text:
\begin{document}

\title{Explicit solutions of division problems for matrices of polynomials}
\author{Elin Götmark}

\maketitle

\begin{abstract}
We present integral representations of solutions to division problems in
$\mathbb{C}^n$ involving matrices of polynomials. We also find estimates
of the polynomial degree of the solutions by means of careful degree
estimates of the so-called Hefer forms which are components of the
representations.
\end{abstract}

\tableofcontents

\section{Introduction}

Let $P(z) = (P_1(z), \ldots, P_m(z))$ be a tuple of polynomials in $\Cn$, and
$\Phi(z)$ another polynomial in $\Cn$ which vanishes on the zero set
$Z$ of $P$. By Hilbert's Nullstellensatz, we can
find $\nu \in \mathbb{N}$ and polynomials $Q(z) = (Q_1(z), \ldots,
Q_m(z))$ such that
\[
\Phi^\nu = P \cdot Q = P_1 Q_1 + \cdots + P_m Q_m.
\]
There has been much research devoted to finding effective versions of this,
i.~e.~determining bounds on $\nu$ and the polynomial degrees of the
$Q_i$'s. These degrees depend not only on
the degrees of $P$ and $\Phi$, but on $Z$ and the singularity of $P$ at
infinity.

A breakthrough in the problem of degree estimates came in \cite{BR}, where
Brownawell
found bounds by a combination of algebraic and analytical methods.
The optimal degrees, which are slightly better than Brownawell's, were
found by Kollár in \cite{KO}, using purely algebraic
methods. There are also classical theorems by M. Nöther and Macaulay
which treat simpler cases where one has conditions on the
singularity.

In \cite{MA1}, the problem of solving
\be \label{zabira}
\Phi = P \cdot Q = P_1 Q_1 + \cdots + P_m Q_m
\ee
with degree estimates is treated by means
of residue currents on $\Pn$ based on the Koszul complex. One can then
recover the Nöther and Macaulay theorems. In the same article are explicit
solutions $Q$ constructed by means of integral representations; one then
has some loss of precision in the degree estimates.

It is natural to look at generalizations of the division problem
(\ref{zabira}) to the case where $P$ is a matrix. More precisely, we
let $P$ be a polynomial
mapping $\Cn \to \textrm{Hom} (\mathbb{C}^m, \mathbb{C}^r)$, i.~e.~an
$r \times m$ matrix $P$ where the entries are polynomials in $n$ complex
variables. Let $\Phi = (\Phi_1, \ldots, \Phi_r)$ be an $r$-column of
polynomials. If we assume that there exists an $m$-column
of polynomials $Q$ such that $\Phi = P Q$, we want to find an explicit
solution $Q$ and get an estimate of its degree.
We can reduce the case $r>1$ to the case $r=1$ by
means of the Fuhrmann trick \cite{FU} (see Remark \ref{martha}),
however, we will then lose precision in the degree estimates.

The case $r>1$ is treated in \cite{MA3}, where an estimate of the
degree of the solution is obtained by means of
residue currents on $\Pn$, based on the Buchsbaum-Rim complex.
The aim of this paper is to present explicit integral representations
of $Q$ for $r \geq 1$. The paper \cite{MA2}
contains a method of obtaining explicit solutions to division problems
$\Phi = P Q$ over open domains $X \subset \Cn$, and we will adapt this
method so that we can use it on $\Pn$. Also, we
present careful degree estimates of so-called Hefer forms, which make it
possible to estimate the degree of the solutions. We will lose some
precision in the degree estimates compared to \cite{MA3}, but in
return we get explicit solutions. We also
point out that in principle, explicit solutions can be obtained which
satisfy the same degree estimates as in \cite{MA3}.

\section{A division formula for tuples of polynomials} \label{tofsvipa}

Instead of solving the problem $\Phi = P Q$ in $\Cn$, we will
homogenize the expression and look at
the corresponding problem in $\Pn$ instead.
Let $\z = (\z_0, \z') = (\z_0, \z_1, \ldots \z_n)$.
Set $\phi =\z_0^{\rho} \Phi(\z'/\z_0)$, where $\rho = \deg \Phi$; the
entries of $\phi$ will then be
$\rho$-homogeneous holomorphic polynomials in $\C^{n+1}$, and can be
interpreted as sections of a line bundle over $\Pn$. Assume that
$\deg(P^j) \leq d_j$, where $P^j$ is column number $j$ in $P$, and let
$p(\z)$ be the matrix with columns
$p^j(\z) = \z_0^{d_j} P^j(\z'/\z_0)$. We will try to solve the division
problem $\phi = p \cdot q$ by means of integral
representation. Generally, this is not always possible, so we will get a
residue term $\phi R^p$ as well. When $\phi$ annihilates the
residue $R^p$, we can solve the division problem. Dehomogenizing $q$ will
give a solution to our original problem.

We will now find integral representation formulas which solve the
division problem in a more general setting, and return to our original
problem in Section \ref{xavier}.
Let $S=\C[z_0, \ldots, z_n]$ be the ring of holomorphic homogeneous
polynomials over
$\C^{n+1}$ with the ordinary grading, and let $S(-k)$ denote the same
ring, but where the grading is given by adding $k$ to the degree of the
polynomial. For example, the constants have degree $k$. Assume
that we have a graded complex $M$ of modules over $\mathbb{C}^{n+1}$
\[
0 \stackrel{f_{N+1}}{\longrightarrow} M_N \stackrel{f_N}{\longrightarrow} \cdots
\stackrel{f_2}{\longrightarrow} M_1 \stackrel{f_1}{\longrightarrow} M_0 \to 0,
\]
where $M_m = S(-d_{m}^1) \oplus \ldots \oplus S(-d_{m}^{r_m})$ has
rank $r_m$ and basis $e_{m1}, \ldots,  e_{mr_m}$. We will take $d_0^i
= 0$ for all $i$. Since the complex is graded, $f_m$ must
preserve the degree of elements of $M_m$. To attain this, we must have
\be \label{pinsent}
f_m = \sum_{i=1}^{r_{m-1}} \sum_{j=1}^{r_{m}} f_m^{ij} e_{m-1,i} \otimes
e_{mj}^\ast,
\ee
where $f_m^{ij}$ is a $d_{m-1}^i-d_m^j$-homogeneous holomorphic
polynomial. Let $M = M_0 + \cdots + M_N$. We define $f:M \to M$ to be
such that $f_{|M_m} = f_m$.

Let $\mathcal{O}(-1)$ denote the tautological line bundle over $\Pn$, and if
$k$ is positive we let $\mathcal{O}(-k) : = \mathcal{O}(-1)^{\otimes k}$ and $\mathcal{O}(k) : =
(\mathcal{O}(-1)^\ast)^{\otimes k}$. Note that sections of $\mathcal{O}(k)$ can be seen
as $k$-homogeneous polynomials in $\C^{n+1}$.
Our complex over $\C^{n+1}$ has an associated complex $E$ over $\Pn$,
namely
\[
0 \to E_N \stackrel{f_N}{\longrightarrow} \cdots
\stackrel{f_2}{\longrightarrow} E_1 \stackrel{f_1}{\longrightarrow} E_0 \to 0,
\]
where
\[
E_m = \sum_{i = 1}^{r_m} \mathcal{O}(d_m^i) \otimes L_{mi}
\]
and the $L_{mi}$ are trivial line bundles with frames $e_{mi}$. We can
then write
\[
f_m = \sum_{i=1}^{r_{m-1}} \sum_{j=1}^{r_{m}} f_m^{ij} e_{m-1,i} \otimes
e_{mj}^\ast,
\]
where $f_m^{ij}$ is the section of $\mathcal{O}(d_{m-1}^i-d_m^j)$ corresponding
to $f_m^{ij}$ in (\ref{pinsent}). Note that from $E$, we can obtain a
complex $E'$ of trivial
vector bundles over $\Cn$, simply by taking the natural trivialization over
the coordinate neighborhood $\{\z_0 \neq 0\}$. We have
\be \label{salieri}
0 \to E'_N \stackrel{F_N}{\longrightarrow} \cdots
\stackrel{F_2}{\longrightarrow} E'_1 \stackrel{F_1}{\longrightarrow}
E'_0 \to 0,
\ee
where $F_m(\z') = f_m(1,\z')$ for $\z' \in \Cn$, and $E'_m$ is a
trivial vector bundle of rank $r_m$ over $\Cn$.

We introduce a $\mathbb{Z}_2$-grading on $E$, by writing  $E = E^+
\oplus E^-$ where $E^+ = \bigoplus E_{2k}$ and $E^-
= \bigoplus E_{2k+1}$, and we say that sections of $E^+$ have even degree and
sections of $E^-$ have odd degree. This induces a grading
$\textrm{End} E = (\textrm{End} E)^+ \oplus (\textrm{End} E)^-$ where
a mapping $f:E \to E$ has odd degree if it maps $E^+ \to E^-$ and $E^-
\to E^+$, and similarly for even mappings. We also get an induced
grading on currents taking values in $E$, that is, $\D'_{\bullet,
  \bullet} (\Pn, E)$, by combining the gradings on
$\D'_{\bullet, \bullet} (\Pn)$ and $E$, so that $\alpha \otimes \phi =
(-1)^{\deg \alpha \cdot \deg \phi} \phi \otimes \alpha$ if $\alpha$ is
a current and $\phi$ a section of $E$. Similarly, we get
an induced grading on $\D'_{\bullet, \bullet} (\Pn, \End E)$, then $\dbar$
is an odd mapping. Since $f$ is odd and holomorphic, it is easy to
show that $\dbar \circ f = - f \circ \dbar$.

