0
$\begingroup$

Let A be an $m\times n$ matrix $(m\geq n)$, $X$ be an $n\times n$ symmetric matrix. Then, $x={\rm vec}(X)\in \mathbb{R}^{n^2}$. The symmetry of $X$ restricts $x$ to an $s$-dimensional subspace of $\mathbb{R}^{n^2}$, where $s=n(n+1)/2$. Let $S^n$ be a matrix whose columns form an orthonormal basis for this subspace. Define $G=(I_n\otimes A)S_n$.Then what is the singular value of $G$.

I have thought the singular value of $G$ is $\{\sigma_1(A),\ldots,\sigma_1(A), \sigma_2(A),\ldots,\sigma_2(A),\ldots,\sigma_n(A),\ldots,\sigma_n(A)\}$. There are $n$ $\sigma_i(A)'s$. But something seems wrong.

The singular values of $G$ appears to be $\sigma(A)=\{\sqrt{\frac{1}{2}(\sigma_i^2(A)+\sigma_j^2(A))}:1\leq i\leq j\leq n\}$. Is there any hint for finding the singular value of $G$?

$\endgroup$
0
$\begingroup$

Let $M_{m,n}$ be the space of $m\times n$ matrices endowed with the usual inner product $\langle P,Q\rangle=\mathrm{tr}(P^TQ)$. This induces an inner product on the space of symmetric matrices $\mathrm{Sym}_n\subseteq M_{n,n}$.

Note that $S_n$ induces an isometry from $\mathbb R^s$ to $\mathrm{Sym}_n$, so we can identify $G$ with the map $\mathrm{Sym}(n)\to M_{m,n}$ given by $P\mapsto AP$. Thus $\lambda\geq0$ is a singular value of $G$ iff there exists a nonzero $P\in\mathrm{Sym}_n$ such that $\langle AQ,AP\rangle=\lambda^2\langle Q,P\rangle$ for all $Q\in\mathrm{Sym}_n$. That is, $$ \mathrm{tr}(QA^TAP)=\lambda^2\,\mathrm{tr}(QP). $$ Conjugating by an orthogonal matrix, we may suppose $$ A^TA=D=\mathrm{diag}(\sigma_1(A)^2,\ldots,\sigma_n(A)^2). $$ If $P=E_{ii}$, we have $DP=\sigma_i(A)^2P$, giving a singular value $\sigma_i(A)$.

If $P=E_{ij}+E_{ji}$ then $DP=\sigma_i(A)^2E_{ij}+\sigma_j(A)^2E_{ji}$, so $$ \mathrm{tr}(QP)=Q_{ji}+Q_{ij}, $$ $$ \mathrm{tr}(QDP)=\sigma_i(A)^2Q_{ji}+\sigma_j(A)^2Q_{ij}. $$ But $Q_{ij}=Q_{ji}$, so $$ \mathrm{tr}(QDP)=\frac12(\sigma_i(A)^2+\sigma_j(A)^2)\mathrm{tr}(QP). $$ This gives a singular value $\sqrt{\frac12(\sigma_i(A)^2+\sigma_j(A)^2)}$.

$\endgroup$
  • $\begingroup$ In second paragraph of the proof, I guess if $\lambda$ is a singular value of $G$ iff $\sqrt{tr(QA^TAP)}=\lambda\sqrt{tr(QP)}$, is this right? $\endgroup$ – Jin Aug 1 '17 at 12:47
  • $\begingroup$ @Jin yes, thanks for catching that. Edited. $\endgroup$ – stewbasic Aug 1 '17 at 21:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.