# Diagonalizability of a certain class of matrices

All matrices below are real and square. Let $A,B$ be diagonal matrices (i.e., off-diagonal entries are zero) with (edit: strictly) positive diagonal entries. Let $P$ be a symmetric positive definite matrix.

1. Is $A+BP$ diagonalizable?
2. Is every eigenvalue of $A+BP$ positive?
• (1) $P=[[1,1],[1,2]]$, $B=[[2,0],[0,-1]]$ and $A=[[-2,0],[0,2+2\sqrt{2}]]$. Commented Sep 27, 2017 at 12:13
• @Hellen : $A$ and $B$ should have positive diagonal entries. Commented Sep 27, 2017 at 12:14
• Counterexample : take $$A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, \; P = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}.$$ $A$ and $B$ are both diagonal matrices with positive entries. $P$ is symmetric positive definite. But: $$A + B P = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$$ is not diagonalizable. Commented Sep 27, 2017 at 12:45
• Thanks for the example! I edited the original to say that the diagonals of $A$ and $B$ are strictly positive. In your example, if we perturb the matrices a little, e.g. $$A = \begin{bmatrix}1 & 0 \\ 0 & \epsilon\end{bmatrix}, \quad B = \begin{bmatrix} \epsilon & 0 \\ 0 & 1\end{bmatrix}$$ then the questions that I asked are true for this example. Commented Sep 27, 2017 at 13:21

## 1 Answer

The answers to both your questions are "yes", because $A+BP = B(B^{-1}A+P)$ is similar to $B^{1/2}(B^{-1}A+P)B^{1/2}$, which is positive definite and diagonalisable.

• @beeflavor Your edit is wrong and I roll it back. $B(B^{-1}A+P)$ is not similar to $B^{1/2}(B^{-1}A+P)B^{\color{red}{-1/2}}$. Rather, it is similar to $B^{-1/2}\left(B(B^{-1}A+P)\right)B^{1/2} = B^{1/2}(B^{-1}A+P)B^{\color{red}{1/2}}$. The latter is positive definite (hence diagonalisable). Commented Sep 27, 2017 at 17:23
• Oops, you are correct. Sorry! Commented Sep 27, 2017 at 18:26