# Operator norm of ABC where B is a PSD matrix

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!

• Welcome to MSE. Please edit and use MathJax to properly format math expressions. – Lee David Chung Lin Aug 14 at 1:34
• What does $\|A\|$ mean in this context? Are we specifically using the operator norm induced by the usual norm on $\Bbb C^n$? – Omnomnomnom Aug 14 at 2:08

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 $$\|ABC\| = \|(AM)(AM)^T\| = \|AM\|^2 \leq \|A\|^2 \|M\|^2 = \|AA^T\| \cdot \|MM^T\| = \|B\| \cdot \|AC\|$$

• Thank you! Then what if $A = C^T$? Does it hold now? – RunStat Aug 14 at 14:36
• Yes, see my latest edit – Omnomnomnom Aug 14 at 14:49
• 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. – Omnomnomnom Aug 14 at 14:50
• Thank you so much for your help! This is my first question. I will follow the standard in the future. – RunStat Aug 14 at 15:18
• @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. – Omnomnomnom Aug 14 at 15:29

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 $$\|ABC\| > 0 = \|B\| \|AC\|$$.

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