For a singular matrix A, if row space = column space, is it always true that $A = \pm A^T$?

Question

Problem:

For a singular matrix A, if row space = column space, is it always true that $A = \pm A^T$

My line of thoughts:

We know A is not full rank, and A is square.

So A's columns/rows are dependent.

Let $v$ be any vector from A's row space, then there exists $x$ so that $v^Tx = 0$

And $v$ is also in column space, for same $x$, we have $x^Tv=0$

Let $v$ be every row of A, we have $Ax=0$. Same let $v$ be every column of A, we have $A^Tx=0$.

\begin{cases} Ax=0 \\ A^Tx=0 \end{cases} Since A is not full rank, $x$ cannot be $0$.

To satisfy above equation for a given $x$,

$A = c \cdot A^T$ is a solution. => by transpose rules we'll find out $c = \pm 1$

Now I'm stuck.

My questions:

1. Is there any other solutions to above equations?
2. How should I better represent the constraint between $A$ and $A^T$. (Now I'm simply treating them as $A$ and $B$ as totally separate matrices. )
3. Can you prove if the statement is true? Or give a counterexample if false.

(Working through Gilbert Strang's linear algebra section 3.6)

Answer 1

Take any nonsingular, non-symmetric and non-skew-symmetric matrix $B$ and let $A=\pmatrix{B\\ &0}$.