}$.
\end{lemma}
\begin{proof}
It suffices to show that the set is linearly indepedent and for that it suffices to show linear independence of the same set but with $e_5$ replaced by $-e_5$. Viewing $\mathcal{O}$ as $D_\gamma(D_\beta(D_\alpha(F)))$ and using the shorthand $A = D_\beta(D_\alpha(F))$ again, we obtain a vectorspace decomposition $\mathcal{O} = A \oplus Ae_2$ as in Definition \ref{DicksonDouble}. By construction $\{1, e_0, e_1, e_3\} \subset A$ and $\{e_2 = 1e_2, e_4 = e_1e_2, -e_5 = e_3e_2, e_6 = e_0e_2\} \subset Ae_2$. The decomposition $\mathcal{O} = A \oplus Ae_2$ reduces linear independence of the set $\{e_0, \ldots e_6\}$ to linear independence of the two subsets $\{1, e_0, e_1, e_3\}$ and $\{e_2, e_4, -e_5, e_6\}$ but we can say something more. Since the linear map $R_{e_2}$ (right multiplication by $e_2$) is invertible (its inverse being $\frac{1}{\gamma}R_{e_2}$) we have that the set $\{e_2 = 1e_2, e_6 = e_0e_2, e_4 = e_1e_2, -e_5 = e_3e_2\}$ is linearly independent if and only if the set $\{1, e_0, e_1, e_3\}$ is. Using the decomposition $A = D_\alpha(F) \oplus (D_\alpha(F))e_1$ we can then use similar reasoning to show that $\{1, e_0, \ldots, e_6\}$ is linear independent if and only the two element set $\{1, e_0\}$ is, and this is the case by definition.
\end{proof}

Having identified the vertex set of our Fano plane we move on to recognizing the lines.

\begin{lemma}\label{linescalars}
Let $a, b \in \im(\mathcal{O})$ be two imaginary elements satisfying $ab \in \im(\mathcal{O})$ and let $\ell = \{a, b, ab\}$. Then for every two distinct elements $p, q \in \ell$ there is scalar $\lambda_{p, q} \in F$ such that $pq = \lambda_{p,q} r$ and $qp = -\lambda_{p,q}r$ where $r$ is the third element in the set (`line') $\ell$.
\end{lemma}
\begin{proof}
The relation between $pq$ and $qp$ follows from (\ref{anticommutative}) so we focus on estabishing the $pq = \lambda r$ for some $\lambda \in F$. The case where $p = a$ and $q = b$ or vice versa is trivial, so we may assume that $q = ab$. For $p = a$ we can use alternativity to move the brackets and conclude $a(ab) = (aa)b = \lambda b$ with $\lambda = a^2 \in F$ as desired. For $p = b$ we have $b(ab) = -b(ba) = -(bb)a = \lambda a$ with $\lambda = -b^2 \in F$.
\end{proof}
The lemma gives a different way of thinking about the Fano-lines. When we view every tripple $\{a, b, c\} \subset \im(\mathcal{O})$ as a `line' whenever it satisfies $ab = \lambda c$ for some $\lambda \in F^\times$ and some labeling of the elements of the triple as $a$, $b$ and $c$, then the lemma states that this definition does in fact not depend on the choice of labeling and that for any pair $\{p, q\}$ of imaginary elements we have that all $r$ making $\{p, q, r\}$ into a line are scalar multiples of each other. (Where of course for some choices of $p, q$ no such $r$ exist.)

Our goal is to establish that (1) every two basis elements $e_i, e_j$ lie on such a line, i.e. have a product that is imaginary, and (2) that the third point on that line (which by the last lemma is unique up to scalar multiplication) can be taken to be the basis element indicated by Table \ref{multtabgeneral}. For four of the seven lines this has already been achieved by the above.

For example: our definition of $e_3$ as the product $e_0e_1$ establishes that the Fano-line through $e_0$ and $e_1$ passes through $e_3$, which by the lemma means concretely that $e_0e_1 \in Fe_3$, $e_3e_0 \in Fe_1$ and $e_1e_3 \in Fe_0$. Similarly we obtain the `lines' $\{e_1, e_2, e_4\}$, $\{e_2, e_3, e_5\}$ and $\{e_0, e_2, e_6\}$. What remains to be done to obtain the full Fano structure is verifying that the `line' through $e_3$ and $e_4$ contains $e_6$ (that is: verifying that $e_3e_4 \in F^\times e_6$) and that the line through $e_0$ and $e_4$ as well as the line through $e_1$ and $e_6$ both pass through $e_5$. This can be done by explicit computation:
\begin{eqnarray*}
e_4e_6 = (e_1e_2)(e_0e_2) = (e_2e_1^*)(e_0e_2) = \gamma(e_1^*e_0)^* = \gamma e_0^*e_1 = -\gamma e_0e_1 = -\gamma e_3 \\
e_0e_4 = e_0(e_1e_2) = e_0(e_2e_1^*) = e_2(e_0^*e_1^*) = e_2(e_0e_1) = e_2e_3 = e_5 \\
e_6e_1 = (e_0e_2)e_1 = (e_0e_1^*)e_2 = e_2(e_0e_1^*)^* = -e_2(e_0e_1)^* = -e_2e_3^* = e_2e_3 = e_5
\end{eqnarray*}
Here the first equation uses (\ref{xi}) and (\ref{pq3}), both with $e_2$ in the role of $i$, followed by (\ref{star}) and (\ref{im2}). The second equation uses (\ref{xi}) and (\ref{pq1}), both with $e_2$ in the role of $i$, followed by (\ref{im2}). The third equation uses (\ref{pq2}) and (\ref{xi}), both with $e_2$ in the role of $i$, followed by (\ref{im2}). Of course all applications of (\ref{im2}) rely on Lemma \ref{imbasis} above.

