How can I show if $AA^T+A^TA=A+A^T$, then $Col(A)=Col(A^T)$? Question: For a square matrix $A$, if $$AA^T+A^TA=A+A^T,$$ prove that $$\text{Col}(A)=\text{Col}(A^T).$$
I want to prove this statement, but it's quite difficult...
I think I should use "$Null(A)$ is orthogonal to $Col(A^T)$"
Any good ideas?
 A: We prove that the null-spaces of $A$ and $A^T$ are equal by showing
$$
Av=0\implies A^Tv=0
\\A^Tv=0\implies Av=0
$$
First, to show $Av=0\implies A^Tv=0$, we assume $Av=0$ for some vector $v$. Then by multiplying both sides of $AA^T+A^TA=A+A^T$ in $v^T$ from left and $v$ from right we obtain $${
v^TAA^Tv+v^TA^TAv=v^TAv+v^TA^Tv\implies
\\
v^TAA^Tv+v^TA^TAv=2v^TAv\implies
\\
v^TAA^Tv+0=0\implies
\\
||A^Tv||^2=0\implies
\\
A^Tv=0
}$$
Similarly the proof holds conversely and the statement is proved $\blacksquare$.
A: Using your observation that the null space is the orthogonal complement to the row space, it suffices to show that $\operatorname{Null}(A) = \operatorname{Null}(A^T)$. Now, decompose $A$ into the sum $A = S + K$, where $S$ is symmetric and $K$ is skew-symmetric. In fact, this implies that $S = \frac 12(A + A^T)$ and $K = \frac 12 (A - A^T)$.
Suppose that $x \in \Bbb R^n$ is such that $Ax = 0$. We have
$$
Ax = 0 \implies (S + K)x = 0 \implies Sx = -Kx.
$$
It follows that $x^TSx = -x^TKx$. Because $K$ is skew-symmetric, we must have $x^TKx = 0$ (why?). Thus, we have $x^TSx = 0$. That is,
$$
\begin{align}
0 = x^TSx &= \frac 12 x^T[AA^T + A^TA]x = 
\frac 12 [x^TAA^Tx + x^TA^TAx]
\\ & = 
\frac 12 [(A^Tx)^T(A^Tx) + (Ax)^T(Ax)] = 
\frac 12 [\|A^Tx\|^2 + \|Ax\|^2].
\end{align}
$$
Thus, we must have $\|A^Tx\| = 0$, so $A^Tx = 0$.
That is, we have $Ax = 0 \implies A^Tx = 0$, so that $\operatorname{Null}(A) \subseteq \operatorname{Null}(A^T)$. Symmetrically, we can conclude that $\operatorname{Null}(A^T) \subseteq\operatorname{Null}(A)$. So, we have $\operatorname{Null}(A) = \operatorname{Null}(A^T)$ as desired.
