Let $A$ be a PSD matrix, and $B$ a diagonal matrix where all diagonal entries are between 0 and 1. Is $A - BAB$ PSD?

Question

Let $A \in \mathbb{R}^{n \times n}$ be a positive semi-definite (PSD) matrix, and $B \in \mathbb{R}^{n \times n}$ a diagonal matrix where all diagonal entries are between 0 and 1. Is it true that $A - BAB$ is PSD? That is, for any vector $v$, that

$$v^T(A - BAB)v\geq 0$$

Answer 1

The best way to address those problems is to look for a counterexample. Take for instance,

$$A=\dfrac{1}{10}\begin{bmatrix}6 & 5\\4 & 4\end{bmatrix},\ B=\dfrac{1}{10}\begin{bmatrix}4 & 0\\0 & 8\end{bmatrix}.$$

The matrix $A$ has eigenvalues $0.9583$ and $0.0417$. However, the matrix

$$A-BAB=\begin{bmatrix}0.504 & 0.340\\ 0.272 & 0.144\end{bmatrix}$$

has eigenvalues $0.6774$ and $-0.0294$.