2
$\begingroup$

Let $A, B, C, D \in \mathbb{R}^{n \times n}$, let $$ M_1 = \begin{bmatrix}A & C \\ B & D\end{bmatrix} \quad M_2 = \begin{bmatrix}A \\ B \\ C \\ D\end{bmatrix} $$ I suspect that: \begin{equation} \sigma_1(M_1) \leq \sigma_1(M_2) \end{equation} where $\sigma_1(M_1)$ is the maximum singular value of $M_1$.

We can show that: \begin{align} \sigma_1(M_2)^2 = \lambda_1(M_2^TM_2) &= \lambda_1(A^TA + B^TB + C^TC + D^TD) \\ \Leftrightarrow \sigma_1(M_2) &= \sqrt{ \lambda_1(A^TA + B^TB + C^TC + D^TD) } \end{align}

This equality for $M_1$ is false and numerical test suggest that $\sigma_1(M_1) \leq \sigma_1(M_2)$ and I haven't been able to find a counter example.

I am having trouble proving it. Thanks!

$\endgroup$
1
  • $\begingroup$ Can you elaborate on your motivation for suspecting this, and on what your have tried so far? $\endgroup$ – Brian Jan 26 '20 at 22:37
2
$\begingroup$

This isn't true. Random counterexample: we have $\|M_1\|_2=2\sqrt{3}=3.46>3.24=1+\sqrt{5}=\|M_2\|_2$ when $$ A=B=\pmatrix{0&1\\ 0&1},\ C=D=\pmatrix{1&1\\ 1&1}, \ M_1=\pmatrix{0&1&1&1\\ 0&1&1&1\\ 0&1&1&1\\ 0&1&1&1\\}, \ M_2=\pmatrix{0&1\\ 0&1\\ 0&1\\ 0&1\\ 1&1\\ 1&1\\ 1&1\\ 1&1}. $$

$\endgroup$
2
  • $\begingroup$ Thanks ! Do you know the kind of hypothesis we could make on A, B , C and D for the inequality to be true ? $\endgroup$ – Alex Jan 27 '20 at 17:33
  • $\begingroup$ @Alex Sorry, no, except the really obvious ones (degenerate cases like $A=B=0$ or $C=D=0$, or trivial cases such as $M_1$ is unitary). $\endgroup$ – user1551 Jan 27 '20 at 20:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.