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
$$\sqrt{\lambda\_{\text{max}}\left(\left(UAU^T-VAV^T\right)^2\right)}\leq \lambda\_{\text{max}}\left((U-V)A(U-V)^T\right)$$
?



largest eigenvalue of the difference between two quadratic forms
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 $b\ne0$.


