Rank, invertibility and solution spaces

Question

Let A and B be two matrices of the same order. Let's assume that ρ(B)=ρ(AB) (ρ=Rank).

Which of the following two statements are correct or incorrect?

1. A is invertible
2. The solution space of $AB\mathbf x=\mathbf0$ is equal to the solution space of $B\mathbf x=\mathbf0$.

My answer is "incorrect" to 1): I see no theoretical reason to assume it's correct, and I have found a counterexample. Yet I believe 2). to be correct. If A were invertible it would be very easy to prove this, but as it is not I am looking for the proper evidence.

Can anyone help out?

Thank you!

Answer 1

You can make use of the following, which is not too hard to prove: $$ \text{N}(B) \subseteq \text{N}(AB) $$ (null space of $B$ is a subset of the null space of $AB$) and combine it with the rank-nullity theorem.

Answer 2

Hint: Try with simple examples. For $2 \times 2$ matrices, you've already figured out that if the rank is $2$, then $A$ is invertible, and the claim is true. The same is true if the rank is zero. That leaves rank 1 matrices to investigate. So look at some examples.

For instance, here's a rank-1 $ 2 \times 2$ matrix $B$: $$ \pmatrix{1 & 0 \\ 0 & 0} $$

If you pick $A = B$, then the ranks of $AB$ and $B$ are the same, and the solution space of $ABx = 0$ is the same as the solution space of $Bx = 0$, which suggests that maybe "2." is true. But that's not a lot of evidence. What other simple $2 \times 2$ rank-1 matrices can you think of? Try a few.

Also: think about what $B$ does as a linear transformation -- I mean geometrically what it does. Then play some more.