Invertibility of Matrix Products

Question

Let $\boldsymbol{A}$ be an $n \times n$ invertible matrix and $\boldsymbol{B}$ be an $n \times k$ matrix with $k < n$ such that $\boldsymbol{B}^\top\boldsymbol{B}$ is invertible. Does this imply that $\boldsymbol{B}^\top \boldsymbol{A}^\top \boldsymbol{A} \boldsymbol{B}$ is invertible?

Answer 1

Yes, it is true. Suppose that $B^TA^TABv=0$ for some vector $v$. Then $v^TB^TA^TABv=0$, too, which means that $\|ABv\|=0$. Since $A$ is invertible, $Bv=0$. Since the product by $B$ is injective, $v=0$. Since the only $v$ for which $B^TA^TABv=0$ is the null vector, $B^TA^TAB$ is invertible.

Answer 2

If $A'$ denotes the transpose, the answer is yes.

The fact that $B'B$ is invertible tells you that $B$ has rank $k$. Then $AB$ has rank $k$, and $B'A'AB$ has rank $k$, so invertible.

The above depends on the fact that the rank of $T'T$ is the same as the rank of $T$. This can be seen for instance from $\ker T=\ker T'T$ and the rank-nullity theorem.

Answer 3

Another take:

We have $\{0\} = \ker (B^TB) = \ker B = \ker (AB) = \ker ((AB)^T AB) = \ker (B^T A^T A B)$.

Since $B^T A^T A B$ is square we see that is must be invertible.