Eigenvalues of a diagonalizable stochastic matrix

Question

Suppose $T$ is a stochastic diagonalizable matrix. Then we can write $T\cdot 1=P^{-1}DP\cdot 1=1$ where $1$ is the column vector with all entries equal to one. Then we have $DP\cdot 1=P\cdot 1$ where $D=diag\left(d_1,d_2,...,d_n\right)$. This means that $d_k\cdot\Sigma_j\left(p_{k,j}\right)= \Sigma_j\left(p_{k,j}\right)$ and eventually $d_k=1$ for any $k$. But this implies that all eigenvalues of $T$ are equal to one so also its determinant. Is this correct?

Answer 1

No, that only means the logical disjunction "$d_k=1$ or $\sum_jp_{kj}=0$" is true. E.g. consider $$ T=\frac14\pmatrix{1&3\\ 2&2}=\pmatrix{-\frac32&1\\ 1&1}\pmatrix{-\frac14\\ &1} \pmatrix{-\frac25&\frac25\\ \frac25&\frac35}=P^{-1}DP, \quad P\mathbf1=\pmatrix{0\\ 1}. $$ The first diagonal entry of $D$ is $-\frac14\,(\ne1)$ and the first entry of $P\mathbf1$ is zero.

In fact, by Perron-Frobenius theorem, $1$ is a simple eigenvalue of $T$ whenever $T$ is irreducible.