We have reached the point that for each two elements from $\{e_0, \ldots, e_6\}$ we know the unique Fano line $\ell$ to which they belong, and for each line $\ell$ we have computed the scalar $\lambda_{a, b} \in F$ in the equality $ab = \lambda_{ab}c$ for at least one of the six ways to asign the labels $a, b, c$ to the elements of $\ell$. This means that we can compute the remaining multiplications of elements on $\ell$ by the rules $ba = -ab = -\lambda_{ab}c$, $ac = \lambda_{a,b}^{-1}a^2b$, $ca = -ac$, $cb = \lambda_{ab}^{-1}ab^2$ and $bc = -cb$. This completes the verification of Table \ref{multtabgeneral}.

\begin{table}\caption{Multiplication table of the general Cayley-Algebra $D_\gamma(D_\beta(D_\alpha(F)))$ with respect to the basis $e_0, \ldots, e_6$ described in Section \ref{OtoFano} \label{multtabgeneral}}
\begin{tabular}{l|rrrrrrrr}
      &  $1$ &          $e_0$ &        $e_1$ &         $e_2$ &              $e_3$ &              $e_4$ &               $e_5$ & $e_6$  \\
      \hline
  $1$ &  $1$ &          $e_0$ &        $e_1$ &         $e_2$ &              $e_3$ &              $e_4$ &               $e_5$ & $e_6$ \\
$e_0$ & $e_0$ &      $\alpha$ &        $e_3$ &         $e_6$ &       $\alpha e_1$ &              $e_5$ &        $\alpha e_4$ & $\alpha e_2$ \\
$e_1$ & $e_1$ &        $-e_3$ &      $\beta$ &         $e_4$ &       $-\beta e_0$ &        $\beta e_2$ &        $-\beta e_6$ & $-e_5$ \\
$e_2$ & $e_2$ &        $-e_6$ &       $-e_4$ &      $\gamma$ &              $e_5$ &      $-\gamma e_1$ &        $\gamma e_3$ & $-\gamma e_0$ \\
$e_3$ & $e_3$ & $-\alpha e_1$ &  $\beta e_0$ &        $-e_5$ &     $-\alpha\beta$ &       $-\beta e_6$ &   $\alpha\beta e_2$ & $\alpha e_4$ \\
$e_4$ & $e_4$ &        $-e_5$ & $-\beta e_2$ &  $\gamma e_1$ &        $\beta e_6$ &     $-\beta\gamma$ &   $\beta\gamma e_0$ & $-\gamma e_3$ \\
$e_5$ & $e_5$ & $-\alpha e_4$ &  $\beta e_6$ & $-\gamma e_3$ & $-\alpha\beta e_2$ & $-\beta\gamma e_0$ & $\alpha\beta\gamma$ & $\alpha\gamma e_1$ \\
$e_6$ & $e_6$ & $-\alpha e_2$ &        $e_5$ &  $\gamma e_0$ &      $-\alpha e_4$ &       $\gamma e_3$ & $-\alpha\gamma e_1$ & $-\alpha\gamma$

\end{tabular}

\end{table}

\subsection{Which fields allow Cayley division algebras?}\label{Serre}
We are interested in the question in the title of this subsection because it are the Cayley division algebra that give rise, in Section \ref{qfano}, to $q$-Fano planes. It was pointed out to me by Matthias Wendt \cite{WendtMO} that an answer in terms of Galois cohomology can be found in Serre's paper \cite{Serre}: combining Theorem 9 of that paper with Theorem 5.7 of Merkurjev-Suslin \cite{MerkurSuslin} as indicated in Section 8.2 of \cite{Serre} we find:

\begin{theorem}\label{Serre2}
The field $F$ of characteristic $\neq 2$ allows Cayley division algebras if and only if the Galois cohomology group $H^3(F, \ZZ/2\ZZ)$ is non-trivial.
\end{theorem}

The definition and other applications of the groups $H^i(F, \ZZ/2\ZZ)$ are beyond the scope of the current paper. Instead we will discuss a class of examples where we can understand quite concretely what is going on.

Since the elements $e_i$ in Table \ref{multtabgeneral} are imaginary we have for a generic element $x = \eta + \xi_0e_0 + \ldots + \xi_6 e_6 \in D_\gamma(D_\beta(D_\alpha(F)))$ that $x^* = \eta 1 - \xi_0e_0 - \ldots - \xi_6 e_6 \in D_\gamma(D_\beta(D_\alpha(F)))$ (equation (\ref{im2})) and hence, by (\ref{taun}) and Table \ref{multtabgeneral} we find that

\begin{equation}\label{generaln}
n(x) = \eta^2 - \xi_0^2 \alpha - \xi_1^2 \beta - \xi_2^2 \gamma + \xi_3^2 \alpha \beta + \xi_4^2 \beta\gamma + \xi_6^2 \gamma\alpha - \xi_5^2 \alpha \beta \gamma.
\end{equation}

We are interested in finding fields $F$ and elements $\alpha, \beta, \gamma \in F$ such that $D_\gamma(D_\beta(D_\alpha(F)))$ is a division algebra. From (\ref{generaln}) and Lemma \ref{splitn0} we see (with some relabeling) that this is the case exactly when the equation
\begin{equation}\label{stroth1}
\lambda_0^2 - \lambda_1^2\alpha - \lambda_2^2\beta - \lambda_3^2\gamma + \lambda_4^2\alpha\beta + \lambda_5^2\beta\gamma + \lambda_6^2\gamma\alpha - \lambda_7^2\alpha\beta\gamma = 0
\end{equation}
has only the trivial solution $\lambda_0 = \ldots = \lambda_7 = 0$.

