Rank theorem of linear algebra Let $U$ be an orthogonal matrix consisting of eigenvectors of $AA^{T}$, and $V$ be an orthogonal matrix consisting of eigenvectors of $A^{T}A$. Here $A$ is an $m\times n$ matrix which has rank $r$.  
I want to show that $\operatorname{rank}(AA^{T})=\operatorname{rank}(A^{T}A)$. It says that since $\dim(N(AA^{T}))=m-r$, we can find independent $u_1,\ldots,u_m$ from $N(AA^{T})$ such that $U=[u_1 \ldots u_m]$ is orthogonal.  
But I can't understand why $\dim(N(AA^{T}))=m-r$ guarantees there exist $m$ independent column vectors and that the matrix $U$ is orthogonal.
I studied rank theorem before but a little confused in here.  
 A: Probably there is some mistake in the question. There can't be, in any case, $m$ independent columns in the matrix $AA^T$. Because, its rank is $r$ which means there should be only $r$ independent columns. But, the first $r$ columns of $U$ (which you have defined) will be a basis for the columns for $AA^T$ and rest of the $m-r$ columns will be a basis for $N(AA^T)$. 
Let $B_1=AA^T$. Now according to rank-nullity theorem, $rank(B_1)+nullity(B_1)=number~of~columns~in~B_1$. Number of columns in $B_1$ is $m$. Rank of $B_1$ is $r$ (rank of A), so from that equation above, $nullity(B_1)=m-r$.  
A: another way to approach:
assume $A\in M_{m\times n}(\mathbb{R})$, $\operatorname{rank}(A)=r$
step 1: $$\operatorname{rank}(AA^T)=\operatorname{rank}(A)$$
obviously, $\operatorname{rank}(AA^T)\leq\operatorname{rank}(A)$
for $\operatorname{rank}(A)=r$,so we can find a subdeterminant
$\begin{vmatrix}
a_{i_1j_1} &\cdots  & a_{i_1j_r}\\ 
\cdots &\cdots  & \cdots\\ 
a_{i_rj_1} &\cdots  &a_{i_rj_r} 
\end{vmatrix}\not=0$
by Binet-Cauchy Formula:
$$AA^T\bigl(\begin{smallmatrix}
i_1 &i_2  &\cdots  &i_r \\ 
i_1 & i_2 &\cdots  &i_r 
\end{smallmatrix}\bigr)=\sum_{1\leq v_1<v_2<\cdots<v_r\leq n}A\bigl(\begin{smallmatrix}
i_1 &i_2  &\cdots  &i_r \\ 
v_1 & v_2 &\cdots  &v_r 
\end{smallmatrix}\bigr)A^T\bigl(\begin{smallmatrix}
v_1 &v_2  &\cdots  &v_r \\ 
i_1 & i_2 &\cdots  &i_r 
\end{smallmatrix}\bigr)=\sum_{1\leq v_1<v_2<\cdots<v_r\leq n}\left[A\bigl(\begin{smallmatrix}
i_1 &i_2  &\cdots  &i_r \\ 
v_1 & v_2 &\cdots  &v_r 
\end{smallmatrix}\bigr)\right]^2>0$$
because at least one term of the above sum $\not=0$,we have found above.the subdeterminant of $A$ of order $r$.
so we have proved $\operatorname{rank}(AA^T) \geq r=\operatorname{rank}(A)$
then we get the conclusion: $\operatorname{rank}(AA^T)=\operatorname{rank}(A)$
step 2:
$$\operatorname{rank}(AA^T)=\operatorname{rank}(A)=\operatorname{rank}(A^T)=\operatorname{rank}(A^TA)$$
