Counter example for the following

Question

Let $A - B$ be positive semidefinite and A and B are positive definite. Is $A^2-B^2$ also positive semidefinite? I have been trying to find an example where it is not using 2x2 matrices but cannot seem to find one.

Answer 1

In general NO: Consider $$A = \begin{pmatrix} 4 & 1 \\ 1 & 1\end{pmatrix} \quad\text{and}\quad B = \begin{pmatrix} 2 & 1 \\ 1 & 1\end{pmatrix}, \quad\text{which satisfy}\quad AB = \begin{pmatrix} 9 & 5 \\ 3 & 2\end{pmatrix}.$$ Then $$A-B = \begin{pmatrix} 2 & 0 \\ 0 & 0\end{pmatrix} \quad\text{is positive semidefinite}$$ yet $$A^2-B^2 = \begin{pmatrix} 12 & 2 \\ 2 & 0\end{pmatrix} \quad\text{is not positive semidefinite.}$$

As an aside: If $A$ and $B$ commute, then YES even for operators: In Chapter VII of the book by F. Riesz and B. Sz.-Nagy on Functional Analysis (Dover), you will find that if $X$ and $Y$ are commuting positive semidefinite matrices, then their product is also symmetric and positive semidefinite. Apply this to $X=A+B$ and $Y=A-B$ which luckily commutes since $XY=YX = A^2-B^2$.