We have two real matrices $A$ and $B$. Let $\sigma_{\max}(A)$ and $\sigma_{\max}(B)$ denote the maximum singular value of matrices $A$ and $B$, respectively. Intuitively, the maximum singular value of summation $A^TA+B^TB$ can be upper bounded by:
$\sigma_{\max}(A^TA+B^TB) \leq [\sigma_{\max}(A)]^2 + [\sigma_{\max}(B)]^2$.
However, I couldn't show this (assuming it is correct). Can you show me any direction?
Thanks in advance.
The maximum singular value of a matrix $A$ is the operator norm $\|A\|$ of $A$, i.e. the maximum of $\|A x\|$ for unit vectors $x$. Since this is a norm, we have the triangle inequality $\|A^T A + B^T B\| \le \|A^T A\| + \|B^T B\|$. Moreover, using the fact that $A^T A$ is positive semidefinite we get $\|A^T A\| = \|A\|^2$ (the $C^*$ identity).