Suppose $A$ is a square $n \times n \ (n > 3)$ matrix. Is it true that if $\operatorname{Im}A = \operatorname{Im}A^T$ then $A = A^T$?
My attempt: $\operatorname{Im}A^T = (\ker A)^{\bot}$ hence $ImA = (\ker A)^{\bot}$ hence $\operatorname{Im}A \ \oplus \ \ker A = \mathbb{R}^n$. Then $\forall \ x \rightarrow x = x_1 + x_2$ where $x_1 \in \operatorname{Im}A, x_2 \in \ker A$. Then consider $x^TAy \ (x,y \in \mathbb{R}^n) \ x^TAy = (x_1+x_2)^TA(y_1+y_2) = x_1^TAy_1 + x_1^TAy_2 + x_2^TAy_1+x_2^TAy_2 = x_1^TAy_1$ because the second and the fourth terms are zero according to $y_2 \in \ker A$ and the third term is zero because $Ay_1 \in \operatorname{Im}A$ but $x_2 \in \ker A$. Now consider $x^TA^Ty = (x_1 + x_2)^TA^T(y_1+y_2) = (Ax_1)^Ty_1 + (Ax_1)^Ty_2 + (Ax_2)^Ty_1 + (Ax_2)^Ty_2 = (Ax_1)^Ty_1$ because the last two terms are zero because $x_2 \in \ker A$ and the second term is zero because $Ax_1 \in \operatorname{Im}A$ while $y_2 \in \ker A$ hence $\forall \ x, y \in \mathbb{R}^n \ x^TAy = x^TA^Ty$ therefore $A = A^T$
Is it a correct proof? Thanks in advance!
The statement is not true.
For any invertible $3\times 3$ matrix, we have $\operatorname{Im}A = \Bbb R^3$, and its transpose is also invertible, so it necessarily has the same image. This does not mean that every invertible matrix is symmetric.
For posterity, the mistake was in the last line. Yes, $x^TAy = x_1^TAy_1$ and $x^TA^Ty = x_1^TA^Ty = (Ax_1)^Ty$, but that doesn't mean that $x_1^TAy_1= (Ax_1)^Ty$.