Does row sum and column sums equal for binary normal matrices?

Question

Given a matrix $A = (a_{ij})_{m \times m}$ we denote the $i$-th row as $a_{i*} = (a_{i,1}, a_{i,2}, \dots a_{i,m})$ and $i$-th column as $a_{*i} = (a_{1,i}, a_{2,i}, \dots, a_{m,i})^t$. We denote the $i$-th row, and column sum as $\sum a_{i*}$, and $\sum a_{*i}$, respectiely. A matrix $A$ is called a normal matrix if $A^t A = A A^t$, where $A^t$ denotes transpose of the matrix $A$. I am looking for a binary $(0,1)$ normal matrix such that $\sum a_{i*} \neq \sum a_{*i}$ for at lest one $i$. I didnot get any such matrix yet.

If you can construct such a matrix please share it with me.

If there exists no binary $(0,1)$ normal matrix $A = (a_{i,j})_{m \times m}$, such that $\sum a_{i*} \neq \sum a_{*i}$ holds, please give me a proof.

Thank you in advance.

Answer 1

For a binary matrix, the row sums are the diagonal elements $(AA^t)_{ii}=\sum_{j}a_{ij}a_{ij}=\sum_{j}a_{ij},$ while the column sums are the diagonal elements $(A^tA)_{ii}=\sum_{j}a_{ji}a_{ji}=\sum_{j}a_{ji}.$ For a normal matrix these must be equal.