Let $Z_i$ be the set where $f_i$ does not attain its highest rank, and set $Z
= \bigcup Z_i$. To construct integral formulas, we need to find
currents $U_k$ with bidegree $(0,k-1)$ and
$R_k$ with bidegree $(0,k)$, with values in $\textrm{Hom}
(E_0,E_k)$, such that if $U = \sum_k U_k$ and $R = \sum_k R_k$, then
\be \label{bone}
(f - \dbar)U = I_{E_0} - R.
\ee
For the reader's convenience, we indicate the construction here, but
see \cite{AW} for details. We first define mappings
$\sigma_i:E_{i-1} \to E_i$ defined outside $Z_i$, such that $f_i
\sigma_i = \textrm{Id}$ on $\textrm{Im} f_i$, $\sigma_i$
vanishes on $(\textrm{Im} f_i)^\perp$, and $\textrm{Im} \sigma_i \perp
\textrm{Ker} f_i$. Now, set $u_1 = \sigma_1$ and
$u_k = \sigma_k \dbar \sigma_{k-1} \w \cdots \w \dbar \sigma_1$. We have $(f -
\dbar)u = \sum_k (f_k u_k - \dbar u_k) = f_1 \sigma_1 + \sum_k
(\dbar u_{k-1} - \dbar u_k) = \textrm{Id}_{E_0}$ outside $Z$.
To extend $u$ over $Z$, we let $U$ be the analytical continuation of
$|f|^{2\lambda} u$ to $\lambda = 0$, then we get
(\ref{bone}). Note that for degree reasons we necessarily have $U_k
= 0$ for $k > \mu + 1$, where $\mu = \min(n,N-1)$.

We will need to recall a proposition
about representation of holomorphic sections of line bundles over
$\Pn$. See for example \cite{EG} for more on this. Let $z = (1,z')$ be
a fixed point, and set
\[
\eta = 2 \pi i \sum_0^n z_i \ddzi.
\]
If we express a projective form in homogeneous coordinates, we can let
contraction with $\eta$ act on it and get a new projective form. In
fact, we will get an operator
\[
\delta_\eta : \mathcal{D}'_{l+1,0}(\Pn,\mathcal{O}(k)) \to
  \mathcal{D}_{l,0}'(\Pn,\mathcal{O}(k-1)),
\]
where $\mathcal{D}'_{l,0}(\Pn,\mathcal{O}(k))$ denotes the
currents of bidegree $(l,0)$ taking values in $\mathcal{O}(k)$.

\begin{prop} \label{hatsuyuki}
Let $\phi(\z)$ be a section of $\mathcal{O}(\rho)$, and let $g(\z,z) =
g_{0,0} + \ldots + g_{n,n}$ be a smooth form, where
$g_{k,k}$ is of bidegree $(k,k)$ and takes values in
$\mathcal{O}(\rho + k)$. Assume that $(\delta_\eta - \dbar)g = 0$
  and $g_{0,0}(z,z) = \phi(z)$. Then we have
\[
\phi(z) = \int_{\Pn}(g \w \alpha^{n+\rho})_{n,n},
\]
where
\[
\alpha = \alpha_0 + \alpha_1 = \frac{z \cdot \bar \z}{|\z|^2} - \dbar
\frac{\bar \z \cdot d\z}{2 \pi i |\z|^2}.
\]
\end{prop}

The idea now is to find a weight $g$ that contains $f_1$ as a factor,
and then apply Proposition \ref{hatsuyuki}. As components of the
weight, we need so-called Hefer forms.
We define $f^z_m:E_m \to E_{m-1}$ to be the mapping
\[
f^z_m = \sum_{i=1}^{r_{m-1}} \sum_{j=1}^{r_{m}} \z_0^{d_{m-1}^i-d_m^j}
f_m^{ij}(z) e_{m-1,i} \otimes e_{mj}^\ast.
\]

\begin{prop} \label{yak}
There exist $(k-l,0)$-form-valued mappings $h_k^l(\z,z): E_k \to E_l$,
such that $h_k^l = 0$ for $k<l$, $h_l^l = I_{E_l}$, and
\[
\delta_\eta h_k^l = h_{k-1}^l f_k- f_{l+1}^z h_k^{l+1}
\]
Moreover, the coefficients of the form $(h_k^l)_{\alpha \beta}$, that
is, the coefficient of $e_{l\alpha} \otimes e_{k\beta}^\ast$, will take values in
$\mathcal{O}(d_{l}^\alpha-d_{k}^\beta+k-l)$, and can be chosen so that they are
holomorphic polynomials in $z'$ of degree $d_{l}^\alpha-d_{k}^\beta-(k-l)$.
\end{prop}

We will prove this in the next section. Now we can define our weight:

\begin{prop} \label{eowyn}
If
\[
g = f_1^z \sum h_k^1 U_k + \sum h_k^0 R_k = f_1^z h^1 U + h^0 R,
\]
then $g_{0,0} {\arrowvert_{\Delta}} = I_{E_0}$ and $\nabla_\eta g = 0$.
\end{prop}

The following proof is identical to the one in Section 5 in
\cite{MA2}, except that our proof will be in $\Pn$ instead of $\Cn$.
We include it for the reader's convenience.
\begin{proof}
By definition, we have
\be \label{signal}
\nabla_\eta g = f_1^z [(h^1 f - f^z h^2)U - h^1 \dbar U]
+ (h^0 f - f^{z} h^1)R - h^0 \dbar R.
\ee
Note that one has to check the total degree of $h^1$ and $h^0$ to get
all the signs correct. Since $\dbar U = fU-R-I_{E_0}$ and $f^2
= 0$, the right hand side of (\ref{signal}) is equal to
\[
f_1^z [h^1 fU - h^1(fU-R-I_{E_0})]+ h^0(fR-\dbar R)-f^{z} h^1R.
\]
Furthermore, we have $h^1I_{E_0} = 0$ and $(f - \dbar)R = (f - \dbar)(R +
I_{E_0}) = (f - \dbar)^2 U = 0$, so it follows that $\nabla_\eta g = 0$.
We also have $g_{0,0} = f^z h_1^1 U_1 = f^zU_1$, so that $g_{0,0}
{\arrowvert_{\Delta}} = fU_1 = I_{E_0}$.
\end{proof}

This is the main result of this section:

\begin{thm} \label{inuvik}
Let $\phi(\z)$ be a section of $\mathcal{O}(\rho) \otimes E_0$. Fix $z =
(1,z')$ and let $\Phi(z') =
\phi(1,z')$, and $F_1(z') = f_1(1,z')$. If $g$ is defined as in
Proposition \ref{eowyn}, we have the following decomposition:
\[
\Phi(z') =  \int_\z g \phi \w \alpha^{n + \rho} =
F_1(z') Q(z') + \int_\z h^0 R \phi \w \alpha^{n + \rho},

\]
where $Q(z') = \sum_1^{r_1} Q_i(z') e_{1j}$ with
\[
Q_i(z') = \int_\z \z_0^{-d_1^i} (h^1 U \phi)_i(\z,z) \w \alpha^{n + \rho}.
\]
If $R \phi = 0$, the second integral will vanish and we
get a solution to our division problem. Moreover, we get the estimate
$\deg_{z'} F_1Q \leq \max_{k,\beta} (\rho - d_k^\beta)$.
\end{thm}

\begin{proof}
First, we note that
\[
g \phi = \sum_{j=1}^{r_0} (g \phi)_j e_{0j},
\]
and according to Proposition \ref{eowyn}, $(g \phi)_j$ is
a weight of the type needed to apply Proposition \ref{hatsuyuki}. Now
we need only check the degree in $z'$. The term in $(g \phi)_j$ of
bidegree $(k,k)$ is $(f_1^z h_{k+1}^1 U_{k+1} \phi)_j$ (if we
disregard the terms containing $R$). We
must pair this with $\alpha_0^{\rho+k} \alpha_1^{n-k}$ to get
something with full bidegree in $z$.

The degree in $z'$ of a term in
\[
\sum_{k, \alpha, \gamma, \beta}

(f_1^z)_{j \alpha} (h_{k+1}^1)_{\alpha \beta}(U_{k+1})_{\beta \gamma} \phi_\gamma \w
\alpha_0^{\rho +k} \alpha_1^{n-k}
\]
is $- d_1^\alpha + (d_1^\alpha - d_{k+1}^\beta - k) + (\rho
+k) = \rho - d_k^\beta$. We get the estimate $\deg F_1 Q
\leq \max_{k,\beta} (\rho - d_{k+1}^\beta)$.
\end{proof}

\section{Constructing Hefer forms} \label{logan}

In this section we construct Hefer forms for the complexes $E$ and
$E'$, and investigate their degrees. We first state and prove a more
general theorem, and obtain as a corollary Hefer forms $H_k^l$ with
specific degrees for the complex $E'$ over $\Cn$. From the $H_k^l$ we
get corresponding Hefer forms $h_k^l$ for the complex $E$ over
$\Pn$, i.\ e.\ we prove Proposition \ref{yak} in the previous section.