From this observation we recognize one class of examples:

\begin{corollary}\label{Fxyz}
Suppose that $F$ contains a subfield $K$ and elements $\alpha, \beta, \gamma$ that are algebraically independent over $K$ such that $F$ is isomorphic to the function field $F(\alpha, \beta, \gamma)$. Then $D_\gamma(D_\beta(D_\alpha(F)))$ is a division algebra.
\end{corollary}
Our proof is adapted from the proof of Bemerkung 5.3.1 in Kristin Stroth's PhD thesis \cite{stroth}. Chapter 5 of that thesis further provides an interesting geometric perspective on Cayley algebras of this type which will not be discussed here.
\begin{proof}
Let $\lambda_0, \ldots, \lambda_7 \in K(\alpha, \beta, \gamma)$ be such that $(\ref{stroth1})$ holds.

Multiplying both sides of the equation with the lowest common multiple of the denominators of the $\lambda_i^2$ we may assume that each $\lambda_i$ is an element of the polynomial ring $K[\alpha, \beta, \gamma]$.

Our goal is to show that (\ref{stroth1}) only has the zero solution over $K[\alpha, \beta, \gamma]$. We'll first show that the same holds for the simpler equation $\nu_0^2 - \nu_1^2 \alpha = 0$.

\begin{claim1}
Let $\nu_0, \nu_1 \in K[\alpha, \beta, \gamma]$ be such that $\nu_0^2 - \nu_1^2 \alpha = 0$. Then $\nu_0 = \nu_1 = 0$.
\end{claim1}
\begin{proof}
Rewriting the equation as $\nu_0^2 = \nu_1^2 \alpha$ and interpreting both sides as polynomials in $\alpha$ with coefficients in $K[\beta, \gamma]$ we see that the left hand side has even degree, while the right hand side has odd degree (adopting the convention that the zero polynomial has every degree). Hence both sides must be zero.
\end{proof}
We want to use the same reasoning to show that the `intermediate level' equation $\mu_0 - \mu_1 \alpha - \mu_2 \beta + \mu_3 \alpha \beta = 0$ also only has the zero solution, but we need an extra step:
\begin{claim2}
Let $\mu_0, \mu_1 \in K[\alpha, \beta, \gamma]$. Then viewed as a polynomial in $\beta$, $\xi \coloneqq \mu_0^2 - \mu_1^2\alpha$ has even degree.
\end{claim2}
\begin{proof}
Let $m_0, m_1$ be the degrees of $\mu_0, \mu_1$ respectively when viewed as polynomials in $\beta$ and let $m = \max(m_0, m_1)$. When $m_0 \neq m_1$ we have that the degree of $\xi = \mu_0^2 - \mu_1^2 \alpha$ equals $2m$ and we are done. When $m_0 = m_1 = m$ define $\nu_0, \nu_1$ as the leading (i.e. degree $m$) terms of $\mu_0, \mu_1$ respectively when viewed as a polynomial in $\beta$. The degree $2m$ term in $\xi$ equals $\nu_1^2 - \nu_2^2 \alpha$. Since $\xi$ clearly contains no terms of degree higher than $2m$ we see that the only scenario in which the degree of $\xi$ does not equal $2m$ is when its degree $2m$ term $\nu_0^2 - \nu_1^2\alpha$ equals zero. But by Claim 1 this only happens when $\nu_0 = \nu_1 = 0$, which, as the $\nu_i$ were defined as the leading terms of the $\mu_i$ implies that the $\mu_i$ are zero. Hence we see that either the degree of $\xi$ is $2m$ or $\xi = 0$, in which case we consider the degree of $\xi$ to be even as well.
\end{proof}
With this result under our belt we can lift the proof of Claim 1 one level higher:
\begin{claim3}
Let $\mu_0, \mu_1, \mu_2, \mu_3 \in K[\alpha,  \beta, \gamma]$ be such that $\mu_0 - \mu_1 \alpha - \mu_2 \beta + \mu_3 \alpha \beta = 0$. Then $\mu_0 = \mu_1 = \mu_2 = \mu_3 = 0$.
\end{claim3}
\begin{proof}
We can rewrite the equation as
$$(\mu_0^2 - \mu_1^2\alpha) = (\mu_2^2 - \mu_3^2\alpha)\beta.$$
From Claim 2 we see that the left hand side has even degree when viewed as a polynomial in $\beta$ while the right hand side has odd degree. It follows that both sides of the equations equal zero, so that $\mu_0^2 - \mu_1^2 \alpha = 0$ and $\mu_2^2 - \mu_3^2 \alpha = 0$. But Claim 1 then implies that $\mu_0 = \mu_1 = 0$ and that $\mu_2 = \mu_3 = 0$.
\end{proof}
From this point on it is clear how we will proceed. Similar to how we derived Claim 2 from Claim 1 we derive from Claim 3 that expressions of the form
$\mu_0^2 - \mu_1^2 \alpha - \mu_2^2 \beta + \mu_3^2 \alpha \beta$ (with $\mu_0, \mu_1, \mu_2, \mu_3 \in K[\alpha, \beta, \gamma]$) have even degree when viewed as polynomials in $\gamma$.

