Show that a given eigenvector belongs to an eigenvalue of 1 for a stochastic matrix

Question

I'm having trouble with this problem and I don't know how to go about it. If you have an $n \times n$ matrix $A$ that satisfies the property that both $A$ and $A^T$ are stochastic, how could I show that the vector $x = \begin{pmatrix}1/n \\\ 1/n \\\ . \\\ . \\\ . \\\ 1/n \end{pmatrix}$ is an eigenvector corresponding to $\lambda = 1$? All I've been able to gather is that $A - \lambda I$ would have columns that add up to $0$, but I'm not sure if that's useful.

Answer 1

Since $A$ and $A^T$ are stochastic matrices, it means the sum of columns and rows would be equal to the all one vector by the definition of the stochastic matrix.

let $e = \begin{bmatrix} 1 \\ \vdots \\ 1\end{bmatrix}$ be the all one vector, $Ae=\sum_{j}A_j$ corresponds to summing up all the columns. Hence the definition of the stochastic matrix, we have $$Ae=e= 1 \cdot e$$

Hence $e$ is an eigenvecector of $A$ with eigenvalue $1$.

Now divide both sides by $\frac1n$ and we can conclude that $\frac{e}n$ is also an eigenvector with eigenvalue $1$.