Let $\delta_{\z-z}$ denote contraction with the vector field
\[
2 \pi i \sum_1^n (\z_j - z_j) \frac{\partial}{\partial \z_j}.
\]

\begin{thm} \label{linea}
Let $\phi(\z,z)$ be a $(l,0)$-form with holomorphic
polynomials of degree $r$ for coefficients. If $l>0$
and $\delta_{\z-z}\phi = 0$, we have $\phi = \delta_{\z-z} \psi$, where $\psi$
is a $(l+1,0)$-form with holomorphic polynomials of
degree $r-1$ for coefficients. If $l = 0$, let $\phi(\z)$ be a
holomorphic polynomial. We can then write $\phi(\z) - \phi(z) =
\delta_{\z-z} \psi$, where $\psi$ satisfies the same conditions as above.
\end{thm}

For the proof of Theorem \ref{linea}, we need a lemma:

\begin{lma} \label{ineffable}
Let $\alpha = (\alpha_1, \ldots, \alpha_{n})$ and $\beta = (\beta_1,
\ldots, \beta_{n})$ be multiindices and $D_R$ the ball with
radius $R$ in $\Cn$. We have
\be \label{blodpudding}
I_{\alpha,\beta,R} := \int_{\partial D_R} w^\alpha \bar w^\beta
\partial|w|^2 \w ( \ddbar |w|^2 )^{n-1} = 0
\ee
if $\alpha \neq \beta$ and (\ref{blodpudding}) is non-zero and proportional
to $R^{2(|\alpha|+n)}$ if $\alpha = \beta$.
\end{lma}

\begin{proof} (of Lemma \ref{ineffable})
The integral is unchanged if we take the pullback of the integrand by
the function $A(w) = (\lambda_1 w_1, \ldots
\lambda_n w_{n})$, where $\lambda_i \in \mathbb{C}$
and $|\lambda| = |(\lambda_1, \ldots \lambda_{n})| = 1$. We have
\[
I_{\alpha,\beta,R} = \int_{\partial D_R} (A w)^\alpha
(\overline{Aw})^\beta \partial|Aw|^2 \w ( \ddbar |A
w|^2 )^{n-1} = \lambda^{\alpha} \bar \lambda^{\beta} I_{\alpha,\beta,R}.
\]
It is now clear that $I_{\alpha,\beta,R} = 0$ if $\alpha \neq
\beta$. A similar calculation gives
\[
I_{\alpha,\alpha,R} = \int_{\partial D} (R w)^\alpha (R \bar w)^\alpha
\partial|R w|^2 \w ( \ddbar |Rw|^2 )^{n-1} = R^{2(|\alpha|+n)}
I_{\alpha,\alpha,1},
\]
which shows that $I_{\alpha,\alpha,R}$ is proportional to
$R^{2(|\alpha|+n)}$.
\end{proof}

\begin{proof} (of Theorem \ref{linea})
The setup here is the same as in Section 4 in \cite{MA2}.
We view $E = T^\ast_{1,0}(\Cn)$ as a rank $n$ vector bundle over $\Cn$ and let
$f(\z) = 2 \pi i(\z - z) \cdot e^\ast$, where $z \in \Cn$ is
fixed, be a section of $E^\ast = T_{1,0}(\Cn)$. We can now express the
contraction
$\delta_{\z-z}$ as $\delta_f$ operating on $E$. Take a section
$\phi(\z,z)$ of $\Lambda^l E$ with holomorphic polynomials of degree
$r$ for coefficients. We want to show that $\phi = \delta_f
\psi$, where $\psi$ is a section of $\Lambda^{l+1} E$ with holomorphic
polynomials of degree $r-1$ for coefficients.

Let $D \subset \Cn_w$ be the unit ball. We let $h = dw \w e^\ast$,
then $\delta_{w-\z} h = 2 \pi i (w-\z) \cdot e^\ast = f(w) - f(\z)$, so
that $h$ will work as a Hefer form. Let
\[
\sigma = \frac{\bar w \cdot e}{2 \pi i(|w|^2 - \bar w \cdot z)},
\]
then $\delta_{f(w)} \sigma = 1$ if $w \neq z$. Set $u = \sum \sigma \w (\dbar
\sigma)^k$, note that $u$ depends holomorphically on $z$. Since $u$ is
integrable, we let $U$ be the trivial extension over $z$. Moreover, we
have $(\delta_{f(w)} - \dbar)U = 1 - R$, where $R_k = 0$ for $k<n$ and
$(\delta_h)_{n} R_{n} = [z]$, where $[z]$ denotes the $(n,n)$-current
point evaluation at $z$. This follows, for example, from
Theorem 2.2 in \cite{MA4}. Let $\phi(\z,z)$ be a section of
$\Lambda^l E$ with $l>0$ which satisfies $\delta_{f(\z)} \phi = 0$. We set
\[
g(w,\z) = \delta_{f(\z)} \sum \delta_h^{k-1} (U_k \phi(w,z)) + \sum \delta_h^k
(R^k \phi(w,z));
\]
a calculation shows that $(\delta_{w-\z} - \dbar)g = 0$ and
$g_{0,0}(\z,\z) = \phi(\z,z)$. Note that the second sum is actually
zero since $R \w \phi = 0$.

Now define
\[
g_1(w,\z) = \chi_D - \dbar \chi_D \w \frac{s}{\nabla_{w-\z}s} = \chi_D - \dbar
\chi_D \w \sum s \w (\dbar s)^k
\]
where
\[
s = \frac{1}{2 \pi i} \frac{\bar w \cdot dw}{|w|^2 - \bar w \cdot \z}.
\]
Note that $(\delta_{w-\z} - \dbar)g_1 = 0$ and
$(g_1)_{0,0}(\z,\z) = 1$. Also, $g_1$ depends holomorphically on $\z$.

We can now take a current $u = u_{1,0} + \ldots + u_{n,n-1}$ that is smooth
outside $\z$ and such that $(\delta_{w-\z} - \dbar) u = 1-[\z]$. We
have $(\delta_{w-\z} - \dbar)(u \w g \w g_1) = g \w
g_1 - [\z]\w g \w g_1 = g \w g_1 -
\phi(\z,z)[\z]$, so
that $\dbar (u \w g \w g_1)_{n,n-1} = \phi(\z,z)[\z] - (g \w
g_1)_{n,n}$. Stokes' theorem gives us $\phi(\z,z) = \delta_{f(\z)}
T\phi$ where
\[
T\phi =  \int_{\partial D_w} \sum_1^{n-l} \frac{\bar w
  \cdot e \w (\delta_h)_{k-1} [ (d \bar w \cdot e)^{k-1} \w \phi] \w
  \partial |w|^2 \w (\frac{i}{2 \pi} \ddbar |w|^2)^{n-k}}
{2 \pi i (1-\bar w \cdot z)^k (1-\bar w \cdot \z)^{n-k+1}}.
\]
This is seen by noting that $\delta_h \sigma \w s = 0$ and that
\[
\sigma \w \dbar \sigma = \sigma \w \frac{d \bar w \cdot e}{|w|^2 - \bar
w \cdot z}.
\]
First, note that $T\phi$ is holomorphic in $\z$ and $z$ and takes
values in $\Lambda^{l+1} E$. To see that the coefficients are
polynomials of degree $r-1$, we write
\[
\frac{1}{1 - \bar w \cdot \z} = \sum_0^\infty (\bar w \cdot \z)^k
\]
and similarly for $1/(1 - \bar w \cdot z)$. The only place in $T\phi$
where $w$'s occur is in $\phi$, and according to Lemma
\ref{ineffable} we need to match those up with a corresponding number
of $\bar w$'s. These can come either from the factor $\bar w \cdot
e$, or from the geometric expansions. It is clear that if we start with a
term in $\phi(w,z)$ which is of degree $r$ in $w$, after we
integrate we shall get a term which is of combined degree $r-1$ in $\z$
and $z$. It is easy to see that the same thing is also true if we
start with a term which is of combined degree $r$ in $z$ and $w$.
The case when $\phi(w)$ takes values in the trivial bundle is proved
in a similar way.
\end{proof}

We will now find Hefer forms for the complex $E'$ by means of
Theorem \ref{linea}.

\begin{prop} \label{kowalski}
There exist $(k-l,0)$-form-valued mappings $H_k^l: E_k \to E_l$,
where the coefficients are holomorphic polynomials in $\z$ and $z$,
such that $H_k^l = 0$ for $k<l$, $H_l^l = I_{E_l}$, and
\be \label{fraser}
\delta_{\z-z} H_k^l = H_{k-1}^l F_k(\z) - F_{l+1}(z) H_k^{l+1}.
\ee
Moreover, the polynomial degree of $(H_k^l)_{\alpha \beta}$ (that is, the
coefficient of $e_{l \alpha} \otimes e_{k \beta}^\ast$) is
$d_{l}^\alpha-d_{k}^\beta-(k-l)$.
\end{prop}

\begin{proof}
The proof will be by induction, and by application of Theorem
\ref{linea}. For $k-l=1$ we must solve
$\delta_{\z-z} H_{l+1}^l = F_{l+1}(\z) - F_{l+1}(z)$. We can apply
Theorem \ref{linea} to $(H_{l+1}^l)_{\alpha \beta}$, and it follows
that $(H_{l+1}^l)_{\alpha \beta}$ is a
$(1,0)$-form whose coefficients are
holomorphic polynomials of degree $d_{l}^\alpha-d_{l+1}^\beta-1$.

Assume that the proposition holds for $H_k^l$ with $k-l=i$. We first prove
that the right hand side of (\ref{fraser}) is $\delta_{\z-z}$-closed. We
have
\[
\delta_{\z-z} (H_{k-1}^l F_k(\z) - F_{l+1}(z) H_k^{l+1}) =
\]
\[
= (H_{k-2}^l F_{k-1}(\z) - F_{l+1}(z) H_{k-1}^{l+1})F_k(\z) - F_{l+1}(z)
(H_{k-1}^{l+1} F_k(\z) - F_{l+2}(z) H_k^{l+2}) = 0,
 \]
since $E'$ is a complex. Now regard the
\[
\delta_{\z-z} (H_k^l)_{\alpha \beta} = (H_{k-1}^l F_k(\z) - F_{l+1}(z)
H_k^{l+1})_{\alpha \beta} =
\]
\[
= \sum_{\gamma = 1}^{r_{k-1}} (H_{k-1}^l)_{\alpha \gamma}
(F_k(\z))_{\gamma \beta} - \sum_{\gamma = 1}^{r_{l+1}} (F_{l+1}(z))_{\alpha \gamma}
(H_k^{l+1})_{\gamma \beta}.
\]
where $k-l=i+1$. The right hand side is a form of bidegree
$(k-l-1,0)$, and it is easy to see by examining the sums that the
coefficients are
polynomials of degree $d_{l}^\alpha-d_{k}^\beta - (k-l-1)$. It follows
from Theorem \ref{linea} that $(H_k^l)_{\alpha \beta}$ exists and is a form of
bidegree $(k-l,0)$ whose coefficients are holomorphic polynomials of degree
$d_{l}^\alpha-d_{k}^\beta - (k-l)$.
\end{proof}

We will now prove Proposition \ref{yak} by adapting the Hefer forms
$H_k^l$ to the complex $E$ on $\Pn$. From now on, we change notation
so that $\z = (\z_0, \z') \in \C^{n+1}$ and $z = (1,z') \in
\C^{n+1}$.

