TITLE: Is every linear transformation with at least two eigenvalues decomposable?
QUESTION [5 upvotes]: (A linear transformation $T : V \rightarrow V$ is decomposable if $V$ can be written as the direct sum of two proper $T$-invariant subspaces)
Some background for this question: I have been going back over linear algebra recently, with the intention of pushing standard theorems and proofs to their bare essentials, so I can get an idea of when a property is "necessary" and when it is only "sufficient".
For example the theorem which states "Every finite-dimensional inner product space $V$ induces a natural isomorphism $V \rightarrow V^*$, given by $v \mapsto <v,->$" has nothing to do with symmetry or the full power of positive definiteness. So this can be generalised from an inner product to a non-degenerate bilinear form. However this theorem does depend on the finite dimensionality of $V$, and so any proof must make use of this property.
Now I have been trying to prove the following theorem (or find a counter example to it). "Given that $V$ is a vector space, and $T : V \rightarrow V$ is a linear transformation with at least two distinct eigenvalues. Then $T$ is decomposable".
This theorem can be proven for the finite dimensional case by considering the Frobenius normal form of some matrix of $T$, but this proof explicitly uses the property of finite dimensionality. I have tried hard to produce either a more general proof that doesn't use finiteness, or a counter example in the infinite case, but I feel like I have hit a brick wall.
Any help would be greatly appreciated, so thank you in advance.

REPLY [1 votes]: We assume that i) or ii) is true
i) $V$ has finite dimension and $T$ has at least $2$ distinct eigenvalues in $K$.
ii) $V$ has infinite dimension and (*) there is $P(x)=(x-a)^r(x-b)^sQ(x)\in K[x]$ s.t. $a,b\in K,P(T)=0,Q(a)\not= 0,Q(b)\not= 0$ and $T$ does not cancel a strict divisor of $P$.
Then $V$ is $T$-decomposable.
Proof. For i), the condition (*) is also satisfied. The polynomials $(x-a)^r,(x-b)^s,Q(x)$ are pairwise prime; then
$V=\ker((T-aI)^r)\oplus[\ker((T-bI)^s)\oplus \ker(Q(T))]$ and we are done.


TITLE: On Assumptions In Induction
QUESTION [0 upvotes]: I'm currently learning mathematical induction from this site: https://www.mathsisfun.com/algebra/mathematical-induction.html
What I'm confused about is how it presents mathematical induction. It says that there are 3 steps to induction:

Show it true for $n=1$.
Assume it is true for $n=k$
Prove it is true for $n=k+1$ (we can use the $n=k$ case as a fact.)

There are many things I am confused about here, about all $3$ steps.
Starting from the first step, why is it necessary to prove it true for $n=1$? I don't get why this step is needed. Second, why choose $1$ of all numbers; can't a number like $2$ be chosen?
Moving on to the second step, why is it legitimate to assume it true for all $n=k$? Is this assumption proved true by the third step, if so, how?
On the final step, first, how can we prove it true for all $n=k+1$ ? Because this prove fundamentally assumes that it is true for $n=k$, but there is no way to verify this. Second, what happens if the set we're doing induction on has only a limited amount of numbers, let's say 100 numbers. So if we go up to the 100th number, then how can $n=k+1$ still be true? Because there is no 101st term for it to be true on; there are only 100 numbers!
Please explain this as simply as possible; I'm still a beginner. I will not be able to understand complicated proof notation.
DETAILS: This question is different from the question above since the question above uses $\lim_{x\to\infty}$ notation and other pieces of calculus knowledge. However, I have not learnt calculus, and nothing about this question suggest prior calculus knowledge. An answer where induction is explained without calculus would benefit me greatly.

REPLY [1 votes]: To start from the end. You ask:

[W]hat happens if the set we're doing induction on has only a limited amount of numbers[?]

