Prove.

$\displaystyle Rank(A^T ) = (Null(A))^\perp$

Where, if V is a subspace,

$\displaystyle V^\perp = \{\vec{w} : \vec{w} $ is perpendicular to $\displaystyle \\\vec{v} \ \ \forall \ \ \vec{v} \ \ \varepsilon \ \ V \} $

Now, I know the rank is number of linearly independent columns in a matrix. So the rank of $\displaystyle A^T$ would be the number of linearly independent rows of A.

I know that the nullspace is the solutions to $\displaystyle A\vec{x}=\vec{0}$

However, the "perpendicular compliment" is a term I am not familiar with. And the simple definition does not really help me. Can someone help me with this proof?