Two Nonpositive Hermite Matrix diagolized simultaneously

Question

Let $A,B$ be two non-negative $n\times n$ Hermitian matrix, and $rank(A)=n-1$. Show that there exists a nonsingular matrix $P$ such that $P^{-1}AP$, $P^{-1}BP$ both are diagonal.

I have no idea about it.

Answer 1

Assuming that your statement is true, i.e., that $P^{-1}AP$ and $P^{-1}BP$ are both diagonal, and using the obvious fact that the diagonal matrices commute, we see that

$$AB = P (P^{-1}AP) (P^{-1}BP) P^{-1} = P (P^{-1}BP) (P^{-1}AP) P^{-1} = BA,$$

which means that $A$ and $B$ commute. In other words, if $A$ and $B$ do not commute, your statement is false, and it is easy to construct non-commuting $A$ and $B$ that conform to your requirements. For example,

$$A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$$

are two non-negative (in the both possible meanings!) Hermitian matrices of rank $n-1$, but they do not commute, which - by what I've shown before - means that they cannot be simultaneously diagonalized.

The reverse (if $A$ and $B$ commute, then they can be simultaneously diagonalized) is true. For that, see the Schur decomposition of a commuting family of matrices in Notes here and apply the fact that your matrices are Hermitian.