Inverse of symmetric matrix $AA^\top$ where $rank(A)=m \leq n$?

Question

Suppose I have a matrix $A\in\mathbb{R^{m\times n}}$ where $m\leq n$ and $rank(A)=m$. Is the matrix $AA^\top$ singular?

My hunch is that the matrix is only non-singular when $n=m$.

Answer 1

Here's the proof that the $m\times m$ matrix $AA^\top$ is nonsingular. Suppose $AA^\top x = 0$ for some vector $x\in\Bbb R^m$. Then $$0=(AA^\top x)\cdot x = A^\top x\cdot A^\top x$$ by the fundamental property of transpose. This means that $A^\top x = 0$. Since $A^\top$ is an $n\times m$ matrix with rank $m$, the only solution of this equation is $x=0$. (For example, the only linear combination of the columns of $A^\top$ that gives the zero vector is the zero linear combination.) This means that $AA^\top$ is in fact nonsingular.

Answer 2

Let $$A= \begin{bmatrix}1 & 0 \end{bmatrix}$$

then we have $AA^T=1$.

In general, $$rank(AA^T)=rank(A)=m,$$

hence, it is nonsingular.

Answer 3

Actually, with your symbols, $A A^T$ is non-singular, it is $m$ by $m.$

The square $A^T A$ is a square matrix of size $n$ by $n$ but rank $m,$ therfore singular when $m < n$