Null space for $AA^{T}$ is the same as Null space for $A^{T}$ $A$ is an $n\times m$ matrix and $AA^{T}$ is a symmetric real matrix. 
Also, we have: $\operatorname{rank}(AA^{T})=r\stackrel{?}{=}\operatorname{rank}(A)$. 
Let $Q= \begin{Bmatrix} q_1,...,q_{n-r} \end{Bmatrix}$ be a basis for the Null space of $AA^{T}$. i.e. $AA^{T}q_i=0$, show that $A^{T}q_i=0$.
I guess one proof can be that the Null space for $A^{T}$ is a subspace for Null space for $AA^{T}$, then the question would be why $\operatorname{rank}(A)=\operatorname{rank}(AA^{T})$?
 A: I have found this proof on the Web. I find it easy to read, and I have not found anything clearer on the waters.
We wish to prove that $null(A^TA) \subset  null(A)$ 
  Take $x\in R^n$ such that $A^T Ax=0$. Then, as $x$ is orthogonal to every row vector of $A^T A$, and since $A^T A$ is symmetric, then x is orthogonal to every column vectors of $A^TA$.
Thus, $x^T A^T Ax=0$ $\implies$ $(Ax)^T Ax=0$.
 This implies $Ax.Ax=0$ and $Ax=0$. Which finishes the proof.
http://mathsci.kaist.ac.kr/~schoi/sumlinsec7_5.pdf 
A: Let $q_i$ be a null vector of $A A^\top$, i.e. $ A A^\top q_i =0 $, then $ 0 = q_i^\top A A^\top q_i = \vert\vert A^\top q_i \vert\vert_2$, and thus $q_i$ is also a null vector of $A^\top$. Thus $\mathrm{rank}(A A^\top) = \mathrm{rank}(A^\top) = \mathrm{rank}(A)$.
A: I like above answer; it says only what it needed. However, I tried the following. It takes a slightly different approach as it does not rely on $||A^T q_i$$||$ properties.
Let $X = Null(A)$, then $\forall x \in X, Ax = 0$. Assume that $Y = Null(A^TA)$.
$\forall y \in Y, A^TAy = 0$. This implies,


*

*$Ay= 0$; or

*$A^TAy=0$ and $Ay \not= 0$


If case (1) is true than we are done: $y$ is in $X$. Now
$A^TAy = 0 \implies y^TA^TA = 0^T \implies y^TA^T = 0^T$ ($A$ is non-zero) $\implies Ay = 0$. And this is same as case (1). Thus $X = Y$.
Now one can argue about their ranks.
A: The Null space $N(A^\top)$ of $A^\top$ contains vectors $x$ which satisfies the equation $A^\top.x = 0$. These vectors then will also satify the equation $AA^\top.x = 0$, which implies that $N(A^\top) \subset N(AA^\top)$. To prove that $N(A^\top) = N(AA^\top)$, we now need to show that $N(AA^\top) \subset N(A^\top)$. In order to do that, let's pick any vector $x \in N(AA^\top)$. So, $$AA^\top.x = 0 \implies x^\top.AA^\top.x = 0 \implies \lVert A^\top. x \rVert^2 = 0 \implies A^\top. x = 0 \implies x \in N(A)$$
So, from the above imlication we can say that $N(AA^\top) \subset N(A^\top)$. So, $N(AA^\top) = N(A^\top)$.