\begin{proof} (of Proposition \ref{yak})
The form $(H_k^l)_{\alpha \beta}$  has bidegree $(k-l,0)$, so we have
\[
(H_k^l)_{\alpha \beta} = \sum_{|I| = k-l} a_I d(\z')_I.
\]
We set
\[
(h_k^l)_{\alpha \beta} (\z,z') = \z_0^{d_{l}^\alpha-d_{k}^\beta+ (k-l)} \sum_{|I| =
  k-l}a_I (\z'/\z_0,z' ) d ( \z'/\z_0)_I.
\]
It is clear that $(h_k^l)_{\alpha \beta}$ is a projective
form, since we are multiplying with a high enough degree of $\z_0$,
and that it takes values in the right line bundle. Now, note that
\[
\delta_\eta d (\z_j/\z_0) = z_j/\z_0 - \z_j/\z_0^2.
\]
We have
\[
\delta_\eta (h_k^l)_{\alpha \beta} = \z_0^{d_{l}^\alpha-d_{k}^\beta + (k-l-1)} \sum_{|I| =
  k-l} \sum_{j \in I} \pm a_I (\z'/\z_0,z') (z_j - \z_j/\z_0
) d (\z'/\z_0)_{I \setminus j} =
\]
\[
= \z_0^{d_{l}^\alpha-d_{k}^\beta + (k-l-1)} \delta_{\z'/\z_0-z'}
(H_k^l)_{\alpha \beta} (\z'/\z_0,z') =
\]
\[
= \z_0^{d_{l}^\alpha-d_{k}^\beta + (k-l-1)} (H_{k-1}^l (\z'/\z_0,z')
f_k(\z'/\z_0)- f_{l+1}(z') H_k^{l+1} (\z'/\z_0,z'))_{\alpha \beta} =
\]
\[
= (h_{k-1}^l f_k- f_{l+1}^z h_k^{l+1})_{\alpha \beta},
\]
so we are finished.
\end{proof}

\section{The Buchsbaum-Rim complex} \label{xavier}

We will now introduce the Buchsbaum-Rim complex, which is a special
case of our complex $E$ over $\Pn$. For more details about this
complex see e.~g.~\cite{MA3}. Let $P: \Cn \to \textrm{Hom}
(\mathbb{C}^m, \mathbb{C}^r)$ be a generically surjective polynomial
mapping and $Z$ the set where $P$ is not surjective. The mapping $P$
will have the role of $F_1$ in the complex (\ref{salieri}). Assume that
$\deg(P^j) \leq d_j$, where $P^j$ is number $j$ in $P$,
and $d_1 \geq d_2 \geq \cdots$.
Let $p(\z)$ be the matrix with columns $p^j(\z) = \z_0^{d_j}
P^j(\z'/\z_0)$. The mapping $p$
will have the role of $f_1$ in (\ref{salieri}). If
$L_{1}, \ldots, L_m$ are trivial line
bundles over $\Pn$, with frames $e_1, \ldots,
e_m$, we can define the rank $m$ bundle
\[
E_1 = (\mathcal{O}(-d_1) \otimes L_1) \oplus \cdots \oplus (\mathcal{O}(-d_n)
\otimes L_m).
\]
Let $E_0$ be a trivial vector bundle of rank $r$, with the frame
$\{\epsilon_j\}$, and let $p_{ij}$ be the $i$:th element in the column
$p^j$. We can then view
\[
p = \sum_{i=1}^m \sum_{j=1}^r p_{ij} \epsilon_i \otimes e_j^\ast
\]
as a mapping from $E_1$ to $E_0$.
Contraction with $p$ acts as a mapping $\delta_{p} : E_1 \otimes E_0^\ast
\to \C$.

Now, note that we can also write $p = \sum p_i \otimes
\epsilon_i$ where $p_i = \sum_j p_{ij} e_j^\ast$ is a section of
$E_1^\ast$. Let
\[
\det p = p_1 \w \ldots \w p_r \otimes \epsilon_1 \w
\ldots \w \epsilon_r.
\]
We now get a complex
\[
0 \to E_{m-r + 1} \stackrel{\delta_{p}}{\longrightarrow} \cdots
\stackrel{\delta_{p}}{\longrightarrow} E_3
\stackrel{\delta_{p}}{\longrightarrow} E_2
\stackrel{\det {p}}{\longrightarrow} E_1
\stackrel{{p}}{\longrightarrow} E_0 \to 0
\]
over $\Pn$, where for $k \geq 2$ we have
\[
E_k = \Lambda^{k+r-1} E_1 \otimes S^{k-2} E_0^\ast \otimes \det E_0^\ast,
\]
and $S^l E_0^\ast$ is the subbundle of $\bigotimes E_0^\ast$
consisting of symmetric $l$-tensors of $E_0^\ast$.

By $f: E \to E$ we mean the mapping which is either $p$, $\det p$ or
$\delta_{p}$, depending on which $E_k$ we restrict it to. We assume
that $r$ is odd, so that $\dbar \circ \det f = - \det f \circ \dbar$ and
consequently $\dbar \circ f = - f \circ \dbar$. If $r$ is even, one has to
insert changes of sign at some places.

Next, we can construct currents $U_k$ with bidegree $(0,k-1)$ and
$R_k$ with bidegree $(0,k)$, with values in $\textrm{Hom}
(E_0,E_k)$, such that $(f - \dbar)U = I_{E_0} -R$, or in other words
\[
p U_1 = I_{E_0}, \quad (\det p) U_2 - \dbar U_1 = -R_1, \quad
\textrm{and} \quad
\delta_{p} U_{k+1} - \dbar U_k = -R_k
\]
for $k \geq 2$. See \cite{MA3} for details of the construction.

Now, recall that we want to solve the division problem $\Phi = P Q$,
where $\Phi = (\Phi_1, \ldots, \Phi_r)$ is an $r$-column of
polynomials of degree $\rho$. The homogenized version of this is $\phi = pq$,
where $\phi$ takes values in $\mathcal{O}(\rho) \otimes E_0$, so that we can
apply Theorem \ref{inuvik}. To do that, we have to
determine the $d_k^j$ of the Buchsbaum-Rim complex.

For $E_0$ we have
$r_0 = r$ and $d_0^j = 0$ for all $j$, and
for $E_1$ we have $r_1 = m$ and $d_1^j = -d_j$. Furthermore, for $E_k$
with $k \geq 2$
we have $r_k = {m \choose k+r-1} r^{k-2} / (k-2)!$, where ${m \choose k+r-1}$ is
the dimension of $\Lambda^{k+r-1} E_1$ and $r^{k-2} / (k-2)!$ is the dimension of
$S^{k-2} E_0^\ast$. However, it is only the factor $\Lambda^{k+r-1}
E_1$ that contributes with any non-trivial line bundles, and and so we
will not be interested in the factors in $S^{k-2} E_0^\ast \otimes \det
E_0^\ast$. We then have $d_k^{I} = -\sum_{i \in I} d_i$ where $|I|=k +
r -1$, by which we mean that the coefficient of the basis vector $e_I$
in $\Lambda^{k+r-1} E_1$ will take values in $\mathcal{O}(d_k^I) = \mathcal{O}(-\sum_{i
  \in I} d_i)$.

Recall that $f_k = \delta_p: E_k \to E_{k-1}$ for $k \geq 2$. Let
$f_k^{IJ}$ be the
coefficent of $e_I \otimes e_J^\ast$, where $|I| = k+r-2$, $|J| = k+r-1$ and
$I \subset J$. Then $f_k^{IJ}$ takes values in
\[
\mathcal{O}(d_{k-l}^I - d_k^J) = \mathcal{O}(\sum_{i \in I}
d_i -\sum_{i \in J} d_i) = \mathcal{O}(d_j)
\]
where $\{j\} = J \setminus I$. In the same way, $H_k^l: E_k \to
E_{l}$, and for $k,l \geq 2$, we have that $(H_k^l)^{IJ}$ is a $(k-l)$-form
whose coefficients take values in $\mathcal{O}(\sum_{i \in J \setminus I} d_i +
k-l)$ and are holomorphic polynomials in $z$ of degree
$\sum_{i \in J \setminus I} d_i - (k-l)$. Also, $(H_k^l)^{IJ} = 0$ if
$I \not \subset J$.

\begin{thm} \label{eliot}
Let $P(\z'): \Cn \to \textrm{Hom}
(\mathbb{C}^m, \mathbb{C}^r)$ be a generically surjective polynomial
mapping where the columns $P^j$ have degree $d_j$, and $\Phi(\z') =
(\Phi_1, \ldots, \Phi_r)$ a tuple of polynomials with degree
$\rho$. We have the following decomposition:
\[
\Phi(z') = P(z')Q(z') +

