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 $A^T$.

There are a few things we need to prove the equivalence here:


*

*the rank of a matrix is the dimension of its image 
$\DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\im}{im}
\DeclareMathOperator{\null}{null}$

*The rank nullity theorem: $\rank(A) + \null(A) = \rank(A^T) + \null(A^T) = n$, where $\null(A)$ means $\dim \ker(A)$

*for any subspace $U$, $\dim(U) + \dim(U^\perp) = n$


From there, we have
$$
\rank(A^T) = \dim(\im(A^T)) = \dim(\ker(A)^\perp) =\\
n - \dim(\ker(A)) = n - \null(A) =\\
n- (n - \rank(A)) = \\
\rank(A)
$$
