largest eigenvalue of the difference between two quadratic forms

Question

Let $U,V\in\mathbb{R}^{4\times n}$ such that $UU^T=VV^T=I$, and $A\in\mathbb{R}^{n\times n}$ be an Hermitian matrix. Is it true that $$\sqrt{\lambda_{\text{max}}\left(\left(UAU^T-VAV^T\right)^2\right)}\leq \lambda_{\text{max}}\left((U-V)A(U-V)^T\right)$$ ?

Answer 1

This is false for the following reason. Say that $n=2$. Choose $$A=\begin{pmatrix} a & b \\ b & 0 \end{pmatrix},\quad U=I_2,\quad V=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$ Then $(U-V)A(U-V)^T=0_2$, while $$UAU^T-VAV^T=\begin{pmatrix} 0 & b \\ b & 0 \end{pmatrix},$$ whose square is non-zero if $b\ne0$.