Why does a markov matrix always have an eigenvalue of 1? According to this lectuere:

A Markov matrix A always has an eigenvalue 1  

because 

For the transpose of matrix $A^T$, the sum of the row vectors is equal to 1. The matrix $A^T$ therefore has the eigenvector [1;1;...;1]

I don't really understand this. Why does the sum of row vectors adding to 1 make the matrix have an eigenvalue of 1?
 A: Assume $A=\begin{pmatrix} a_{11}&a_{12}&...&a_{1n}\\a_{21}&a_{22}&...&a_{2n}\\.\\.\\.\\a_{n1}&a_{n2}&...&a_{nn}\end{pmatrix}$ is a matrix with all rows summing up to 1 and $v=\begin{pmatrix}1\\1\\.\\.\\.\\1\end{pmatrix}$. Therefore:$$Av=\begin{pmatrix} a_{11}&a_{12}&...&a_{1n}\\a_{21}&a_{22}&...&a_{2n}\\.\\.\\.\\a_{n1}&a_{n2}&...&a_{nn}\end{pmatrix}\begin{pmatrix}1\\1\\.\\.\\.\\1\end{pmatrix}=\begin{pmatrix}a_{11}+a_{12}+...+a_{1n}\\a_{21}+a_{22}+...+a_{2n}\\.\\.\\.\\a_{n1}+a_{n2}+...+a_{nn}\end{pmatrix}=\begin{pmatrix}1\\1\\.\\.\\.\\1\end{pmatrix}=v$$where the latter equality is true according to our assumption. So $\begin{pmatrix}1\\1\\.\\.\\.\\1\end{pmatrix}$ is an eigenvector corresponding to eigenvalue 1.
A: A more general statement is available. 

If $A$ is an $n$-by-$n$ matrix with complex entries, then $A$ has row sums equal to one if and only if $Ae = e$, in which $e$ denotes the all-ones vector. 

Proof. Notice that 
\begin{align*}
Ae = e \Longleftrightarrow \sum_{j=1}^n a_{ij}=1,~i=1,\dots,n.
\end{align*}
A: If you multiply out A*[1;1;...;1], you get the vector [$\sum_{j=1}^n a_{1j}$;$\sum_{j=1}^n a_{2j}$;...;$\sum_{j=1}^n a_{nj}$]. If all these numbers are the same, then [1;1;...;1] was an eigenvector.
