# Projection matrix product under operator norm

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.

• Which norm do you have in mind with $\| \cdot\|$? Mar 29 at 16:53
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 $$\|\cdot\|$$.