Exercise 2.7.14

Solution Both are not possible. Suppose $A$ is $m\times n$ and Ran $A=$ Ker $A^T$. Then Ran $A\subset$ Ker $A^T$, i.e., $A^{T}Av=0$ for any $v\in {ℝ}^n$. This holds only when $A^{T}A=0_{n\times n}$. Then $A=0_{m\times n}$. (Use the row vectors of $A^T$ and check the diagonal entries of $A^{T}A$ equal to 0. It will lead to the conclusion that the row vectors are all-zero vectors.)

On the other hand, if Ran $A=$ Ker $A^T$, Ker $A^T\subset$ Ran $A$. i.e., if $A^{T}b=0$, then the function $Ax=b$ has a solution. But we have $A=0_{m\times n}$, then for arbitrary $b\in {ℝ}^m$, $A^{T}b=0$ holds. But for $b\ne 0$, $Ax=b$ does not have a solution. This is contradictory. So it is not possible for the real or complex matrix $A$ that Ran $A=$ Ker $A^T$.