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!


1 Answer 1


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 $x_1^TAy_1= (Ax_1)^Ty$.

  • $\begingroup$ Thanks a lot, I've found a mistake in my considerations. $\endgroup$
    – D F
    Feb 14, 2018 at 10:31
  • $\begingroup$ @DF You can post an answer to your question pointing out the mistake for future viewers - or editing it in directly in your question $\endgroup$
    – Ant
    Feb 14, 2018 at 10:45

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