We then use this to conclude that the left and right hand sides of the following equivalent reformulation of (\ref{stroth1}) have different degrees as polynomials in $\gamma$ and hence are both zero:

\begin{equation}\label{stroth2}
(\lambda_0^2 - \lambda_1^2\alpha - \lambda_2^2\beta + \lambda_4^2 \alpha\beta) = (\lambda_3^2 - \lambda_5^2\beta - \lambda_6^2\alpha + \lambda_7^2\alpha\beta)\gamma.
\end{equation}

In summary we find that

\begin{eqnarray}
\lambda_0^2 - \lambda_1^2\alpha - \lambda_2^2\beta + \lambda_4^2 \alpha\beta &= 0 \label{stroth3} \\
\lambda_3^2 - \lambda_5^2\beta - \lambda_6^2\alpha + \lambda_7^2\alpha\beta &= 0 \label{stroth4}
\end{eqnarray}
and Claim 3 then tells us that $\lamba_0 = \lambda_1 = \lambda_2 = \lambda_4 = 0$ and that $\lambda_3 = \lambda_5 = \lambda_6 = \lambda_7 = 0$, as we wanted to show.
\end{proof}

\begin{remark}
The proof that over the fields $K(\alpha, \beta, \gamma)$ equation (\ref{stroth1}) has no non-zero solutions relies on the \emph{multiplicative} properties of squares, notably that the product of two squares is a square and the product of a square and a non-square is a non-square. At the other extreme we find the subfields of $\RR$ where we can get to the same conclusion by looking at the \emph{additive} properties of squares over such field. Since in a subfield of $\mathbb{R}$ a sum of squares is always non-negative, we find that over such field equation (\ref{stroth1}) has no non-zero solutions whenever $\alpha < 0, \beta < 0, \gamma < 0$.
\end{remark}

It is tempting to see if we can construct further examples that use some mixture of additive and multiplicative properties e.g. by somewhat loosening the condition that $\alpha, \beta, \gamma$ are algebraically independent over $K$ in Corollary \ref{Fxyz} while being careful about what it means to be square in the resulting field. We won't pursue that direction here, however.

\subsection{Some structure theory}\label{nilp}
So far we have focussed on the similarities rather than differences between split and divsion Cayley-Dickson algebras. However, since the method explored in Section \ref{qfano} that constructs $q$-Fano planes from Cayley division algebras fails to do so in the split case (for reasons explored in depth in Section \ref{split}) there must also be considerable differences between the two cases. Informally speaking the most important difference (to us) is that, unlike their division algebra counterparts, the split Cayley-Dickson algebras contain \emph{subalgebras} that are quite different from the Cayley-Dickson algebras of either type. The prototypical example of such a subalgebra is the algebra $U_3$ of upper triangular 2-by-2-matrices sitting inside the split quaternion algebra $M_4$ of all two-by-two matrices. Being three-dimensional this algebra cannot be a Cayley-Dickson algebra, even if it is closed under the star operator defined in Example \ref{Mat2F} and hence is a quadratic algebra with a strong involution. In this subsection we recall some notions from the structure theory of alternative algebras that enable us to articulate the differences between this algebra and the Cayley-Dickson algebras.

\begin{definition}
A two-sided \emph{ideal} in an  alternative algebra $A$ is a linear subspace $I$ such that $ai \in I$ and $ia \in I$ for every $a \in A, i \in I$. An algebra $A$ is called \emph{simple} if its only two sided ideals are $\{0\}$ and $A$ itself.
\end{definition}
The definition of ideal ensures that when $I$ is an ideal in $A$ then the multiplication in $A$ descends to a well-defined multiplication on the quotient vector space $A/I$. When $A$ is alternative or even associative then so is the quotient algebra $A/I$. It is easy to see that all division algebras are simple. Perhaps more surprising is that the split 4- and 8-dimensional Cayley-Dickson algebras are simple as well. (We'll come back to that at the end of this section). The algebra $U_3$ of upper triangular matrices is \emph{not} simple as its subspace consisting of strict upper triangular matrices is an ideal. The elements of this ideal also serve as an example to the following concept, that will be important in the sequel:

\begin{definition}
An element $x$ of an alternative algebra $A$ is called \emph{nilpotent} if there exists a natural number $n$ such that $x^n = 0$.
\end{definition}

\begin{lemma}\label{nilpotent}
Let $x \neq 0$ be an element of an alternative Cayley-Dickson algebra $O$. Then the following are equivalent:
\begin{enumerate}
\item $x$ is nilpotent
\item $x^2 = 0$
\item $x$ is a zero-divisor and $x \in \im(O)$
\end{enumerate}
\end{lemma}
\begin{proof}
$(1) \Rightarrow (2).$ Let $x$ be nilpotent and let $k$ be the smallest number such that $x^k = 0$. Since clearly $0$ is the only element for which $k = 1$, and clearly $0^2 = 0$ we will assume in the sequel that $k \geq 2$. From multiplicativity of the function $n$ (equation \ref{nmult}) we find that $n(x)^k = 0 \in F$ and hence $n(x) = 0$. Multiplying both sides of (\ref{taun}) by $x^{k-2}$ we find that $2\tau(x)x^{k-1} = 0$. By minimality of $k$ this implies that $\tau(x) = 0$. Plugging this back into the original equation (\ref{taun}) we find that $x^2 = 0$.

$(2) \Rightarrow (3).$ Assume $x^2 = 0$. It is clear that $x$ is a divisor of zero. The fact that $x \in \im(O)$ follows from (\ref{im}).

