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If B is a PSD matrix, I intuitively think it is true that $\|ABC\| \leq \|B\|\, \|AC\|$, but I can not prove it. Anybody can help? Thank you so much!

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  • $\begingroup$ Welcome to MSE. Please edit and use MathJax to properly format math expressions. $\endgroup$ – Lee David Chung Lin Aug 14 at 1:34
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    $\begingroup$ What does $\|A\|$ mean in this context? Are we specifically using the operator norm induced by the usual norm on $\Bbb C^n$? $\endgroup$ – Omnomnomnom Aug 14 at 2:08
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Your statement is false. For a counterexample, consider $$ A = C = \pmatrix{0&1\\0&0}, \quad B = \pmatrix{1&1\\1&1} $$


The statement will hold in the case where $C = A^T$, assuming that $\|\cdot\|$ is the operator norm induced by the usual (Euclidean) norm on $\Bbb R^n$. One proof is as follows.

Because $B$ is PSD, there exists an $M$ such that $B = MM^T$. With that, we have $$ \|ABC\| = \|(AM)(AM)^T\| = \|AM\|^2 \leq \|A\|^2 \|M\|^2 = \|AA^T\| \cdot \|MM^T\| = \|B\| \cdot \|AC\| $$

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  • $\begingroup$ Thank you! Then what if $A = C^T$? Does it hold now? $\endgroup$ – RunStat Aug 14 at 14:36
  • $\begingroup$ Yes, see my latest edit $\endgroup$ – Omnomnomnom Aug 14 at 14:49
  • $\begingroup$ In the future, it would be best if you posted changes like this to your question as a new question, so that the answers given remain valid and complete. $\endgroup$ – Omnomnomnom Aug 14 at 14:50
  • $\begingroup$ Thank you so much for your help! This is my first question. I will follow the standard in the future. $\endgroup$ – RunStat Aug 14 at 15:18
  • $\begingroup$ @RunStat You're welcome. If you feel that your question is completely answered, click the $\checkmark$ below the arrows on the left to "accept" an answer. $\endgroup$ – Omnomnomnom Aug 14 at 15:29
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No. Let $$ A = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}, \quad C = \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}.$$

Then $AC = 0$, but $ABC = \begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix}$, so $\|ABC\| > 0 = \|B\| \|AC\|$.

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  • $\begingroup$ Thank you! Then what if $A = C^T$? Does it hold now? $\endgroup$ – RunStat Aug 14 at 14:36

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