I've come across a question that is similar to this one, but somewhat "reversed" in the role of the unitary matrix.
Suppose I have matrix product $A\, P\, B$ where $A$, $B$ are generic real matrices and $P$ is an $ n \times n $ projection matrix.
In this case, does a relationship of the kind
$$ \lVert A\, P\, B \rVert \leq \lVert A\, B\rVert $$
hold?
Of course, if $A^\top \equiv B \equiv x \in \mathbb{R}^n$ then this becomes a quadratic form, so the inequality is valid. I am quite unsure whether this makes sense in the general case of $A \not= B$ and both $A$ and $B$ being (compatible) matrices.
Thanks in advance!
Best Answer
The answer is no. As a simple counterexample, consider $$ A = \pmatrix{1&-1\\0&0}, \quad P = \pmatrix{1&0\\0&0}, \quad B = \pmatrix{1&0\\1&0}. $$ We find that $APB = P$ and $AB = 0$. Clearly, $\|APB\| > \|AB\|$.
On the other hand, for the particular case of $B = A^T$, the answer is yes. To see that this is the case, it suffices to note that the difference $AA^T - APA^T$ is positive semidefinite. Indeed, we find that $$ AA^T - APA^T = A(I - P)A^T, $$ and the fact that $I - P$ is positive semidefinite implies that $A(I - P)A^T$ is positive semidefinite. Thus, $AA^T$ and $APA^T$ are positive semidefinite with $AA^T \succeq APA^T$ relative to the Loewner order, which implies that $\|AA^T\| \geq \|APA^T\|$, where $\|\cdot\|$ could be the spectral norm, the Frobenius norm, or any other orthogonally invariant norm $\|\cdot\|$.