Prove $\Xi (I - P)$ has eigenvalues in the non-negative real half-plane.

Question

Let $P$ be a stochastic matrix (square, non-negative,rows sum to one). Let $\Xi$ be a diagonal matrix with any left principal eigenvector of $P$ on the diagonal and zeros elsewhere (stationary distribution if $P$ is ergodic).

Prove or give a counterexample for the following.

$\Xi (I - P)$ has eigenvalues with non-negative real parts.

I am not sure if this can be called `positive semi-definite' since the matrix is not symmetric.

Answer 1

By the Gershgorin's theorem , for every eigenvalue $\lambda$ of $\Xi(Id-P)$ exists $j$ such that $|\Xi_{jj}(1-P_{jj})-\lambda|\leq\Xi_{jj}(\sum_{i\neq j }|-P_{ji}|)$

Now, notice that for every $j$, $\Xi_{jj}(1-P_{jj})=\Xi_{jj}(\sum_{i\neq j }P_{ji})=\Xi_{jj}(\sum_{i\neq j }|-P_{ji}|)\geq 0$. Thus, $|\Xi_{jj}(\sum_{i\neq j }P_{ji})-\lambda|\leq\Xi_{jj}(\sum_{i\neq j }P_{ji})$.

If the real part of $\lambda$ is negative then the real part of $\Xi_{jj}(\sum_{i\neq j }P_{ji})-\lambda$ is bigger then $\Xi_{jj}(\sum_{i\neq j }P_{ji})$ and $|\Xi_{jj}(\sum_{i\neq j }P_{ji})-\lambda|>\Xi_{jj}(\sum_{i\neq j }P_{ji})$, which is a contradiction.