The first thing to note is that the axiom of induction is useful for proving that a statement holds for infinitely many objects. To show that something is true only a finite number of times, one often uses other means -- and, as a last resort, brute calculation (a famous proof that uses this as one of its major steps is the Four-Colour Theorem).
You also say:

[H]ow can we prove it true for all $n=k+1$? Because this prove fundamentally assumes that it is true for $n=k$, but there is no way to verify this.

The point is not verification. We are assuming that the statement holds for the $k$th item in order to prove that it would then hold for the next, namely the $k+1$th item. Indeed, a contemporary way to look at the practice of mathematics is as trying to deduce implications of axioms and definitions (which no one can guarantee to be true or not, or to even have any specific, intrinsic meaning -- in other words, mathematics does not concern the truth of axioms per se, but correct implications of whatever axioms and conventions we make; this is what we mean by saying a statement is true in mathematics -- it follows logically from the axioms, definitions, conventions and other theorems).
Then:

[W]hy is it legitimate to assume it true for all $n=k$? Is this assumption proved true by the third step, if so, how?

First, the assumption is not proved by the third step. As to why it is legitimate to make the assumption. Well, that depends on what you mean by legitimate. I mean, that's what happens in mathematics all the time: We assume that some hypotheses hold, then we deduce some implications from this assumption.
Finally:

[W]hy is it necessary to prove it true for $n=1$? I don't get why this step is needed. Second, why choose $1$ of all numbers; can't a number like $2$ be chosen?

You need to understand the goal of this principle. Again, we want to show that a statement holds for an infinite list of objects. To do this we have to label the objects to talk about them. We use the positive integers $1,2,3,4,...$ for this, usually. However, if it is convenient for you, you could start with any other positive integer -- indeed with any integer at all. But usually however, it is always easiest to start with $n=1$ (and sometimes $n=0$). However there is no rigidity about this. The thing is, whatever integer you start with, whether you count either forwards or backwards matters little (however, we're most accustomed to counting forwards from $1$; hence we index our set of objects using the positive integers).
A final word: The goal of this method of proof is to show that if we are guaranteed that whenever $P(k)$ holds, then $P(k+1)$ must follow, then if we can only demonstrate that $P(n)$ is true for the first object, then it would follow that it holds for the next, namely the second, which would imply that it holds for the third, etc., from which it follows that it holds for them all, infinite though they be -- since the next is always true.


TITLE: Relationship between continuous spectra of operators and the spectra of their adjoints
QUESTION [3 upvotes]: Suppose that $A$ is a bounded linear operator on a Hilbert space $\mathcal{H}$ over the complex numbers $\mathbb{C}$, and that $A^*$ denotes its adjoint. The residual spectrum of $A^*$, $\sigma_r(A^*)$, has a nice relationship with the point spectrum of $A$ in that $\sigma_r(A^*) \subseteq \overline{\sigma_p(A)}$, where
$$
\begin{align*}
\overline{\sigma_p(A)} = \left\{\overline{\lambda}:\lambda\in\sigma_p(A)\right\}
\end{align*}
$$
Is there a similar relationship between the continuous spectrum of $A^*$ and the spectrum of $A$? I know that $\sigma(A^*) = \overline{\sigma(A)}$ so the conjugates of the elements of the continous spectrum of $A^*$ are contained somewhere in $\sigma(A)$, but can we say more than that?

REPLY [1 votes]: Here is an answer that I believe works: it can be shown that
$$
\sigma_c(A) = \overline{\sigma_c\left(A^*\right)}
$$
For a bounded linear operator $A$ on a Hilbert space, we have the following facts:

$\rho(A) = \overline{\rho(A^*)}$ (Hunter and Nachtergaele 239)
$\sigma_r(A)\subseteq\overline{\sigma_p(A^*)}$ (Hunter and Nachtergaele 223)
$\lambda\in\sigma_p\left(A^*\right) \Longrightarrow \overline{\lambda}\notin\sigma_c(A)$ (can't find a source that directly states this but it comes from a question in the same textbook as the previous two citations)

