TITLE: Range of A and null space of the transpose of A
QUESTION [9 upvotes]: So I'm complete stuck with something. I know it the following statements are true (or at least the seem to be from the results that I got from messing around with it a bit on MATLAB), but I don't understand why they are true or how to show so.
Let $A$ be and $m$X$n$ matrix. Show that:
a) if $x \in N(A^TA)$ then $Ax$ is in both $R(A)$ and $N(A^T)$. 
For this one I messed around with it with my own examples and I got $Ax=0$, therefore satisfing the statement, but I don't understand what's actually going on. 
b) $N(A^TA)=N(A)$
again, makes sense when I see the results in MATLAB, but don't undestand why it works.
c) $A$ and $A^TA$ have the same rank
d) If $A$ has linearly independent columns, the $A^TA$ is nonsingular.
For the last two I have no idea on how to even start showing the relationship. I feel like I'm missing some crucial relationship between $A$ and $A^TA$, but I'm just not seeing it. 
I would greatly appreciate any help of sugestions on how to show that these statements are true. 
Thank you very much.

REPLY [2 votes]: a) By definition $Ax\in R(A)$; on the other hand, $A^TAx=0$, by assumption, so $Ax\in N(A^T)$.
b) It is clear that $N(A)\subseteq N(A^TA)$. Suppose $x\in N(A^TA)$; then $x^TA^TAx=0$ as well, so $(Ax)^T(Ax)=0$, which implies $Ax=0$.
c) The rank-nullity theorem says that, if $B$ is an $m\times n$ matrix, then $\dim R(B)+\dim N(B)=n$. Let $A$ be $m\times n$ and apply the rank nullity theorem to $A$ and $A^TA$:
\begin{align}
n&=\dim R(A)+\dim N(A) \\
n&=\dim R(A^TA)+\dim N(A^TA)
\end{align}
Since by part b we have $\dim N(A)=\dim N(A^TA)$, we conclude $\dim R(A)=\dim R(A^TA)$.
d) If $A$ has linearly independent columns, its rank is $n$; so also $A^TA$ has rank $n$, hence it is invertible.