\int_\z H^0 R \phi \w \alpha^{n + \rho},
\]
where $Q(z') = \sum_1^{m} Q_i(z') e_{j}$ with
\[
Q_i(z') = \int_\z \z_0^{d_i} (h^1 U \phi)_i \w \alpha^{n + \rho}.
\]
If $R \phi = 0$, the second integral will vanish and we
get a solution to the division problem. Moreover, the polynomial
degree of $PQ$ is $\rho + d_1 + \ldots + d_{\mu + r}$, where
$\mu = \min(n,m-r)$.
\end{thm}

\begin{proof}
From Theorem \ref{inuvik}, we have
\[
\deg P Q \leq \max_{k,I} (\rho - d_k^I).
\]
Since $d_k^{I} = -\sum_{i \in I} d_i$ with $|I| = k+r-1$,
the degree is obviously largest when $k$ is maximal and when we choose
$I = (d_1, \ldots, d_{k+r-1})$. We recall that $U_k = 0$ if $k
> \mu+1$, where $\mu = \min(n,m-r)$, and the theorem follows.
\end{proof}

\begin{preex}
\emph{If $r=1$, the Buchsbaum-Rim complex will simply be the Koszul
complex. To obtain Hefer forms, we first find a
$(1,0)$-form $h(\z,z') = \sum h_{ij} \epsilon_i \otimes e_j^\ast$
taking values in $\mathcal{O}(1) \otimes \Hom (E_1,E_0)$, such that $\delta_\eta
h = p - p^z$. It is then easy to show that $h_k^l = (\delta_h)^{k-l}$
will act as a Hefer form. We get the decomposition }
\[
\Phi(z') = P(z') Q(z')
+ \int_{\Pn} \sum_k \delta_h^k (R^k \phi(\z)) \w \alpha^{n+\rho},
\]
\emph{where $Q(z') = \sum_1^{m} Q_i(z') e_{j}$ with
\[
Q_i(z') = \int_\z \z_0^{d_i} \sum_k \delta_h^{k-1} U_k \phi(\z) \w
\alpha^{n + \rho}.
\]
Furthermore, $\deg P Q \leq \rho + d_1 + \cdots + d_{\mu+1}$, where $\mu =
\textrm{min}(m-1,n)$.}
\end{preex}

\begin{preremark} \label{martha}
\emph{Following Hergoualch \cite{HE}, who uses an idea of Fuhrmann
\cite{FU}, we can reduce the case $r > 1$ to the
case $r=1$, but we will then lose precision in the degree
estimates. If we wish to solve $P Q = \Phi$, as before, we can
instead set $p' = p_1 \w \ldots \w p_r$ and solve $p'
\cdot q_j' = \phi_j$. Note that we will then get a solution $q_j'$ taking
values in $E_1' = \Lambda^r E_1$. Since the vector space $E_1'$ has
dimension $m' = {m \choose r}$, the previous example will give $\deg
P' Q_j' \leq \rho + d_1' + \cdots + d_{\mu'+1}'$, where $\mu' =
\textrm{min}({m \choose r}-1,n)$. We have $d_1' = d_1 + \cdots + d_r$,
so we can make the estimate
$\deg P' Q_j' \leq \rho + (\mu'+1)(d_1 + \cdots + d_r)$.}

\emph{Now, we can set
\[
q_j = (-1)^j \delta_{p_r} \cdots \delta_{p_{j+1}}
\delta_{p_{j-1}} \cdots \delta_{p_1} q_j'.
\]
Note that $q_j$ takes values in $E_1$ and that $p_j q_j =
\phi_j$. Let $q = q_1 + \cdots q_r$, then we have $P Q = \Phi$, with
$\deg P Q \leq \rho + (\mu'+1)(d_1 + \cdots + d_r)$.}
\end{preremark}

\begin{preremark}
\emph{It is possible to get a slightly better result than Theorem~\ref{eliot}
by means of solving successive $\dbar$-equations in the Buchsbaum-Rim
complex, as is done in \cite{MA3}. We will sketch this approach.
First, we modify the Buchsbaum-Rim complex by taking tensor products
with $\mathcal{O}(\rho)$:
\[
0 \to E_{m-r + 1} \otimes \mathcal{O}(\rho) \stackrel{\delta_{p}}{\longrightarrow} \cdots

\stackrel{\det p}{\longrightarrow} E_1 \otimes
\mathcal{O}(\rho)\stackrel{p}{\longrightarrow} E_0 \otimes \mathcal{O}(\rho) \to 0,
\]
If $\phi$ is a section of $E_0 \otimes \mathcal{O}(\rho)$ and $R^f \phi = 0$, it
is clear that $v = \phi U$ solves $\nabla_{p} v = \phi$. As in the
classical Koszul complex method of solving division problems, we need
to solve a succession of equations $\dbar w_k = v_k + \delta_{p}
w_{k+1}$. The holomorphic solution will then be $v_1 + \delta_{p} w_2$.
In order to solve the $\dbar$-equations, we need to assume that
\[
m \leq n + r - 1 \ \ \ \textrm{or} \ \ \ \rho \geq \sum_1^{n+r} d_j - n,
\]
where $\deg \Phi
\leq \rho$. With this method, one proves that there exist polynomials
$Q_1, \ldots, Q_m$ such that $\deg P Q \leq \rho$ (Theorem 1.8 in
\cite{MA3}). }

\emph{Actually, by Proposition 5.1 in \cite{EG} we can solve
  $\dbar$-equations for $(0,q)$-forms taking values in $\mathcal{O}(l)$
  explicitly. It is thus possible to obtain explicit
solutions with these better degree estimates as well, although these
solutions will be
more complicated than the ones from Theorem~\ref{eliot}, since they
will involve $\mu$ integrations. We do not know if it is possible to
find a weight such that the method in the present article yields
explicit solutions $Q_1, \ldots, Q_m$ such that $\deg P Q \leq \rho$.}
\end{preremark}

\begin{preremark}
\emph{In the case of the Buchsbaum-Rim complex we can at
  least partially find
explicit Hefer forms. As in the example above, we find $h$ such that
$\delta_\eta h = p - p^z$,

and we let $h_k^l = \delta_h^{k-l}$ for $k > l \geq 2$. }

\emph{

We also have $h_1^0 = h$ and }
\[
h_2^1 = \sum_{k=1}^r (-1)^{k+1} p^1 \w \ldots \w p^{k-1} \w h^k \w
(p^{k+1})^z \w \ldots \w (p^r)^z \otimes \epsilon_1 \w \ldots \epsilon_r,
\]
\emph{where $h^k$ is the $k$:th column of $h$. We refrain from trying
  to find explicit expressions for $h_k^0$
for $k \geq 1$ and $h_k^1$ for $k \geq 2$.}
\end{preremark}

We can get several corollaries to Theorem \ref{eliot}, that give
conditions which ensure that $R^f \phi = 0$, and state what the estimate of the
degree is in that case. First, there is a classical theorem by
Macaulay \cite{MAC} which says that if $r = 1$, $\Phi = 1$ and $p^1,
\ldots, p^m$ lack common zeroes in $\Pn$, then we have the estimate
$\deg P^i Q_i \leq \sum_1^{n+1} d_j -n$. By means of residue currents,
one can get a generalization to the case $r>1$:

\begin{prop}
Assume that $Z$ is empty. Then there exists a matrix $Q$
of polynomials
such that $PQ  = I_r$, and $\deg P Q \leq \sum_1^{n+r} d_j - n$.
\end{prop}
\noindent
This is Corollary 1.9 in \cite{MA3}. By using integral
representations, we can get instead

\begin{cor} \label{mountie}
Assume that $Z$ is empty. Then Theorem \ref{eliot} yields an explicit matrix $Q$
of polynomials
such that $PQ  = I_r$, and $\deg P Q \leq \sum_1^{n+r} d_j$.
\end{cor}

\begin{proof}
We first note that necessarily $\mu = n$ since $\textrm{codim} Z =
m-r+1$. The result now
easily follows by several application of Theorem \ref{eliot} where one
chooses the columns of $I_r$ as $\Phi$.
\end{proof}

If $Z$ is not empty, we would like some condition which guarantees
that $R^f \phi = 0$. We define the pointwise norm $\| \phi \|^2 = \det
(pp^\ast) |p^\ast (pp^\ast)^{-1} \phi |^2$.

\begin{cor}
If
\be \label{bryhnia}
\| \phi \| \leq C \det (p p^\ast)^{\min(n,m-r+1)},
\ee
then Theorem \ref{eliot} yields explicit $Q$ such that $PQ = \Phi$ with $\deg P
Q \leq \rho + d_1 + \ldots + d_{\mu + r}$, where $\mu = \min(n,m-r)$.
\end{cor}

This follows from Proposition 1.3 in \cite{MA3}, which says that if
(\ref{bryhnia}) holds, then $R^f \phi = 0$. Just as for the previous
corollary, this is not the optimal result, which states that there
exists a solution $Q$ such that $\deg P Q \leq \rho$ (see Corollary
1.10 in \cite{MA3}).

\begin{preremark}
\emph{Another possible application for Theorem \ref{inuvik} is the
Eagon-Northcott complex. This is used if we want to solve the equation
$(\det p) \cdot q = \phi$, where $\phi$ is a scalar function. The complex is
given by
\[
\cdots\stackrel{\delta_{p}}{\longrightarrow} E_3
\stackrel{\delta_{p}}{\longrightarrow} E_2
\stackrel{\delta_{p}}{\longrightarrow} E_1
\stackrel{\det {p}}{\longrightarrow} \C \to 0
\]
where
\[
E_k = \Lambda^{k+r-1} E_1 \otimes S^{k-1} E_0^\ast \otimes \det E_0^\ast.
\]
We have already solved this equation in Remark \ref{martha}, but by applying
Theorem \ref{inuvik} to the Eagon-Northcott
complex we can obtain a sharper degree estimate. Note that in this
division problem we actually have $r=1$ and so we could solve it by
means of the Kozsul complex, but we can improve the degree
estimate by choosing a complex that takes advantage of the fact that
our mapping $\det p$ is constructed from another mapping $p$.}
\end{preremark}


TITLE: Is it ever not possible to split a joint density into its marginals?
QUESTION [2 upvotes]: If I have a joint probability density $f(x,y)$, are there any regularity conditions that need to be met in order to allow $f(x,y) = f(y|x)f(x)$, or is this always possible? The statistics books I consulted state this as something obviously true without elaborating further.

[Abramovich & Ritov, p. 186] Given the joint distribution of $X$ and $Y$ one can always obtain their marginal distributions. Thus, for the continuous case, if $f_{XY}(x,y)$ is the joint density of $X$ and $Y$, their marginal densities $f_X(x)$ and $f_Y(y)$ are $f_X(x) = \int_{-\infty}^{\infty} f_{XY}(x,y)dy$ and $f_Y(y) = \int_{-\infty}^{\infty} f_{XY}(x,y)dx$. Two random variables X and Y are called independent if their joint density is a product of the marginal densities $f_{XY}(x,y) = f_X(x) \cdot f_Y(y)$ ...

[Spanos, p. 139]: It should come as no surprise to learn that from the joint distribution one can always recover the marginal (univariate) distributions of the individual random variables involved.

