The entries of a column of a stochastic, idempotent matrix all have the same value Let $A \in [0,1]^{n\times n}$,  $A = A^2$ and $\sum _{j=1}^n a_{ij} = 1 \;\forall i \in \{1,..,n\}$ (sum of row values equals $1$ for every row). I want to show, that the column values of $A$ are the same for every row ($ a_{ij} = a_{kj} \;\forall i,j,k$).
I am pretty disconnected from linear algebra right now and a direct proof doesn't work for me right now. I would also be thankful for a hint.
EDIT: I forgot one condition; every entry of the matrix is $ > 0$.
 A: $A^2-A=0$ thus $A$ is diagonalizable in $\mathbb{R}$ and its eigenvalues are $0$ or $1$. If $A=(a_{i,j})_{1\leqslant i,j\leqslant n}$ where we suppose $a_{i,j}>0$ for all $i,j\in[\![1,n]\!]$, then $(a_{i,j}-\delta_{i,j})_{1\leqslant i,j\leqslant n-1}$ is invertible because for all $i\in[\![1,n-1]\!]$, we have
$$ \sum_{\substack{1\leqslant j\leqslant n-1 \\j\neq i}}|a_{i,j}-\delta_{i,j}|=\sum_{\substack{1\leqslant j\leqslant n-1 \\j\neq i}}a_{i,j}=1-a_{i,i}-a_{i,n}>|a_{i,i}-1| $$
and thus ${\rm rank}(A-I_n)\geqslant n-1$. In the general case where we only have $a_{i,j}\geqslant 0$, let $\varepsilon\in ]0,1[$ and $A_{\varepsilon}=(1-\varepsilon)A+\frac{\varepsilon}{n}U$ where the coefficients of $U$ are all $1$. Then $A_{\varepsilon}$ verifies the same hypothesis as $A$, and $(1-\varepsilon)a_{i,j}+\frac{\varepsilon}{n}>0$ for all $i,j\in[\![1,n]\!]$ and thus, because of what said above, ${\rm rank}(A_{\varepsilon}-I_n)\geqslant n-1$. But since $\lim\limits_{\varepsilon\rightarrow 0}(A_{\varepsilon}-I_n)=A-I_n$, we have ${\rm rank}(A-I_n)\geqslant {\rm rank}(A_{\varepsilon}-I_n)$ for all $\varepsilon$ small enough and thus ${\rm rank}(A-I_n)\geqslant n-1$. If we had ${\rm rank}(A-I_n)=n$, then $A-I_n$ would be invertible and thus all the eigenvalues of $A$ would be $0$ which is not because $A\neq 0$. Thus ${\rm rank}(A-I_n)=n-1$ and $1$ is an eigenvalue of $A$ of multiplicity $1$. Since all the other eigenvalues of $A$ are $0$, ${\rm rank}(A)=1$ and, if $i_0\in[\![1,n]\!]$ is such that $C_{i_0}\neq 0$ where $C_1,\ldots,C_n$ are the columns of $A$, then for all $j\in[\![1,n]\!]$, there exists $\lambda_j\in\mathbb{R}$ such that $C_j=\lambda_j C_{i_0}$. Summing all the coefficients of $C_j$ with $j$ fixed gives $\lambda_j=1$ and thus all the columns of $A$ are identical.
EDIT : Since $a_{i,j}$ is supposed to be $>0$ by the OP, we can get rid of the $A_{\varepsilon}$ trick.
