Kernel and Image of matrices

Question

Let $A$ and $B$ be matrices of the same size. Then: How can we prove that

$Im(A) = Im(B)$ holds if and only if $Ker(A^T) = Ker(B^T).$

I know that for matrix $A \in \mathbb{C^{m\times n}},$

$Ker(A)=\{x|Ax=0\}$ and $Im(A)=\{Ax|x\in \mathbb{C^{m}}\}$. But, I have no clue how to proceed.

Any hint please.

Answer 1

You need to know that $Ker(A)$ and $Row(A)$ (row space of $A$) are orthogonal complements, i.e $\perp (Row(A))=Ker(A)$. Also $Row(A)=Im(A^T)$

If $Im(A)=Im(B)$, then we have $\perp (Im(A))= \perp(Im(B)) \implies Ker(A^T)=Ker(B^T)$

With the same reasoning, you can go backward

Answer 2

Hint: $\text{Ker}(A^T) = \text{Im}(A)^\perp = \{v\; :\; v^* u = 0 \ \text{for all}\ u \in \text{Im}(A)\}$

Answer 3

Suppose that $A,B:E\rightarrow F$ are linear. $Ker A^T=\{f\in F^*: f\circ A=0\}$ so $Ker A^T= (Im A)^0$ is the annilator of $Im A$. $(Im A)^0=(Im B)^0$ is equivalent to $Im A=Im B$.