I just wondered if there's more to it.
And, a related question: could there be a situation where $f(x,y) = f(y|x)f(x)$ is possible, but $f(x,y) = f(x|y)f(y)$ is not?

REPLY [1 votes]: As is typically abused in Statistics, I write $f(x,y), f(x \mid y), f(x),$ etc. for the densities of $(X,Y), X \mid Y = y, X,$ etc. I assume all of these are continuous.
You start with $f(x,y).$ Given this, you construct $f(y) = \int dx\ f(x,y)$ and then $f(x \mid y) = \frac{f(x,y)}{f(y)}.$ Note that if $f(x,y) = 0$ on the line $\{x\} \times \mathbf{R}$ then $f(y) = 0$ and reciprocally, if $f(y) = 0,$ then $f(x,y) = 0$ for all $x$ (since we are assuming $f$ is continuous). Then, by definition, $f(x,y) = f(y) f(x \mid y)$ (when  $f(x,y) = f(y) = 0,$ the value of $f(x \mid y)$ can be arbitrarily defined as any number). So, you are asking whether or not $f(x,y) = f(x \mid y) f(y);$ the answer is yes, always, by definition.
Addendum: the question as to why $f(x \mid y)$ is a density for $X \mid Y = y$ is somewhat difficult and usually out of scope in statistics books. It has to do with the definition of conditional expectation and regular conditional probabilities.


TITLE: Find geo coordinate by a coordinate and an angle
QUESTION [0 upvotes]: I need some help with this problem.
I have a GPS coordinate and an angle in degrees. I need a new GPS coordinate x km away from the point I already have.
Degree is counted clockwise and y-axis is pointing at north.

Anybody that can help me out with a formula?

REPLY [0 votes]: Except near the poles, for a distance as short as 5 km, one can treat the coordinate system as orthogonal, though not orthonormal (to stretch the meaning of these terms a bit).  Denote the known point's coordinates by $\phi$ and $\lambda$.  Then the unknown point's coordinates (taking north and east as positive) are
$$
\phi' = \phi - \frac{900 \sin 55}{\pi R \cos \lambda}
$$
$$
\lambda' = \lambda - \frac{900 \cos 55}{\pi R}
$$
where $R \doteq 6370$ is the Earth's radius in km.  The $900/\pi R$ comes from $5/R$ generating the subtended angle in radians, multiplied by $180/\pi$ to get degrees.  So, for instance, if one starts at $\phi = 2, \lambda = 49$ (near Paris), and travels $5$ km at heading $235$ (as in the diagram you provided), one ends up at
$$
\phi' \doteq 2 - 0.05615 = 1.94385
$$
$$
\lambda' \doteq 49 - 0.02580 = 48.97420
$$
Is that good enough?  Near the poles it's a bit more involved.


TITLE: Relationship Between Maurer-Cartan Forms on a Lie Group and its Central Extension
QUESTION [2 upvotes]: Given a Lie group $G$ (or, if you like, reformulate everything below in terms of the Lie algebra $\mathfrak{g}$ of $G$), one can define a distinguished 1-form $\omega$ on $G$, the Maurer-Cartan form, as the pushforward of left multiplication. Furthermore, one may be able to find a non-trivial central extension $\tilde{G}$ of $G$, which would then have its own Maurer-Cartan form $\tilde{\omega}$.

Question: what is the explicit relationship between $\omega$ and $\tilde{\omega}$, if any?

I've looked around in the references I have, but have not come up with anything along these lines. I would be nice to have some sort of explicit discussion of this to help develop a bit of intuition for this topic... if anything, I'd imagine there must be some discussion somewhere of an important case (important for physicists, at least) like the Galilean group and its central extension.

REPLY [0 votes]: I also suppose this is a standard result in mathematics, but indeed, Google returns very little. The one reference where I know this to be done is section 3.3 of arXiv:1703.06142, though the specific example considered there is the Virasoro group.
It's actually easy to compute the Maurer-Cartan form for any centrally extended group. Here goes. Let $\hat G$ be a central extension of some Lie group $G$ (say $\hat G=G\times\mathbb{R}$ as a set). Let its elements be written as pairs $(f,a)$ where $f\in G$ and $a\in\mathbb{R}$, and let its group operation read
\begin{equation}
(f,a)\cdot(g,b)
=
\big(fg,a+b+C(f,g)\big)
\end{equation}
where $C$ is some real-velued 2-cocycle on $G$. Then, let $(f(t),a(t))$ be a path in $\hat G$, defining a tangent vector $(\dot f(t),\dot a(t))$ at time $t$. The left Maurer-Cartan form $\hat\Theta$ of $\hat G$ acting on that vector is, by definition,
\begin{equation}
\hat\Theta_{(f(t),a(t))}(\dot f,\dot a)
\equiv
\frac{\partial}{\partial\tau}\bigg|_{\tau=t}
(f(t),a(t))^{-1}(f(\tau),a(\tau))
=
\Big(
\frac{\partial}{\partial\tau}\bigg|_{\tau=t}\big(f(t)^{-1}f(\tau)\big),
\dot a(t)+\frac{\partial}{\partial\tau}\bigg|_{\tau=t}C(f(t)^{-1},f(\tau))
\Big)
\end{equation}
where we used the group operation written above. On the right-hand side of the last equation, the first entry of the parenthesis is the left Maurer-Cartan form $\Theta$ of the non-extended group $G$, acting on $\dot f$. The second entry involves the derivative of the cocycle with respect to its second argument. To rewrite that more `intrinsically', one may think of $C(f,\cdot):G\to\mathbb{R}:g\mapsto C(f,g)$ as a map depending parametrically on $f$; the differential of that map at $g\in G$ is a linear map
\begin{equation}
\big(\text{d}C(f,\cdot)\big)_g:
T_gG\to
T_{C(f,g)}\mathbb{R}=\mathbb{R}.
\end{equation}
Thus, with this notation, the Maurer-Cartan form of $\hat G$ is given by
\begin{equation}
\hat\Theta_{(f,a)}(V,v)
=
\Big(\Theta_f(V),v+\big(\text{d}C(f^{-1},\cdot)\big)_g(V)\Big)
\end{equation}
where we renamed $\dot f$ into $V\in T_gG$ and $\dot a$ into $v\in T_a\mathbb{R}=\mathbb{R}$.
Disclaimer: I'm a physicist, not a mathematician, so I was happy to formulate the notion of `central extension' in terms of an explicit cocycle function $C$. A proper mathematician would presumably prefer to formulate everything in more intrinsic terms, using short exact sequences and sections for the definition of central extensions and cocycles. I have avoided that formulation, for simplicity.
Also, I have written everything in terms of the left Maurer-Cartan form. The construction of the right Maurer-Cartan form is similar.


TITLE: Permutations with special restrictions and combinations
QUESTION [1 upvotes]: How many permutations are there for $10$ football games were every game could be either $H, V$ or $T$ (which means that either the home team wins, the visitor team wins or tie accordingly)? The games are distinct and from that I assume that the order matters and thus we talk about permutations.
Now let's put the restriction of $4$ games being $T$ and $4$ games being $V$. How many cases are there?
Please note that the games are distinct and thus it is different for the first $4$ games to come a tie from the last $4$ games to come a tie.
What I have thought is:
The number of cases of $4$ games being tied out of $10$ is $P(10,4)$ and then there are $6$ games left so the number of cases for $4$ games being $V$ out of $6$ is $P(6,4)$. Finally the last 2 games can have either one of the three outcomes so this is $3^2$. The final result is their product:
$P(10,4)*P(6,4)*3^2=5040*360*9 = 16329600$ cases.
Is my solution correct or am I missing something? These seem like a lot of cases.

REPLY [1 votes]: Given the confusion in the question, I am adding both cases.
Case1 : When the outcomes of $10$ games are at least $4V$ and $4T$ which means other two games can have any of $3$ possible outcomes $(H, V, T)$.
Possible outcomes in this case -
$\{4V, 4T, 2H\}, \{6V, 4T\}, \{4V, 6T\}, \{5V, 5T\}, \{5V,4T,1H\}, \{4V,5T,1H\}$
So total permutations $ = 10! (\frac{1}{4!4!2!} + \frac{2}{6!4!} + \frac{1}{5!5!} + \frac{2}{5!4!1!}) = 6342$
Case 2: Outcomes are exactly 4 wins for V and 4 wins for T (so outcome of other two have to be $H$).
Just take the first term of the above $\{4V, 4T, 2H\}$ and that gives you $3150$ permutations.


TITLE: trouble calculating sum of the series $ \sum\left(\frac{n^2}{2^n}\right) $
QUESTION [1 upvotes]: Find out the sum of the series $\displaystyle \sum\limits_{n=1}^{\infty} \dfrac{n^2}{ 2^n}$. I have checked the convergence, but how to calculate the sum?

REPLY [4 votes]: Note $\sum\limits_{n\geqslant 0}  t^n = \dfrac{1}{ 1-t}$ gives $\sum\limits_{n\geqslant 1}  nt^n = \dfrac{t}{( 1 - t)^2}$ after differentiation and multiplication by $t$, which in turn gives $\sum\limits_{n\geqslant 1}  n^2t^n = \dfrac{t(t+1)}{( 1 - t)^3}$ by the same token.


TITLE: Is the book: A Course of Pure Mathematics (G H Hardy) still relevant today?
QUESTION [8 upvotes]: A very intelligent family member of mine has recently decided to send me a large number of books and other resources on mathematics, include Hardy's book. Since this was published in 1921, is it still a useful textbook from which to learn pure mathematics?

REPLY [5 votes]: Congratulations to such generous and wise members in your family. My answer is: Yes, of course. This text is great to read and will  help  to  gain  a deep and  proper understanding of the material.

Hint:  In order to better evaluate this treasure, the following might help to come to a decision by yourself:

Review 1 by Paramanand Singh, one of the members of this platform.

Review 2 from MAA.

Proposal: Select two, three sections from Hardy's book which you might find interesting, read them and compare it with corresponding sections from books you are familiar with.


