Operator norm of ABC where B is a PSD matrix

Question

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!

Answer 1

Your statement is false. For a counterexample, consider $$ A = C = \pmatrix{0&1\\0&0}, \quad B = \pmatrix{1&1\\1&1} $$

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

Answer 2

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