$A$, $B$ PSD matrices. Then $\|A-B\| \leq \sqrt{\|AA^T-BB^T\|}$?

Question

Let $A$ and $B$ be real, symmetric, PSD matrices. Then is $\|A-B\| \leq \sqrt{\|AA^T-BB^T\|}$? Here, $\|\cdot\|$ is the spectral norm.

Answer 1

Yes. Let $C=A-B$ and $\lambda$ be an eigenvalue of $C$ such that $|\lambda|=\|C\|$. Let $u$ be a unit eigenvector of $C$ corresponding to $\lambda$. If $\lambda\ge0$, then $$ u^T(A^2-B^2)u=u^T(BC+CB+C^2)u=2\lambda u^TBu+\lambda^2\ge\lambda^2. $$ If $\lambda\le0$, then $$ u^T(A^2-B^2)u=u^T(AC+CA-C^2)u=2\lambda u^TAu-\lambda^2\le-\lambda^2. $$ So, in either case, $$ \|A^2-B^2\|\ge|u^T(A^2-B^2)u|\ge\lambda^2=\|C\|^2=\|A-B\|^2. $$