Relate the singular values of $A$ and $\frac{A^T+A}{2}$

Consider a square matrix with real entries $A\in\mathbb{R}^{n\times n}$ such that the symmetric matrix $\frac{A^T+A}{2}$ is positive definite.

Is it possible to find a relationship linking together the singular values of $\frac{A^T+A}{2}$ with the singular values of $A$?

The relationship can either be an equality or an inequality.

EDIT: I've tried some numerical examples in some $2 \times 2$ matrices and it appears that when $(A^T+A)/2$ is positive definite then the two singular values are perturbed by the same amount. For example:

Example 1: $$ G=\begin{bmatrix} 100 &10\\4 & 5\end{bmatrix} $$ Singular values of $G$: $$ 4.5726253611256148668420314380021\\ 100.59866349662300975905062166134$$ $$ \frac{G^T+G}{2}=\begin{bmatrix} 100 &7\\7& 5\end{bmatrix} $$ Singular values of $\frac{G^T+G}{2}$: $$ 4.4869809322513025538957048883323 \\ 100.51301906774869744610429511167$$

The difference is: $$0.085644428874312312946326549669858\\ 0.085644428874312312946326549669858$$

Example 2: $$ G=\begin{bmatrix} 20 &8\\2 & 7\end{bmatrix} $$ Singular values of $G$: $$ 5.6287069525109683834660548089479\\ 22.02992641936769354654076803634$$ $$ \frac{G^T+G}{2}=\begin{bmatrix} 20 &5\\5& 7\end{bmatrix} $$ Singular values of $\frac{G^T+G}{2}$: $$ 5.2993902665716374184626433863038\\ 21.700609733428362581537356613696$$

The difference is: $$ 0.32931668593933096500341142264419\\ 0.32931668593933096500341142264419$$

Solution 1:

$\sigma_k(A)$, the $k$-th largest singular value of $A$, is always greater than or equal to $\sigma_k\left(\frac{A+A^T}2\right)$.

Proof. WLOG we may assume that $D=\frac{A+A^T}2$ is a positive diagonal matrix such that its diagonal entries are arranged in a decreasing order. Then $K=A-D$ is a skew symmetric matrix. By Courant-Fischer minimax principle, $$ \sigma_k(A)=\max\limits_{\dim S=k}\ \min\limits_{x\in S,\, \|x\|_2=1} \|Ax\|_2. $$ In particular, if $\tilde{A}$ denotes the leading principal $k\times k$ submatrix of $A$, $e_i$ denotes the $i$-th vector in the canonical basis and $V=\operatorname{span}\{e_1,\ldots,e_k\}$, we have $$ \sigma_k(A)\ge\min\limits_{x\in V,\, \|x\|_2=1} \|Ax\|_2 \ge \min\limits_{y\in \mathbb{R}^k,\, \|y\|_2=1} \|\tilde{A}y\|_2.\tag{1} $$ Let $\langle\cdot,\cdot\rangle$ denotes the usual dot product on $\mathbb{R}^k$ and $\tilde{D}$ denotes the leading principal $k\times k$ submatrix of $D$. For any unit vector $y$, the orthogonal projection of $\tilde{A}y$ on the span of $y$ is given by $\langle \tilde{A}y,y\rangle y$. Since $\langle \tilde{A}y,y\rangle=y^T\tilde{D}y\ge D_{kk}$, it follows that $\|\tilde{A}y\|_2\ge D_{kk}$. Therefore, from $(1)$, we conclude that $\sigma_k(A)\ge D_{kk}=\sigma_k(D)$.