Row space and column space of $A^T A$ and $A A^T$ Suppose we have a $n \times m$ rectangular matrix $A$ with row space $\text{row}(A)$ and column space $\text{col}(A)$. 
What are the row spaces and the column spaces of $A^T A$ and $A A^T$?
 A: Since $A^TA$ and $AA^T$ are symmetric, we have
$$
\operatorname{row}(A^TA) = \operatorname{col}(A^TA) \\
\operatorname{row}(AA^T) = \operatorname{col}(AA^T)
$$
Note that each column of $A^TA$ is in $\operatorname{row}(A)$, thus $\operatorname{row}(A^TA) = \operatorname{col}(A^TA) \subseteq \operatorname{row}(A)$. Since $\operatorname{rank}(A^TA) = \operatorname{rank}(A)$, thus in fact
$$
\operatorname{row}(A^TA) = \operatorname{col}(A^TA) = \operatorname{row}(A)
$$
A similar derivation shows that
$$
\operatorname{row}(AA^T) = \operatorname{col}(AA^T) = \operatorname{col}(A)
$$
A: If $Ax=0$, then $A^TAx=0$, which means $N(A)\subset N(A^TA)$, $N(A)$ is the null space of $A$.
On the other hand, if $A^TAx=0$, then 
$$
x^TA^TAx=0, \hspace{3 mm}\text{or} \hspace{3 mm} \|Ax\|=0
$$ 
which means $Ax=0$, and thus
$$N(A^TA)\subset N(A) \hspace{3 mm}\text{and} \hspace{3 mm} N(A^TA)=N(A)
$$
Since $\operatorname{rank}(A) = n-N(A)$, there is 
$$
\operatorname{rank}(A) = \operatorname{rank}(A^TA)
$$
Suppose $A=[\alpha_1,\cdots,\alpha_n] \hspace{2 mm} (\alpha_i $ is the column vector of $A$), then
$$
A^TA=A^T[\alpha_1,\cdots,\alpha_n]=[A^T\alpha_1,\cdots,A^T\alpha_n]
$$
For each column of $A^TA$
\begin{align}
A^T\alpha_i&=[\beta_1  \cdots \beta_n]\alpha_i \hspace{20 mm} (\beta_i \text{ is the column of } A^T \text{ and row of }A)
\\
&=[\beta_1  \cdots \beta_n]\left[ \begin{array}{}
   a_{i1}  \\
   \vdots \\
   a_{in} \\
  \end{array}  \right]
\\
&=\sum \limits_{j=1}^{n}a_{ij}\beta_j
\end{align}
So column of $A^TA$ is the linear combination of rows of $A$, or
$$
\operatorname{col}(A^TA) =\operatorname{row}(A)
$$
Obviously $\operatorname{rank}(A^T)=\operatorname{rank}(A)$, so 
$$
\operatorname{row}(A^TA) =\operatorname{col}(A^TA) =\operatorname{row}(A)
$$ 
Similarly we have 
$$
\operatorname{row}(AA^T) =\operatorname{col}(AA^T) =\operatorname{row}(A^T)=\operatorname{col}(A)
$$ 