$(3) \Rightarrow (1)$ Let $x$ be an imaginary divisor of zero. By the latter property there exist a $y$ such that $xy = 0$ or $yx = 0$. In the latter case we have that $0^* = (yx)^* = x^*y^* = -xy^* = x(-y^*)$, so after replacing $y$ with $-y^*$ if needed we see that there exist non-zero $y$ such that $xy = 0$. Now by alternativity we can move the brackets in the following expression to obtain $0 = x(xy) = (x^2)y$. By $(\ref{im})$, $x^2 \in F1$ so the right hand side is a scalar multiple of the non-zero element $y$ and hence we find that $x^2 = 0$ proving (2) and hence (1).
\end{proof}

\begin{definition}\label{Jacobson}
An element $x$ in an alternative algebra $A$ is called \emph{strongly nilpotent} if $xy$ is nilpotent for all $y \in A$ or, equivalently, if $yx$ is nilpotent for all $y \in A$. The set of all strongly nilpotent elements is called the \emph{Jacobson radical} of $A$ and will be denoted $J(A)$.
\end{definition}

When $A$ is associative, it is clear that $J(A)$ is a two-sided ideal in $A$. In fact this is already true, though highly non-obviously so, when $A$ is merely alternative. A proof of this fact (originally due to Zorn \cite{Zorn}) can be found in Chapter 3 of Schafer's book \cite{Schafer} (though unfortunately only in the 1966 edition and not in the 1961 edition that is freely available online), as can the proof of the following two theorems:

\begin{theorem}\label{semisimple}
Let $A$ be an alternative algebra. The following are equivalent:
\begin{enumerate}
\item $J(A) = \{0\}$
\item $A$ is isomorphic as an algebra to a direct sum of simple alternative algebras.
\end{enumerate}
\end{theorem}
The algebra direct sum $A_1 \oplus \ldots \oplus A_n$ of algebras $A_i$ is the vector space direct sum $A_1 \oplus \ldots \oplus A_n$ equiped with the \emph{pointwise} multiplication. In other words: multiplication of two elements in the same summand $A_i$ is just the ordinary multiplication in the algebra $A_i$ while the multiplication of elements from different summands equals zero. This multiplication defined on elements of the summands is then extended linearly to all of $A$. It is clear that this construction preserves alternativity and associativity.

\begin{definition}
An algebra satisfying the equivalent conditions of Theorem \ref{semisimple} is called \emph{semi-simple}.
\end{definition}

An example of a semi-simple algebra which is not simple is the two-dimensional split Cayley Dickson algebra which we met in Proposition \ref{CD2}. This algebra is isomorphic to the algebra $D_2$ of 2-by-2 diagonal matrices which sits as a subalgebra inside the algebra $U_3$ introduced at the beginning of this section. Since it is not hard to show that $J(U_3)$ equals the ideal of strict upper triangular matrices we find that $U_3$ decomposes (as a vector space, not as an algebra) as a sum of subalgebras $U_3 = D_2 \oplus J(U_3)$. It turns out that this is part of a more general pattern:

\begin{theorem}\label{SJ}
Let $A$ be an alternative algebra with Jacobson radical $J(A)$. Then $A$ contains a semi-simple subalgebra $S$ such that as a vectorspace $A = S \oplus J(A)$ and such that the resulting vectorspace isomorphism $S \isom A/J(A)$ is in fact an isomorphism of algebras.
\end{theorem}
The fact $A/J(A)$ even \emph{is} an algebra follows from Zorn's result quoted above. The fact that the resulting quotient algebra is semi-simple is then a straightforward consequence of the definitions. The really surprising part is that $A$ already had this semi-simple quotient within it all along.
\begin{notation}
The subalgebra $S$ of Theorem \ref{SJ} will be called the \emph{semi-simple part} of $A$.
\end{notation}

In the cases we are interested in we can give an alternative characterisation of the Jacobson radical:

\begin{lemma}\label{stronglyimaginary}
Let $A$ be an alternative algebra with a strong involution (e. g. any subalgebra of a Cayley algebra) and let $x \in A$. Then $x$ is strongly nilpotent if and only if it is \emph{strongly imaginary}, that is: if and only if $xy \in \im(A)$ for every $y \in A$. (And, equivalently, $yx \in \im(A)$ for every $y \in A$.)
\end{lemma}
\begin{proof}
The `only if' direction is immediate from implication $1 \Rightarrow 3$ of Lemma \ref{nilpotent}. For the `if' direction let $x$ be strongly imaginary. Let $y \in A$. We first prove the parenthetical statement that $yx \in \im(A)$. Since $x1 \in \im(A)$, we see that strongly imaginary elements are imaginary themselves. Applying (\ref{im2}) to the imaginary elements $x$ and $xy^*$ we find that $(yx)^* = -yx$: $(yx)^* = x^*y^* = -xy^*$ $= (xy^*)^* = yx^* = -yx$. This implies that $yx$ is imaginary by (\ref{im2}).

Next we establish that $x$ itself is nilpotent. Since $x$ is strongly imaginary, $x^2 \in \im(A)$. But since $x$ is `ordinarily' imaginary, $x^2 \in F1$. It follows that $x^2 \in F1 \cap \im(A) = \{0\}$ and hence $x^2 = 0$.

Finally to see that $xy$ is nilpotent as well, we compute $(xy)^2 = -(xy)^*(xy) = -y^*x^*xy = y^*xxy = y^*0y = 0$.
\end{proof}

Using this lemma it is easy to show inductively:
\begin{theorem}\label{CDsemisimple}
Let $A$ be a Cayley-Dickson algebra of dimension 1, 2, 4 or 8, then $A$ is semi-simple.
\end{theorem}

We already saw that the same is not true for all \emph{subalgebras} of the split quaternion and Cayley algebras.

