Exercise 1.3.4

Suppose for matrix $A$, its rank is $r$, since $N(A)=N(A^T)$, we know that the dimension of $N(A)$ and $N(A^T)$ spaces are equal, i.e. $m-r=n-r$, so we have $m=n$, $A$ is a square matrix.

For any given vector $x$ in the nullspace $N(A)$, it is also in the left-nullspace $N(A^T)$, so we have $\mathit{Ax}=0$ and $A^{T}x=0$, combine both equations, we have $(A-A^T)x=0$, so unless $x$ is zero, we have $A=A^T$. So if the nullspace has dimension $>0$, we can say $A$ is symmetric.

If, however, the nullspace of $A$ has dimension 0, i.e. $r=m=n$, then we may not have a symmetric matrix $A$.

For example, let $A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 3\hfill \\ \hfill 2\hfill & \hfill 4\hfill \end{array}\right]$, this matrix has a rank of 2, and row space = column space, since both span the full $R^2$ plane, but clearly, $A$ is not symmetric.