There is a n by n probability matrix $Q = Q_{ij}$ and $Q_{ij} \ge 0 \quad(i, j = 1,2,...,n)$ with $\sum_{j=1}^nQ_{ij} = 1 \quad (j = 1,2,...,n)$. Furthermore, there is a probability vector p = $ p_i$ and $p_i \ge 0 \quad (i = 1,2,...,n)$ with $\sum_{i=1}^np_i = 1$
I would like to prove that $p^TQ$ is a probability vector.
I have written two proofs:
1-1.
$$ \begin{align*} p = \begin{pmatrix} p_1 \\\ p_2 \\\ \vdots \\\ p_n \end{pmatrix}, Q = \begin{pmatrix} Q_1 \\\ Q_2 \\\ \vdots \\\ Q_n \end{pmatrix} \\ \text{where } Q_i \text{ is a row vector with } ||Q_i|| = 1 \quad (i = 1,2,...,n) \end{align*} $$
$$ \begin{align*} ||p^TQ||_{i} & = \begin{pmatrix} p_1 & p_2 & \cdots & p_n \end{pmatrix} \begin{pmatrix} Q_1 \\\ Q_2 \\\ \vdots \\\ Q_n \end{pmatrix} \\\ & = ||p_1Q_1||_{1} + ||p_2Q_2||_{1} + ... ||p_nQ_n||_{1} \\\ & = p_1||Q_1||_{1} + p_2||Q_2||_{1} + ... + p_n||Q_n||_{1} \\\ & = p_1 + p_2 + ... p_n = 1 \quad (i = 1,2,...,n) \end{align*} $$
1-2.
$Qx = λx$. Since $Q$ is a probability matrix, $λ = 1$ must be the largest eigenvalue of $Q$, such that $Qx = x$. Therefore, $p^TQx = p^Tx$ is a probability matrix.
The first proof is based on the given vector and matrix, and after the multiplication, I sum all the rows up to confirm it indeed is a probability vector.
Now, I would like to know if there is a way that uses eigenvalues to prove it like 1-2 is written. To me, I am not sure if it makes sense to write this way.