Assume $ p \in$ null(A). Show that if $a$ is any vector such that $a^{T}p \neq 0$, then $a$ must be linearly independent of the rows of A. I am pretty sure that the converse holds, i.e., assume $a$ is linearly indepent of the the rows of matrix $A$. 

$a$ can be decomposed as $a_{N}+a_{R}$ where $a_{N} \in$ null(A) and $a_{R} \in$ range($A^{T}$).  Thus $a_{n} \neq 0$ since if it was, $a$ would be in the range($A^{T}$), thus $a$ would be linearly independent of the rows of A.   

We can construct a $p$ such that it is in the null(A) and $a^{T}p \neq 0.  For example:

$p=a_{N}$ 

Then $a^{T}p = |a_{N}|^{2} \neq 0$.   (since $a_{n} \neq 0$).

So if $a$ is linearly independent of the rows of $A$ and $p \in $ null(A), then $a^{T}p \neq 0$.

Can someone help me go the other way with this proof.
Hint: $p$ must be linearly independent, (in fact, orthogonal) of the span of the rows of $A$.
Hint: $p$ must be linearly independent, (in fact, orthogonal) of the span of the rows of $A$.