# What is the singular value of $G=(I_n\otimes A)S_n$

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 $G$?

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 $\sqrt{\frac12(\sigma_i(A)^2+\sigma_j(A)^2)}$.
• 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? – Jin Aug 1 '17 at 12:47