We conclude this section by proving of the following variant of Theorem \ref{CDsemisimple}:

\begin{theorem}\label{CDsimple}
Let $A$ be a Cayley-Dickson algebra of dimension 1, 4 or 8, then $A$ is simple.
\end{theorem}
\begin{definition}
The \emph{center} of an algebra $A$ is the vector space of all elements $c \in A$ that commute and associate with every element of $A$. Clearly $F1$ is contained in the center of any algebra $A$. When $F1$ equals the entire center of $A$ then $A$ is called \emph{central}.
\end{definition}
Since the center of a direct sum of algebras contains the direct sum of the centers of the summands, it follows that a semi-simple algebra can only be central when it consists of a single summand and thus is simple. We conclude that Theorem \ref{CDsimple} follows from Theorem \ref{CDsemisimple} together with the following proposition:
\begin{proposition}
All Cayley-Dickson algebras (not necessarily alternative) of dimension $\neq 2$ are central.
\end{proposition}
\begin{proof}
For the Cayley-Dickson algebra $F$ the statement is clearly true so we can restict our attention to algebras of the form $D_\gamma(A)$, the Dickson double of a Cayley-Dickson algebra $A$. Let $c$ be in the center of $D_\gamma(A)$. We can write $c = a + ib$ with $a, b \in A$ and $i$ as in $(\ref{pq1} - \ref{xistar})$. Expanding both sides of the equality $ci = ic$ into the standard form $r + is$ using $(\ref{pq1} - \ref{pq3})$ and comparing terms along the vectorspace decomposition $D_\gamma(A) = A \oplus iA$, we find that $a = a^*$ and $b = b^*$. It follows by Corollary \ref{strong} that $a \in F1$ and $b \in F1$.

Since $c$ and $a$ are both in the center of $D_\gamma(A)$ so is $c - a = ib$. We distinguish two cases: $b = 0$ and $b \neq 0$. First suppose that $b \neq 0$. Since $b \in F1$ we conclude that $b$ is invertible and that, since $ib$ is in the center of $D_\gamma(A)$, so is $i$. This means in particular that for any $x \in A$ we have that $x = x^*$ by (\ref{xi}). By Corollary \ref{strong} this implies that $A = F1$ and hence that $D_\gamma(A)$ is two-dimensional.

Conversely when $D_\gamma(A)$ is not two-dimensional we conclude that $b = 0$. But this implies that $c = a \in F1$. Since $c$ was an arbitrary element of the center of $D_\gamma(A)$, we conclude that $D_\gamma(A)$ is central.
\end{proof}

\begin{remark}
It is known that the dimension of any central simple \emph{associative} algebra is a square. Hence the smallest examples beyond $F$ are 4-dimensional and in the literature on associative algebras these are called \emph{quaternion algebras}. We just proved that a quaternion algebra in our sense, i.e. a 4-dimensional Cayley-Dickson algebra, is indeed central and simple. The converse also holds: any 4-dimensional central simple algebra $Q$ contains, as a subalgebra, a quadratic field extension $C$ of $F$ such that $Q$ is isomorphic to a Dickson double of $C$. (The proof of this latter fact can be found in any text on associative algebras, e.g. \cite{Pierce}, \cite{Vigneras}.) Thus the two notions of quaternion algebra are equivalent.
\end{remark}

config: arxiv


TITLE: How find this $\det(M_{n})$
QUESTION [0 upvotes]: Let $n$ be a postive integer, let $a,b,c$ be real numbers, with $a\neq b$, and let $M_{n}$ denote the $2n\times 2n$ matrix whose $(i,j)$ entry $m_{ij}$ is given by
$$m_{ij}=\begin{cases}
x & \text{if } \ i=j\\
a&\text{if }\ i\neq j,i+j=2k,\\
b&\text{if } \ i\neq j,i+j=2k+1
\end{cases}$$
Find the $\det (M_{n})$
I fell this matrix is very ugly, and I can't find it. Thank you

REPLY [5 votes]: Let
$$\begin{align}
  u_1 &= \begin{pmatrix}1\\1\\\vdots\\1\\1\end{pmatrix}
& u_2 &= \begin{pmatrix}1\\-1\\\vdots\\1\\-1\end{pmatrix}
\end{align}$$
with $2n$ rows each.
Furthermore, let
$$ L_n = \frac{a+b}{2}u_1^~ u_1^\top + \frac{a-b}{2}u_2^~ u_2^\top = \begin{pmatrix}a&b&a&b&\cdots&a&b\\b&a&b&a&\cdots&b&a\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\a&b&a&b&\cdots&a&b\\b&a&b&a&\cdots&b&a\end{pmatrix}$$
Then
$$M_n = L_n + (x-a)I\tag{*}$$
where $I$ is the $2n\times2n$ identity matrix.
Note that $u_1$ and $u_2$ are orthogonal eigenvectors of $L_n$ and $M_n$,
in particular,
$$\begin{align}
L_n u_1 &= n(a+b) u_1 & L_n u_2 &= n(a-b) u_2
\end{align}$$
and all other vectors orthogonal to both $u_1$ and $u_2$ are mapped to zero by $L_n$.
Therefore $L_n$ has eigenvalues
$$\begin{align}
\lambda_1 &= n(a+b) & \lambda_2 &= n(a-b) & \lambda_3 = \lambda_4 = \ldots = \lambda_{2n} = 0
\end{align}$$
And by (*), $M_n$ has eigenvalues
$$\begin{align}
\mu_1 &= n(a+b)+(x-a) & \mu_2 &= n(a-b)+(x-a) & \mu_3 = \mu_4 = \ldots = \mu_{2n} = (x-a)
\end{align}$$
The determinant is the product of all eigenvalues, therefore we get
$$\det M_n = \left(n(a+b)+(x-a)\right)\left(n(a-b)+(x-a)\right)(x-a)^{2n-2}$$

