Prove Null($A^t A$) = Null($A$) I couldn't find anything like this (though it may have been bad searching), so:
Given that $A$ is an m x n matrix, prove that:
Null($A^t A$)= Null($A$)
I'm not sure how to prove this or which properties to use.

Assume $x\in\operatorname{Null}(A)$ so we have $Ax=0$ and therefore $A^TAx=A^T\cdot 0=0$ and $x\in \operatorname{Null}(A^TA)$ and we have proven that $\operatorname {Null}(A)\subset \operatorname{Null}(A^TA)$
Now assume that $A^TAx=0$ so we have $\langle A^TAx,x\rangle=0$ where $\langle , \rangle$ is the Euclidean inner product of the space. The property of transpose leads to $\langle Ax,Ax\rangle=0$ and because the inner product is definite we have $Ax=0$ and therefore $\operatorname {Null}(A^TA)\subset \operatorname{Null}(A)$
