$AB-AA^TABB^TA^Tvv^TAB$ is not full rank

Question

My question is very simple.

$A \in \mathbb R^{2 \times 3}, B \in \mathbb R^{3 \times 3}, v \in \mathbb R^{2 \times 1}$

I'm looking for matrices $A,B,v$ of the dimensions above, such that:

a) $AB$ has full row rank.

b) The matrix $AB-AA^TABB^TA^Tvv^TAB$ does not have full row rank.

I've been looking for an example like that for a while, beginning to think maybe one doesn't exist.

Anyone care to try find an example?

Answer 1

Notice that $AB-AA^TABB^TA^Tvv^TAB = (I-AA^TABB^TA^Tvv^T)AB$ and the eigenvalues of $AA^TABB^TA^Tvv^T$ are $0$ and $v^TAA^TABB^TA^Tv$, so it is enough that $v^TAA^TABB^TA^Tv = 1$

In order to find the simplest example, take $B=I$ and $$ A = \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0 \end{pmatrix} $$ so in this case, $v^TAA^TABB^TA^Tv = v^Tv$ and any norm one $v$ is sufficient.

Answer 2

Example: $$ A = \pmatrix{1&0&0\\0&1&0}, \quad B = \pmatrix{1&0&0\\0&1&0\\0&0&1}, \quad v = \pmatrix{1\\0}. $$ We find that $AB$ has full rank, but $$ AB-AA^TABB^TA^Tvv^TAB = A - AA^Tv(A^Tv)^T = \pmatrix{0&0&0\\0&1&0} $$ does not.