REPLY [2 votes]: Let $c=(a+b)/2$ and $d=(a-b)/2$.  Then
$$
M=(x-a)I+cJ+duu^T=A+duu^T,
$$
where $I$ is the $2n\times 2n$ identity matrix, $J$ is the $2n\times 2n$ all-ones matrix, $A=(x-a)I+cJ,$ and
$$u^T=\begin{bmatrix}1 & -1 & 1 & -1 & \ldots & 1 & -1\end{bmatrix}.$$
The inverse of $A$ and determinant of $A$ are straightforward to compute: $A^{-1}$ has the form $rI+sJ,$ and you can solve for $r$ and $s.$  The determinant is $(x-a)^{2n-1}(x-a+2nc).$  This follows from the fact that the determinant is the product of the eigenvalues, and that, for a  circulant, the eigenvalues are the components of the discrete Fourier transform of the generating row of the matrix.
Once you have $A^{-1}$ and $\det A$, you can use the matrix determinant lemma to compute $\det M.$


TITLE: Pairwise non-integral numbers
QUESTION [0 upvotes]: I have a set of complex numbers a_1 through a_n which are said to be "pairwise non-integral numbers".  Could someone explain to me exactly what this means?  Thanks.
From comment below:
I should also say the exact wording is "Pairwise different non-integral numbers".

REPLY [1 votes]: Given the alternate wording provided in comments, that seems just to mean a sequence of distinct complex numbers which are not integers.


TITLE: Implicit finite difference scheme for parabolic PDE
QUESTION [2 upvotes]: Let $u=u(t,x,y):[0,\infty)\times\mathbb{R}\times[0,\infty)\to\mathbb{R}$ be sufficiently smooth and consider the parabolic PDE $$u_t+u_{xx}+u_y=0.$$
Using forward/central finite differences, we obtain
\begin{align*}
\frac{u_{i+1,j,k}-u_{i,j,k}}{\Delta t} + \frac{u_{i,j+1,k}-2u_{i,j,k}+u_{i,j-1,k}}{(\Delta x)^2} + \frac{u_{i,j,k+1}-u_{i,j,k-1}}{2\Delta y}=0.
\end{align*}
Thus,
\begin{align*}
u_{i+1,j,k} = \frac{\Delta t}{2\Delta y} u_{i,j,k-1} -\frac{\Delta t}{2\Delta y} u_{i,j,k+1}-\frac{\Delta t}{(\Delta x)^2} u_{i,j-1,k} + \left(1+2\frac{\Delta t}{(\Delta x)^2}\right) u_{i,j,k} -\frac{\Delta t}{(\Delta x)^2} u_{i,j+1,k}
\end{align*}
This is an implicit finite difference scheme.
I've got boundary conditions on $u$ for $t\to\infty$, $x\to\pm\infty$, $y\to0$ and $y\to\infty$.
Thus, I seek $u(0,x,y)$.
To numerically solve that PDE, I need to implement a nested for loop going (backwards) through the time dimension, the $x$ dimension and finally through the $y$ dimension.
I do not know what equation system needs solving. In one dimension, $u_t+u_{xx}=0$, it's a tridiagonal matrix but I do not know how to generalize the setting to the above PDE. Could you please guide me in what system $Ax=b$ needs solving?

REPLY [2 votes]: Is it possible in your situation to make the following change of variables:
$$
\tilde{t} = -t
$$
So that:
\begin{aligned}
-u_{\tilde{t}} + u_{xx} + u_y = 0\\
u(-\infty, x, y) = f(x,y)
\end{aligned}
Thus, your finite difference scheme will be explicit.


TITLE: Computation and Visualisation of Type I and Type II Error
QUESTION [0 upvotes]: [Undergraduate : Introductory Level Statistics - Hypothesis Testing]

There is a general consensus that 80% of all economists believe inflation will increase for 2018. Call this proportion the null hypothesis. However, there is an alternative view that says that less than 80% of them believe inflation will increase for 2018. Call this the alternative hypothesis. Suppose we follow this decision rule: if we randomly poll 100 economists and find that less than 71% of them believe that inflation will increase in 2018, then reject the null hypothesis. Otherwise, we cannot reject the null hypothesis.
(i) Calculate the probability of committing a type I error for the above decision rule.
(ii) By maintaining the above decision rule, calculate the probability of committing a type II error when in fact only 60% of all economists believe inflation will increase for 2018.

I understand the general idea of the question, however, my difficulty lies in comprehending what the errors look like on a standard normal curve. My thinking is that as a result of the Central Limit Theorem, $\hat{p} \sim N (0.8\, , \, 0.04^2)$, and therefore the test statistic can be calculated as -2.25. How do I interpret this test-statistic though in terms of the first question? Is $P(Z<-2.25) = 0.0122$ the required probability? My lecturer said that this was the method but I do not understand what it means when the probability is 0.0122.
With regards to the second part, I am unable to compute this as well. How do I visualise / compute the type II probability on a standard normal curve? Thank you in advance.

REPLY [1 votes]: As per the definition that you surely will find in your lectures, Type I error is "the probabilty to reject $H_0$ when is is true", say
$$\mathbb{P}\left[ \text{Reject}H_0|p=0.80 \right]$$
thus
$$0.0122=\mathbb{P}[Z<-2.25]$$
is the error that you will commit in rejecting your null hypothesis under the condition that it is true.
In the same manner, using the definiton, Type II error is the error that you commit if you Accept (better, you do not Reject) the null hypothesis under the condition that the alternative is true, say
$$\mathbb{P}\left[ \hat{p}>0.71|p=0.6 \right]=\mathbb{P}\left[Z>\frac{0.71-0.6}{\sqrt{\frac{0.6\cdot0.4}{100}}}  \right]=1-\Phi(2.245)\approx0.0124$$

Visualization

the left picture's purple area is Type I error, $1.22\%$ while the right picture's purple area is Type II error, $1.24\%$


TITLE: Continuous function to describe a shifted Gaussian curve
QUESTION [0 upvotes]: We have a set of numerical data that strictly follows Gaussian function (e.g. Fig. 1).  Suppose if we shift the left half of the Gaussian curve to to its right end (see Fig. 2), trend lines based on Gaussian function do not provide best-fits.
Figure Links: Fig. 1 Fig.2
Is there a an elegant continuous function to capture the trend shown in Fig. 2? Note that the Y-axis can vary though the X-axis remains the same (0-360).

