Row sum and column sum of a square matrix If $A$ is a real square matrix with all row sums equal to 1 and $A$ commutes with its transpose then show that column sums of $A$ is also equal to 1 
I tried to solve using the fact that $A1=1$ where $1$ is the column vector of proper size with all it's entries 1 and using the hypothesis. But couldn't proceed.
Please help me.
Thanks in advance.
 A: It becomes convenient to introduce complex numbers here.  
$A^T A = AA^T = AA^* = A^*A$ 
so $A$ commutes with its conjugate transpose given by $A^* = A^T$, which means $A$ is normal and thus unitarily diagonalizable. So, selecting unitary matrix $U$, this implies  
$U^{-1} A U  =U^* A U = D$
or equivalently
$U^* A  = DU^*$ 
your problem about row sums being equal to one implies an eigenvector equation, i.e. $A\mathbf 1 = \mathbf 1$.  Your problem then wants you to confirm that $\mathbf 1^T A = \mathbf 1^T$. But this is implied by the above unitary diagonalization.  
You may select (/assume WLOG) that $\mathbf u_1 = \mathbf 1$. I.e. the diagonal matrix $D$ is given by
$D = \pmatrix{1 & \mathbf 0^*\\\mathbf0& D_0}$ 
putting this all together
$\pmatrix{\lambda_1  \mathbf u_1^*  \\*} = \pmatrix{ \mathbf 1^T  \\*} = \pmatrix{1 & \mathbf 0^*\\\mathbf0& D_0}U^* = DU^* = U^* A$ 
thus
$\lambda_1 \mathbf u_1^* = 1 \cdot \mathbf 1^T = \mathbf 1^T = \mathbf 1^T A$
i.e. $\mathbf 1$ is a left eigenvector for $A$ 
A: Since $A^TA = AA^T$ and $AJ = J$ then
\begin{align}
A(A^TJ)= A^T(AJ) = A^TJ
\end{align}
i.e. $A^TJ$ is an eigenvector of $A$ with eigenvalue $1$. This
also means $A^TJ-J$ is an eigenvector of $A$ with eigenvalue $1$ or
equivalently, 
\begin{align}
A^TJ-J \in E(1)= \text{Nul}(A-I) = \text{Row}(A-I)^\perp
\end{align}
But, we also know that $A^TJ-J \in \text{Row}(A-I)=\text{Col}(A^T-I)$. Hence it follows $A^TJ-J=0$ which is the desired conclusion.
A: I got the answer.
Let $x=A^T1-1$. 
We can show that $x=0$ by showing that $x^Tx=0$. 
Use $A1=1$ and $AA^T=A^TA$.
