1

I was given the question following question:

Suppose $A$ and $B$ are $n\times n$ matrices with entries in $\mathbb R$. Show that $\lVert AB\rVert_{HS} \le \lVert A\rVert_{HS}\lVert B\rVert_{HS}.$ [Hint: you may need Cauchy-Schwarz]

Given the definition of the Hilbert-Schmidt inner product and the hint, I was able to arrive at $\lVert AB\rVert_{HS}^2 \le \lVert A^{\top}A\rVert_{HS}\lVert BB^{\top} \rVert_{HS}$ which narrows the problem down to showing $\lVert A^{\top}A\rVert_{HS} \le \lVert A\rVert_{HS}^2$ (the case for $BB^{\top}$ is similar). Given that $A^{\top}A$ is symmetric, this can be shown if I can prove that all the real eigenvalues of that matrix are non-negative, however I'm unsure if this is true, or how to approach that proof if it is true.

For context, I'm not in a functional analysis class, so I would expect the intended solution to use relatively basic results from linear algebra.

Ian S.
  • 131
  • 3
  • 2
    Here is a link to an outline of how to solve this problem for the Frobenius norm, which is the same as the HS norm for finite dimensions (see problem 4): https://www.math.miami.edu/~armstrong/510fa22/510fa22hw2.pdf – David Raveh Feb 19 '23 at 05:33
  • 2
    Assume that $\lambda \in \mathbb{R}$ is an eigenvalue of $A^TA$ with eigenvector $v$, then we have $$ \Vert A v \Vert^2 = \langle Av, Av \rangle = \langle v, A^T A v \rangle = \langle v, \lambda v \rangle = \lambda \Vert v \Vert^2. $$ Thus, we get $v= \Vert Av \Vert^2/\Vert v \Vert^2 \geq 0$. So, yes, all the eigenvalues are nonnegative. – Severin Schraven Feb 19 '23 at 05:36
  • 1
    Does this answer your question? Is $AA^T$ a positive-definite symmetric matrix? The question is slightly different but the answers solve your problem. – Anne Bauval Feb 19 '23 at 08:14

0 Answers0