REPLY [1 votes]: If $f$ is your Gaussian function in figure 1 then the function in your second figure is
$$g(x)=\begin{cases} f(x) & 0 \leq x \leq 180 \\
f(360-x) & 180 \leq x \leq 360 \end{cases}.$$


TITLE: First time two independent Markov chains reach same state
QUESTION [2 upvotes]: Let $\{X_n\}$ and $\{Y_n\}$ two independent Markov chains with the
same state space $\{0,1,2\}$ and same transition probability matrix:
$$ P=\begin{pmatrix} 0.5&0.25&0.25\\ 0.25&0.25&0.5\\ 0&0.5&0.5 \end{pmatrix} $$
Define $T=\inf\{n: X_n=Y_n\}$. Assuming $X_0=0$ and $Y_0=2$, find
$\mathbb{E}[T]$ and $\mathbb{P}[X_T=2]$.

I think I know how to compute the probability, but I'm having trouble computing the expectation of $T$.
My attempt: I thought it would be easy if I consider
$$\mathbb{E}[T] = \sum_{i=0}^{2} \mathbb{E}[T\vert X_T = i]\cdot \mathbb{P}[X_T = i]$$
since the probability I need to compute in the first exercise is involved. So the only thing is left to compute is $\mathbb{E}[T\vert X_T = i]$ for $i=0,1,2$. I know how to solve this expected value if there's only one Markov chain involved, let's say if we want to know the first time $X_t = 2$ since $X_0 = 0$, using a recurrence argument, but I'm stuck when the problem involves two Markov chains.
On the other hand, I found this other question solving the analogous expected value for only one Markov chain, in a more direct way, without using conditional expected value, but I'm having trouble trying to apply a similar argument in this case.
Any help would be appreciatted.

REPLY [4 votes]: Hint: $ET=\sum P(T>n)=\sum P(X_i \neq Y_i: 1\leq i \leq n)$.


TITLE: Series representations of powers of logarithms, i.e. $\ln^{m}(n)=\sum\cdots$?$\;\;\;n\ge2$
QUESTION [1 upvotes]: As you can guess from the tittle, just to satisfy my curiosity, I've been looking for series representations of powers of logarithms, something formally expressed as
$$
\ln^{m}(n)=\sum_{...}^{\infty}\cdots
$$
where $n\ge2 \;\wedge\;n \in \mathbb{N}$.
So thats it. Have you ever found something like this? Could you post it here or point to reffreneces?
Thanks.

EDIT:
I think that I was not very clear in my question, So I'm asking for a series valid for $n\ge 2\;\wedge\;n \in \mathbb{N} $.

REPLY [2 votes]: From this we have:
$$\ln ^k(x)=\sum _{n=k}^{\infty } \frac{(-1)^n (1-x)^n k! S_n^{(k)}}{n!}$$
where: $ S_n^{(k)}$ is Stirling number of the first kind.


TITLE: Length of linearly independent list is less than or equal to length of spanning list
QUESTION [0 upvotes]: This is the proof of the theorem given in Sheldon Axler's Linear Algebra done right:
Proof of the theorem
The Linear Dependence lemma is:
Linear Dependence Lemma
In the proof you can see that in each step, one $u$ is added to the list of spanning vectors at the expense of one $w$, removed according to linear dependence lemma.
At the end,

After step $m$, we have removed all the $u$'s and the process stops. At each step, as
we add a $u$ to $B$, the Linear Dependence Lemma implies that there is some $w$
to remove. Thus there are at least as many $w$’s as $u$’s.

Here is my question: How can you guarantee that some $w$'s will remain in the spanning list $B$ waiting to be removed by linear dependence lemma (note that linear dependence lemma can only remove $w$'s, not $u$'s) till you are done with the $m^{th}$ step? Doesn't the proof already assume that the number of linearly independent vectors is less the number of spanning vectors?

REPLY [1 votes]: Sure, the author assumes that we can carry on the process up to the $m^{th}$ step, but doesn't seem to provide an explanation why we wouldn't run out of $w$'s by that step (which is a possibility). As given by @egreg, here is why all $w$'s won't be deleted by the $k^{th}$ step, $k < m$.
Suppose all $w$'s are gone by the $k^{th}$ step, we will have $\{ u_k, u_{k-1}, u_{k-2}, \cdots ,u_{1}\}$ in the spanning list, whereas $u_{k+1}, \cdots u_{m}$ are left back in list containing the linearly independent vectors.
But $\{ u_k, u_{k-1}, u_{k-2}, \cdots ,u_{1}\}$ spanning vector space $V$ means that the remaining linearly independent vectors are expressible as their linear combinations, which is contradiction to the fact that those vectors are linearly independent.
Therefore so long as there are $u$'s left in list $B$, there would be $w$'s waiting to be dropped from the spanning list. Because linear dependence lemma replaces one $w$ by one $u$ in list $A$, length of list of spanning vectors $\geq$ length of list of linear independent vectors.