Let $\lambda\in\sigma_c(A)$. As a corollary to (1), $\sigma(A) = \overline{\sigma(A^*)}$, so $\overline{\lambda}\in\sigma(A^*)$. By (3), $\overline{\lambda}\notin\sigma_p(A^*)$, and so $\overline{\lambda}\in\sigma_c(A^*)\cup\sigma_r(A^*)$. But $\overline{\lambda}\in\sigma_r(A^*)\Longrightarrow\overline{\overline{\lambda}}=\lambda\in\sigma_p(A^{**}) = \sigma_p(A)$ by (2), contradicting $\lambda\in\sigma_c(A)$.
It follows that $\lambda\in\sigma_c(A)\Longrightarrow\overline{\lambda}\in\sigma_c(A^*)$, and since $\overline{\lambda}\in\sigma_c(A)\Longrightarrow\overline{\overline{\lambda}}=\lambda\in\sigma_c(A^{**}) = \sigma_c(A)$, the continuous spectrum of $A$ is the same as the conjugate of the continuous spectrum of its adjoint, i.e. $\lambda\in\sigma_c(A)\Longleftrightarrow\overline{\lambda}\in\sigma_c(A^*)$, QED.
Can anybody verify this?
References:

Hunter, John K., and Bruno Nachtergaele. Applied Analysis. World Scientific, 2007.


TITLE: Distance in metric space p_{1}
QUESTION [0 upvotes]: I need to evaluate distance of point [6,6] and circle $x^2 + y^2 = 25$ in metric space $p_{1}(x,y) = ∑|x_k-y_k|$ (sum metric). I know that I need to count $inf(p_{1}([6,6],X), X $ are points from circle, but I: 1) don't know how to write this as function; 2) can't derivate it to find minimum; 3) don't know if I am thinking in the right way.

REPLY [0 votes]: Well, it's clear the closest point is in the positive quadrant. So we need to minimise $d(x,y) = (6-x) + (6-y)$ (the distance from $(6,6)$ to $(x,y)$ on the circle) under the constraint $x^2 + y^2 = 25$. Of course we can just maximise $x+y$ instead. Intuitively it's clear this should happen on the line $y = x$, which is confirmed by applying the Lagrange multiplier method, e.g. So the point $(\sqrt{\frac{25}{2}},\sqrt{\frac{25}{2}})$ assumes the minimum (which exists by compactness).


