How can we use $(im(A))^\perp = \ker(A^T)$ to prove $rank(A)=rank(A^T)$? Why is it that these two statements are essentially equivalent? $(im(A))^\perp$ represents all vectors orthogonal to the $im(A)$. Yet I'm not sure what this being equal to $\ker(A^T)$ exactly means, nor how it relates to the rank of $A$ and $A^T$. I feel like some of my confusion may be stemming from my shoddy understanding of $A^T$.

Notice that in finite dimensional space $E$ we have
$$(im(A))^\perp\oplus im(A)=E$$
so with the rank-nullity theorem the result is straightforward.
