TITLE: Rank of product of a matrix and its transpose
QUESTION [34 upvotes]: How do we prove that
$\operatorname{rank}(A) = \operatorname{rank}(AA^T) = \operatorname{rank}(A^TA)$ ?
Is it always true?

REPLY [24 votes]: Here is a common proof.
All matrices in this note are real. Think of a vector $X$ as an $m\!\times\!1$ matrix. Let $A$ be an $m\!\times\!n$ matrix.
We will prove that $A A^T X = 0$ if and only if $A^T X = 0$.
It is clear that $A^T X = 0$ implies $AA^T X = 0$.
Assume that $AA^T X = 0$ and set $Y = A^T\!X$. Then $X^T\!A\, Y = 0$, and thus $(A^T\!X)^T Y = 0$. That is $Y^T Y = 0$. Setting $Y = [y_1 \cdots y_n]^\top$ we obtain $0 = Y^T Y = y_1^2 + \cdots + y_n^2$. Since the entries of $Y$ are real,  for all $k \in \{1,\ldots,n\}$ we have $y_k^2 \geq 0$. Therefore, $Y^T Y = 0$ yields $y_k = 0$ for all $k \in \{1,\ldots,n\}$. Thus $Y = A^T X = 0$.
We just proved that the $m\!\times\!m$ matrix $AA^T$ and the $n\!\times\!m$ matrix $A^T$ have the same null space. Consequently, they have the same nullity. The nullity-rank theorem states that
$$
\operatorname{Nul} AA^T + \operatorname{Rank} AA^T = m = \operatorname{Nul} A^T + \operatorname{Rank} A^T. 
$$
Hence $\operatorname{Rank} AA^T = \operatorname{Rank} A^T$.