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 $A^T$.

$$\text{rank }A 
= n - \dim \ker A
$$
Now using 
$$
\ker A = \text{im }A^\perp\\
\dim S^\perp = n-\dim S
$$
you can conclude.