TITLE: Verification on proof of $\sigma(kU)=k^{2n+1}\sigma(U)$?
QUESTION [1 upvotes]: Let $U$ be a bounded open subset of an $n$-dimensional euclidean space endowed with the usual topology and the usual metric $d$. Now, let
$$\sigma(U)=\iint_U d(x,y)dxdy$$
Suppose we scale the space by a factor of $k$ from the origin. Let $k U$ be the new shape.
I think I have a proof of $\sigma(kU)=k^{2n+1}\sigma(U)$, but I don't feel confortable with it, as it seems too easy. Anyway, here is the proof:

Each of its points of $U$ will have coordinates $x=(x_1,\ldots,x_n)$. Then the new shape $kU$ will have points with coordinates $x'=(x_1',\ldots,x_n')$ with the relations
$$x_i'=kx_i$$
$$dx_i'=kdx_i$$
Then, because $x_i=x_i'/k$, the shape $kU$ gets transformed to $\{x\mid \exists x'\in kU:x'=kx\}=U$:
\begin{align}
\sigma(kU) &= \iint_{kU}d(x',y')dx'dy' \\
&= \iint_U d(kx,ky)kdx\ kdy \\
&= \int_{x_1^0}^{x_1^1}\cdots\int_{x_n^0}^{x_n^1}\int_{y_1^0}^{y_1^1}\cdots\int_{y_n^0}^{y_n^1} \sqrt{(kx_1-ky_1)^2+\cdots+(kx_n-ky_n)^2}kdx_1\ldots kdx_nkdy_1\ldots kdy_n \\
&= \int_{x_1^0}^{x_1^1}\cdots\int_{x_n^0}^{x_n^1}\int_{y_1^0}^{y_1^1}\cdots\int_{y_n^0}^{y_n^1} k\sqrt{(x_1-y_1)^2+\cdots+(x_n-y_n)^2}k^{2n}dx_1\ldots dx_ndy_1\ldots dy_n \\
&= k^{2n+1}\int_{x_1^0}^{x_1^1}\cdots\int_{x_n^0}^{x_n^1}\int_{y_1^0}^{y_1^1}\cdots\int_{y_n^0}^{y_n^1} \sqrt{(x_1-y_1)^2+\cdots+(x_n-y_n)^2}dx_1\ldots dx_ndy_1\ldots dy_n \\
&= k^{2n+1}\iint_Ud(x,y)dxdy \\
\sigma(kU)&= k^{2n+1}\sigma(U)
\end{align}
Is this correct? Thanks.

REPLY [1 votes]: One thing : the shape $U$ can be pretty complicated so there's no reason it can be written as nicely as $$\int_{x_1^0}^{x_1^1}...\int_{x_n^0}^{x_n^1}\int_{y_0^0}^{y_0^1}...\int_{y_n^0}^{y_n^1}d(x,y)\,dx\,dy.$$
But that's fine because what you only need is that $d(kx,ky)=kd(x,y)$ for all $x,y\in U$ (which is certainly true) and that $d(kx)d(ky)=k^{2n}dxdy$ which is true as well for the reason you mention.


TITLE: How to calculate kinetic friction coefficient given velocity and height?
QUESTION [0 upvotes]: The problem I am dealing with is about an object sliding down a ramp. I am only given the height of the top of the ramp to the floor, initial and final velocity, and weight of the object (in lbs). How do I calculate the coefficient of kinetic friction with only this information?

REPLY [0 votes]: Conservation of energy (or work-kinetic energy theorem). You know the work by kinetic friction is $W=-\mu_k N d$ where $N$ is the normal force (which you must find using the dimensions of the incline) and $d$ is the length of the incline.
You're likely given enough information to find the angle the incline makes with the horizontal (which is required to find $N$).


TITLE: Given $r^2=x^2+y^2$ and function $Z=f(r)$ prove that the function fulfills $yZ_x=xZ_y$
QUESTION [1 upvotes]: Given $r^2=x^2+y^2$ and function $Z=f(r)$ prove that the function fulfills $yZ_x=xZ_y$, $Z(x,y)$. We can suppose that we can use chain rule. Also a hint is given that essentially we need to prove that any function which is only dependent on the distance from the origin fulfills $yZ_x=xZ_y$.

I don't quite understand this problem.
1) Is $Z=f(r)$ a single variable function now?
2) Is there any connection between the $r$ from $f(r)$ and $r^2=x^2+y^2$? Can $r$ be anything? Can it be a complicated expression of any form?

REPLY [2 votes]: Think of $f$ as any function of one variable differentiable at $(0,+\infty)$. Then $Z(x,y) = f(\sqrt{x^2+y^2})$ and as you can see, $Z$ is function of two variables and is given as a composition of $f$ and function $(x,y)\mapsto\sqrt{x^2+y^2}$. Now use chain rule:
$$Z_x(x,y)=f'\left(\sqrt{x^2+y^2}\right)\frac \partial{\partial x}\sqrt{x^2+y^2},$$
$$Z_y(x,y)=f'\left(\sqrt{x^2+y^2}\right)\frac \partial{\partial y}\sqrt{x^2+y^2}.$$
Can you finish from here?


TITLE: Question about characteristics and classification of second-order PDEs
QUESTION [43 upvotes]: I am currently reading through the book 'Computational Techniques for Fluid Dynamics', by C.A.J. Fletcher. Chapter 2 discusses classification of PDEs by finding the number and nature of their characteristics. However, there is a section about finding characteristics of second-order PDEs (2.1.3), which I am a little confused about.
They give a generalized second-order PDE as:
$$A\frac{\partial^2u}{\partial x^2}+B\frac{\partial^2u}{\partial x\partial y}+C\frac{\partial^2u}{\partial y^2}+H=0\tag{1}$$
where $A$, $B$ and $C$ are functions of $x,y$ and $H$ contains all the first-derivative terms. They then introduce some new variables to simplify the notation:
$$P=\frac{\partial u}{\partial x}, Q=\frac{\partial u}{\partial y}, R=\frac{\partial^2 u}{\partial x^2}, S=\frac{\partial^2 u}{\partial x\partial y}, T=\frac{\partial^2 u}{\partial y^2}$$
They then state that a curve K is introduced and along a tangent to K, the differentials for $P$ and $Q$ satisfy:
$$dP=Rdx+Sdy\tag{2}$$
$$dQ=Sdx+Tdy\tag{3}$$
Using the substitutions above, the original PDE can be written:
$$AR+BS+CT+H=0$$
Equations (2) and (3) are then used to eliminate $R$ and $T$ in the above equation, which is then re-arranged to give the following:
$$S\Bigl[A\Bigl(\frac{dy}{dx}\Bigr)^2-B\Bigl(\frac{dy}{dx}\Bigr)+C\Bigr]-\Bigl\{\Bigl[A\Bigl(\frac{dP}{dx}\Bigr)+H\Bigr]\frac{dy}{dx}+C\frac{dQ}{dx}\Bigr\}=0\tag{4}$$
It then states that if:
$$A\Bigl(\frac{dy}{dx}\Bigr)^2-B\Bigl(\frac{dy}{dx}\Bigr)+C=0\tag{5}$$
then that eliminates the left-hand term in equation (4), which yields a simpler relationship between $\frac{dP}{dx}$ and $\frac{dQ}{dx}$. The solutions of equation (5) define the characteristic curves for the PDE.
So, here is where I am confused: why did they choose to split up equation (4) in that manner? As I understand it, the goal of finding characteristic curves is to reduce a PDE to a total differential, so that it can be more easily solved. However, how has this goal been met, when the terms $P$, $Q$ and $H$ on the right-hand side of equation (4) still contain partial differentials? Given that it still contains partial differentials, how is the reduced form of the equation more useful and why does equation (5) provide the characteristic curves?
Thanks in advance!

