Show that if $AA^t = A^tA$, then $A=A^t$ Suppose $A$ is a matrix with non-negative real entries. If $A^tA = AA^t$, show that $A=A^t$.
My proof says: $AA^t = A^tA = (AA^t)^t$. I can't seem to get to the point of $A=A^t$
Edit: What if $A$ is a $2x2$ matrix?
 A: If we restrict ourselves to $2\times 2$ matrices, let $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}.$$ Then $A^TA=AA^T$ implies that 
$$A^TA = \begin{bmatrix}a^2+c^2& ab + cd\\ab + cd& b^2 + d^2\end{bmatrix} = \begin{bmatrix}a^2 + b^2& ac + bd\\ ac + bd & c^2+d^2\end{bmatrix} = AA^T.$$
Hence,
$$b^2 = c^2$$ and $$ac+bd = ab+cd.$$
Since $A$ has nonnegative entries, the first equation implies that $b=c$, and hence that $A=A^T$.
A: I don't think the statement is true to begin with. A normal matrix is not necessarily symmetric. Consider $$A = \begin{bmatrix}
    1 & 1 & 0 \\
    0 & 1 & 1 \\
    1 & 0 & 1 \\
\end{bmatrix}$$
The entries of $A$ consist of only 0 and 1, which are clearly non-negative. We have $$AA^t=\begin{bmatrix}
    2 & 1 & 1 \\
    1 & 2 & 1 \\
    1 & 1 & 2 \\
\end{bmatrix}=A^tA$$
But $A_{31} \neq A_{13}$ so $A^t \neq A$.

For a $2\times2$ matrix, suppose $$A = \begin{bmatrix}
    a & b \\
    c & d \\
\end{bmatrix}$$
Then $AA^t=A^tA$ implies $b^2=c^2$ (top-left entry of product matrix), and since $b$ and $c$ are non-negative, it follows that $b=c$ and thus $A=A^t$.
A: With regards to $2 \times 2$ matrices:
The Perron-Frobenius theorem implies that any non-negative matrix has a positive real eigenvalue.  Since the complex eigenvalues of real matrices come in pairs, we can conclude that every non-negative $2 \times 2$ matrix has only real eigenvalues.
We then note that if a matrix has real eigenvalues, then it is normal (satisfies $AA^T = A^TA$) if and only if it is symmetric (satisfies $A = A^T$). 
It follows that a non-negative $2\times 2$ matrix is symmetric if and only if it is normal, which was the desired result.
