Convert stochastic matrix to act on diagonal matrices

Question

Given a stochastic matrix $Q$ that converts probability distributions $p$ to other probability distributions, $q = Qp$, how do I adapt this to work for diagonal matrices?

To be more specific, I have probability vectors $p$ of size $N\times 1$ and the stochastic matrix of $Q$ of size $M\times N$ acts on it to produce an output probability vector $q$ of size $M\times 1$. Now, I am given instead the input probability vector as an $N\times N$ diagonal matrix and I would like my output to be the $M\times M$ diagonal matrix corresponding to $q$.

I do not want to convert the $N\times N$ diagonal matrix into a vector and simply apply the original stochastic matrix and then transform the output $M\times 1$ vector into a $M\times M$ diagonal matrix.

Is there a direct way to transform $Q$ to achieve this?

EDIT: As pointed out by Eric, it is impossible if I insist that the transformed $Q$ acts in the same way as $Q$. But assuming I will be doing $\hat{Q}\hat{p}\hat{Q}^T$, (where $\hat{p} = diag(p)$) what is the correct way of getting $\hat{Q}$ from $Q$?

Answer 1

No.

Minimal example: Let $p = \begin{pmatrix}p_1\\p_2 \end{pmatrix}$ and $Q = \begin{pmatrix}q_1 & q_2\\q_3 & q_4\end{pmatrix}$. Then $$ q = Qp = \begin{pmatrix}q_1 & q_2\\q_3 & q_4\end{pmatrix} \cdot \begin{pmatrix}p_1\\p_2 \end{pmatrix} = \begin{pmatrix}q_1 p_1 + q_2 p_2 \\ q_3 p_1 + q_4 p_2 \end{pmatrix} \text{.} $$

You want a $\hat{Q} = \begin{pmatrix}a & b \\ c & d \end{pmatrix}$ taking $\hat{p} = \begin{pmatrix}p_1 & 0 \\ 0 & p_2\end{pmatrix}$ to $\hat{q} = \begin{pmatrix}q_1 p_1 + q_2 p_2 & 0 \\ 0 & q_3 p_1 + q_4 p_2 \end{pmatrix}$. But, $$\hat{Q}\hat{p} = \begin{pmatrix}a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix}p_1 & 0 \\ 0 & p_2\end{pmatrix} = \begin{pmatrix}a p_1 & b p_2 \\ c p_1 & d p_2 \end{pmatrix} \text{,} $$ requiring $b = c = 0$, and leaving no possible assignments for $a$ and $d$. Also, \begin{align*} \hat{Q}\hat{p}\hat{Q}^T &= \begin{pmatrix}a & b \\ c & d \end{pmatrix} \cdot \begin{pmatrix}p_1 & 0 \\ 0 & p_2\end{pmatrix} \cdot \begin{pmatrix}a & c \\ b & d \end{pmatrix} \\ &= \begin{pmatrix}a p_1 & b p_2 \\ c p_1 & d p_2 \end{pmatrix} \cdot \begin{pmatrix}a & c \\ b & d \end{pmatrix} \\ &= \begin{pmatrix}a^2 p_1 + b^2 p_2 & a c p_1 + b d p_2 \\ a c p_1 + b d p_2 & c^2 p_1 + d^2 p_2 \end{pmatrix} \text{,} \end{align*} requiring $ac = bd = 0$ to get the off-diagonals. For $ac = 0$, either $a^2 = q_1 = 0$ or $c^2 = q_3 = 0$ (or both). For $bd = 0$, either $b^2 = q_2 = 0$ or $d^2 = q_4 = 0$. Consequently, for this to have any chance of working, $Q$ is half zeroes.