REPLY [21 votes]: If $A\neq0$ then the equation (5) is a quadratic equation, and unless your PDE is elliptic ($B^2-4AC<0$), locally it has two solutions $y_1(x)$ and $y_2(x)$ which satisfy Vieta's formula,
$$
A\left(\frac{dy_1}{dx}+\frac{dy_2}{dx}\right)=B,\quad A\frac{dy_1}{dx}\frac{dy_2}{dx}=C.
$$
If your PDE is parabolic ($B^2-4AC=0$) then $y_1=y_2$. It follows that
$$
A\partial_x^2+B\partial_x\partial_y+C\partial_y^2=A\left(\partial_x^2+\left(\frac{dy_1}{dx}+\frac{dy_2}{dx}\right)\partial_x\partial_y+\frac{dy_1}{dx}\frac{dy_2}{dx}\partial_y^2\right)
$$
$$
=A\left(\partial_x+\frac{dy_1}{dx}\partial_y\right)\left(\partial_x+\frac{dy_2}{dx}\partial_y\right)+\mbox{first order terms}.
$$
Therefore (1) now becomes
$$
A\left(\partial_x+\frac{dy_1}{dx}\partial_y\right)\left(\partial_x+\frac{dy_2}{dx}\partial_y\right)u+\mbox{lower order terms}=0,
$$
which is the expected simplification.
This is the high school level answer to the question why characteristics matter. The scientific answer is much longer and involves microlocal analysis (propagation of singularities), quantum field theory (trajectories of classical point particles corresponding to a linear quantum field), etc.
This is, however, not a complete answer to the question posted, for I have no suggestion as to why (4) is relevant.
Edit: By no means could I read the author's mind when they wrote equation (4), but let me elaborate on what could be done along those lines (at OP's request).
(5) is a first order ODE, of which the solutions depend on one parameter (initial data). For every $(x_0,y_0)$ in the domain let $y_1(x;x_0,y_0)$ and $y_2(x;x_0,y_0)$ be the two solutions of (5) satisfying
$$
y_1(x_0;x_0,y_0)=y_0,\quad y_2(x_0;x_0,y_0)=y_0.
$$
In other words, $y_1(x;x_0,y_0)$ and $y_2(x;x_0,y_0)$ are the two characteristics through the point $(x_0,y_0)$. Fix $(x_0,y_0)$ for a moment, and omit their mention for brevity. Let us do with the characteristic $y_1(x)$ first. Consider $P$ and $Q$ as independent functions, then their restriction to the characteristic $P(x,y_1(x))$ and $Q(x,y_1(x))$ as unknown functions of $x$. If $H$ contains no zero order terms then $H(P,Q,x,y_1(x))$ is a function of $P$,$Q$ and $x$. This renders (4) a first order ODE relating two unknown functions $P$ and $Q$ of one variable $x$,
$$
\left[A(x,y_1(x))\frac{dP}{dx}+H(P,Q,x,y_1(x))\right]y_1'(x)+C\frac{dQ}{dx}=0.
$$
If you can solve this equation in one way or another then you obtain
$$
F_1(P,Q,x)=C_1,
$$
where $F_1$ is a function and $C_1$ is a number, both depending on the choice of the characteristics, i.e., $(x_0,y_0)$. Now if you do the same thing with the characteristic $y_2(x)$ you end up with
$$
F_2(P,Q,x)=C_2.
$$
On the other hand, if your $H$ contains zero ordet terms, $H=H(P,Q,x,y,u)$, then
$$
u(x,y_1(x))=\int^x\left[P+y_1'(t)Q\right]dt,
$$
and (4) becomes an integral-differential equation. Good luck solving it.
Now the final solution of (1) depends on the specification of the boundary (Cauchy) data. Suppose that you have a Cauchy curve $\Sigma$ that intersects every characteristic exactly once, and suppose that $P$, $Q$, $R$, $S$, $T$ and $u$ are known along $\Sigma$. Consider only those $(x_0,y_0)\in\Sigma$, and let $p(x_0,y_0)$, $q(x_0,y_0)$ be the values of $P$ and $Q$ at $(x_0,y_0)$. We now recall all dependences upon $(x_0,y_0)$ above. If $(x,y)$ is any point in the domain then let $(x_1,y_1)$ and $(x_2,y_2)$ be the intersections of $y_1(x)$ and $y_2(x)$ with $\Sigma$, respectively,
$$
y=y_1(x;x_1,y_1),\quad y=y_2(x;x_2,y_2).
$$
This brings us to the system
$$
\begin{cases}
y=y_1(x;x_1,y_1)\\
y=y_2(x;x_2,y_2)\\
F_1(P,Q,x;x_1,y_1)=F_1(p(x_1,y_1),q(x_1,y_1),x_1;x_1,y_1)\\
F_2(P,Q,x;x_2,y_2)=F_2(p(x_2,y_2),q(x_2,y_2),x_2;x_2,y_2)
\end{cases}
$$
This is the parametric solution of the problem (1) with given boundary data $p$ and $q$.
If we want an explicit solution, we make a few more steps. The curve $\Sigma$ can be locally parameterized by one variable $s$, so that $(x_1,y_1)=\eta(s_1)$ and $(x_2,y_2)=\eta(s_2)$. Therefore the above is a system of 4 equation for 4 variables $P$, $Q$, $s_1$ and $s_2$. Eliminating $s_1$ and $s_2$ we find the desired solution
$$
P(x,y),\quad Q(x,y).
$$
Finally,
$$
u(x,y)=u(x_0,y_0)+\int_{(x_0,y_0)}^{(x,y)}\left[Pdx+Qdy\right]
$$
along your favourite curve joining $(x_0,y_0)$ and $(x,y)$. The point $(x_0,y_0)\in\Sigma$ therefore we know $u(x_0,y_0)$ from the boundary data.
This is how, I think, you would solve (1) if you want to start from (4). Generally, the method of characteristics always ends with a parametric solution, and for an explicit solution you need to exliminate the dependence upon the characteristic.


TITLE: Linear Algebra by Hoffman and Kunze, Section 3.4, Exercise Problem 12
QUESTION [5 upvotes]: I am trying to work out problems from Linear Algebra, by Hoffman and Kunze and came across this problem in the exercise of Section 3.4, I have a difficulty solving the (c) part of the problem.
Problem 12
If $V$ is an $n$ - dimensional vector space over the field $F$, and let $B = \{\alpha_1,\alpha_2,\alpha_3, \cdots \alpha_n\}$ be an ordered basis for $V$.
(a) There is a unique Linear Operator $T$ on $V$ such that :
$$
\begin{align}
T(\alpha_j) = \alpha_{j+1}, \quad j=1, \cdots ,n-1 \quad T(\alpha_n)=0
\end{align}
$$
What is the matrix $A$ of $T$ in the ordered basis $B$?
(b) Prove that $T^n=0$, but $T^{n-1} \neq 0$
(c) Let $S$ be any linear operator on $V$ such that $S^n=0$, but $S^{n-1} \neq 0$ . Prove that there is an ordered basis $B^{'}$ for $V$ such that the matrix of $S$ in the ordered basis $B^{'}$ is the matrix $A$ of part (a).
(d) Prove that if $M$ and $N$ are $n \times n$ matrices $F$ such that $M^n = N^n = 0$, but $M^{n-1} \neq 0 \neq N^{n-1}$, then $M,N$ are similar.

However the solution for the problem (c) can be obtained assuming (d) is true, in the following way:
Let $[S]_B$ represent the matrix corresponding to the linear operator $S$ under some basis $B$, and we want to show that this operator has the matrix $[S]_{B^{'}}=A$ under some basis $B^{'}$.
If the two nilpotent matrices of order $n$ are similar, then there must be an invertible matrix $P$ such that $[S]_{B^{'}} = P^{-1}[S]_BP$. The matrix $P$ can be used to prove the existence of another basis $B^{'}$ such that:
$$
\begin{align}
[\alpha]_{B} = P[\alpha]_{B^{'}}
\end{align}
$$
and therefore, there indeed exists a basis $B^{'}$ such that
$$
\begin{align}
[S]_{B^{'}} = P^{-1}[S]_BP
\end{align}
$$

I am also having a problem solving part (d), however, I was wondering if one could prove (c) without using (d) explicitly ? I will be thankful for any hints to solve the problem!

REPLY [5 votes]: (c) Let $S$ be any linear operator on $V$ such that $S^n=0$, but $S^{n-1} \neq 0$. Prove that there is an ordered basis $\mathfrak{B}'$ for $V$ such that the matrix of $S$ in the ordered basis $\mathfrak{B}'$ is the matrix $A$ of part (a).

Since $S^{n-1} \neq 0$, there exists $v \in V$ such that $S^{n-1} v\neq 0$. Let$$\mathfrak{B}' = \{v, Sv, S^2v, \dots, S^{n-1}v\}.$$If we can show that $\mathfrak{B}'$ is a basis, the definition of $\mathfrak{B}'$ makes it clear that $[S]_{\mathfrak{B}'} = A$.
Since there are $n$ vectors in $\mathfrak{B}'$ and $V$ is an $n$-dimensional vector space, it is enough to show $\mathfrak{B}'$ is linearly independent. Suppose$$c_1v + c_2Sv + \dots + c_nS^{n-1}v = 0.$$Apply $S^{n-1}$ to both sides. Since $S^n = 0$, all the terms except the first one vanish, and we have $c_1 S^{n-1} v = 0$, and hence $c_1 = 0$ because $S^{n-1} v \neq 0$. Now, we can similarly apply $S^{n-2}$ to show that $c_2 = 0$, and so on (this may be formalized by induction, if desired), and we conclude that all the $c_j$ are $0$ and $\mathfrak{B}'$ is linearly independent, and hence a basis.

(d) Prove that if $M$ and $N$ are $n \times n$ matrices over $F$ such that$$M^n = N^n = 0$$but$$M^{n-1} \neq 0 \neq N^{n-1},$$then $M$ and $N$ are similar.

Let $U$ be the linear operator on $V$ whose matrix with respect to the ordered basis $\mathfrak{B}$ is $M$. Then $U$ satisfies the conditions of part (c), so there exist an ordered basis for which the matrix of $U$ is $A$. Hence, $M$ is similar to $A$. Similarly, $N$ is similar to $A$, so since similarity is an equivalence relation, $M$ and $N$ are similar.


TITLE: Why all norms define the same relative interior?(Convex Optimization, Stephan Boyd)
QUESTION [0 upvotes]: When I was reading 'Convex Optimization', Stephan Boyd, I was stopped by Example 2.2.
Before Example 2.2 is started, following definition is coming.

If the affine dimension of a set $C\subseteq\mathbf{R}^n$ is less than $n$, then the set lies in the affine set $\mathbf{aff}\;C\neq$. We define the $\;relative\;interior$ of the set $C$, as its interior relative to $\mathbf{aff}\;C:$
$$\text{relint}\;C=\{x\in C\,|\,B(x,r)\cap \text{aff}\;C \subseteq C\;\text{for some}\; r>0\},$$
where $B(x,r)=\{y\,|\,\|y-x\|\le r\}$, the ball of radius $r$ and center $x$ in the norm $\|\bullet\|$. (Here $\|\bullet\|$ is any norm; all norms define the same relative interior.)
$\mathbf{Example\;2.2}$ Consider a square in the $(x_1,x_2)$-plane in $\mathbf{R}^3$, defined as
$$C=\{x\in\mathbf{R}^3\,|\,-1\le x_1\le 1,\, -1\le x_2\le 1,\,  x_3=0\}$$
Its affine hull is the $(x_1,x_2)$-plane, $i.e.,\;\text{aff}\;C=\{x\in\mathbf{R}^3\;|\;x_3=0\}$. The interior of $C$ is empty, but the relative interior is

$$\text{relint}\;C=\{x\in\mathbf{R}^3\,|\,-1\lt x_1\lt 1,\, -1\lt x_2\lt 1,\,  x_3=0\}$$
I wondering why all norms define the same relative interior. And how we know relint $C$ in $\mathbf{Example\;2.2}$ which is not determined any ball's information(center $x$ and radius $r$)

REPLY [0 votes]: In a finite dimensional space, all norms are equivalent.
In particular, in Example 2.2 you can use any norm you want to.