How to verify the following?

Question

If $x \in \mbox{Ker}(A)$, then can we prove that $x \in \mbox{Ker}(AB)$, where $A$ and $B$ are matrices of compatible orders and $B$ is invertible?

Answer 1

We can't verify that because it is not true in general. Suppose that$$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}\text{ and that }B=\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$Then, if $v=\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)$, then $v\in\ker A$, but $v\notin\ker(AB)$.

Answer 2

$B^{-1}x \in \operatorname{ker}(AB)$ as $(AB)B^{-1}x = A(BB^{-1})(x) = Ax = 0$. José already gave an example where your direct statement fails. We do have that the dimensions of the kernel of $A$ and $AB$ are the same.