Decomposition of a stochastic matrix

Question

Let $T$ be a square row (not doubly) stochastic matrix. Assume that $T$ is diagonalizable. Then write $T$ as $P^{-1}DP$. Consider now the matrices $D_1=P^{-1}D^{\frac{1}{2}}P$ and $D_2=P^{-1}D^{\frac{1}{2}}P$ whose product is the initial stochastic matrix $T$. Can you prove that $D_i$ is stochastic for $i=1,2$? Note that the converse is true, that is if $D_i$ were stochastic for $i=1,2$ then also $T$ would be stochastic.

Answer 1

The statement here is False.

Let $T$ be any $2\times 2$ stochastic matrix such that $\text{trace}\big(T\big) \lt 1$. Then $T$ has $\lambda_1 =1$ and $\lambda_2 \lt 0$, both real and distinct so $T$ is diagonalizable. $T=P^{-1}DP$

It is true that $\mathbf 1$ is still an eigenvector of
$T^\frac{1}{2}= D_1 =P^{-1}D^\frac{1}{2}P$
(and I assume $\sqrt{\lambda_1}=1$)
that is $T^\frac{1}{2}\mathbf 1 = \mathbf 1$

What is also true is that a stochastic matrix is defined as having real non-negative components. But $T^\frac{1}{2}$ has a natural eigenvalue of $\sqrt{\lambda_1} = 1$ and an unnatural eigenvalue given by $\sqrt{\lambda_2}$, which is non-zero and purely imaginary, irrespective of branch chosen. Thus for some $b\in \mathbb R-\{0\}$ we have
$1 +b\cdot i\notin\mathbb R=\text{trace}\big(T^\frac{1}{2}\big)\implies T^\frac{1}{2}\text{has non-real diagonal}\implies T^\frac{1}{2}\text{not stochastic}$