TITLE: Cluster decomposition $\stackrel{?}{=}$ Translation invariance
QUESTION [0 upvotes]: In Weinberg Volume 1 (section 4.4), Weinberg argues for a certain structure of the interaction Hamiltonian by demanding that it produce an $S$-matrix satisfying cluster decomposition. The proposed structure is the following (flavor/spin indices suppressed)
$$ H = \sum_{n,m \geq 0}\left[\int\prod_{i=1}^n{\rm d}\vec p’_i \prod_{j=1}^m{\rm d}\vec p_j \right] \delta^3 \left(\sum_{i=1}^n \vec p’_i - \sum_{j=1}^m \vec p_j\right) \widetilde h\big(\vec p’_1,\cdots,\vec p’_n, \vec p’_1,\cdots,\vec p’_m\big) a^\dagger(\vec p’_1) \cdots a^\dagger(\vec p’_n) \, a(\vec p_1) \cdots a(\vec p_m)$$
where $\widetilde h$ is a smooth function.
It seems to me that one could more easily arrive at the above structure simply by postulating that the Hamiltonian is translation invariant. Indeed, since the translation group acts on the ladder operators as $U^\dagger(\vec b) a(\vec p) U(\vec b) = e^{-i\vec p \cdot \vec b} a(\vec p) $ (for translation by $\vec b \in \mathbb{R}^3$), we find that the above Hamiltonian is translation invariant:
\begin{align}
U^\dagger(\vec b)H U(\vec b) & = \sum_{n,m \geq 0}\left[\int\prod_{i=1}^n{\rm d}\vec p’_i \prod_{j=1}^m{\rm d}\vec p_j \right] \delta^3 \left(\sum_{i=1}^n \vec p’_i - \sum_{j=1}^m \vec p_j\right) \widetilde h\big(\vec p’_1,\cdots,\vec p’_n, \vec p’_1,\cdots,\vec p’_m\big) \, U^\dagger(\vec b) a^\dagger(\vec p’_1) U(\vec b)\cdots U^\dagger(\vec b)a^\dagger(\vec p’_n) U(\vec b) \, U^\dagger(\vec b)a(\vec p_1)U(\vec b) \cdots U^\dagger(\vec b)a(\vec p_m)U(\vec b) \\
& = \sum_{n,m \geq 0}\left[\int\prod_{i=1}^n{\rm d}\vec p’_i \prod_{j=1}^m{\rm d}\vec p_j \right] \delta^3 \left(\sum_{i=1}^n \vec p’_i - \sum_{j=1}^m \vec p_j\right) \widetilde h\big(\vec p’_1,\cdots,\vec p’_n, \vec p’_1,\cdots,\vec p’_m\big) \exp\left[i \left(\sum_{i=1}^n \vec p’_i - \sum_{j=1}^m \vec p_j\right) \cdot\vec b\right] a^\dagger(\vec p’_1) \cdots a^\dagger(\vec p’_n) \, a(\vec p_1) \cdots a(\vec p_m) \\
& = \sum_{n,m \geq 0}\left[\int\prod_{i=1}^n{\rm d}\vec p’_i \prod_{j=1}^m{\rm d}\vec p_j \right] \delta^3 \left(\sum_{i=1}^n \vec p’_i - \sum_{j=1}^m \vec p_j\right) \widetilde h\big(\vec p’_1,\cdots,\vec p’_n, \vec p’_1,\cdots,\vec p’_m\big) a^\dagger(\vec p’_1) \cdots a^\dagger(\vec p’_n) \, a(\vec p_1) \cdots a(\vec p_m) \\
& = H
\end{align}
where in the second-to-last equality we used the fact that the delta function forces $\exp\left[i \left(\sum_{i=1}^n \vec p’_i - \sum_{j=1}^m \vec p_j\right) \cdot\vec b\right] = 1$.
Conversely, starting with the requirement of translation invariance, I think we can deduce the above form of Hamiltonian.
Given that the translation group is a subgroup of Poincare, why does Weinberg bother introducing cluster decomposition as an additional postulate if it can be deduced already from translation invariance?

REPLY [1 votes]: I completely agree with the answers already provided by hft and Andrew. Thanks.
Here is an explicit counter-example (closely related to Andrew’s) but expressed in the language of ladder operators.
Consider
$$H = \int {\rm d}\vec p’_1 \, {\rm d}\vec p’_2 \, {\rm d}\vec p_1\, {\rm d}\vec p_2 \, \delta^3 \left(\vec p’_1 - \vec p_1\right) \delta^3\left(\vec p’_2 - \vec p_2\right) \, f(\vec p’_1,\vec p’_2, \vec p_1,\vec p_2) \, a^\dagger(\vec p’_1) \, a^\dagger(\vec p’_2) \, a(\vec p_1) \, a(\vec p_2)$$
with $f$ a smooth function.
Acting with translation operator,
\begin{align}
U^\dagger H U
& = \int {\rm d}\vec p’_1 \, {\rm d}\vec p’_2 \, {\rm d}\vec p_1\, {\rm d}\vec p_2 \, \delta^3 \left(\vec p’_1 - \vec p_1\right) \delta^3\left(\vec p’_2 - \vec p_2\right) \, f(\vec p’_1,\vec p’_2, \vec p_1,\vec p_2) \, \exp\left[i\big(\vec p’_1+\vec p’_2-\vec p_1 - \vec p_2\big)\cdot \vec b\right] a^\dagger(\vec p’_1) \, a^\dagger(\vec p’_2) \, a(\vec p_1) \, a(\vec p_2)
\end{align}
The presence of two delta functions again forces $\exp\left[i\big(\vec p’_1+\vec p’_2-\vec p_1 - \vec p_2\big)\cdot \vec b\right] =1$.