rank(AB) = rank(A) if B is invertible

Question

If $B$ is invertible, show that rank($AB$) = rank($A$).

I've seen this question asked elsewhere but all had answers I didn't understand. I know how to solve the following problem

If $A$ is invertible, then rank($AB$) = rank($B$)

Because if $Bx=0$, then $ABx = A0 = 0$, and when $ABx=0$ then $Bx=0$ because $A$ is invertible, so null($AB$)=null($A$), and by the rank-nullity theorem, rank($A$) = rank($AB$).

However when $B$ is invertible, as in the problem we have to tackle, I don't know how to use that fact. $ABx = 0$, but $B$ is in the middle so we can't simply get rid of it to get a meaningful expression.

Does someone know how to tackle this?

Answer 1

The rank is the dimension of the column space. The column space of $AB$ is the same as the column space of $A$.

Answer 2

You already have: If $A$ is invertible, then rank($AB$) = rank($B$)

But I am going to write it out anyway:

if $\mathbf u$, is in the kernel of $B \implies \mathbf u$ is in the kernel of $AB$ that is $AB\mathbf u = A\mathbf 0 = \mathbf 0$

if $\mathbf v$ not in the kernel of $AB, B\mathbf v = \mathbf x,\mathbf x \ne \mathbf 0$ and since $A$ is non-singular $x \ne \mathbf 0 \implies A\mathbf x \ne \mathbf 0$

Now a similar argument can be used for rank($BA$) = rank($B$) but it is a little bit more work.

For every $\mathbf u$ in the kernel of $B$ there exists an $\mathbf x = A^{-1} \mathbf u$ such that $BA \mathbf x = \mathbf 0$

For every $\mathbf v$ not in the kernel of $B$ there exists an $\mathbf y = A^{-1} \mathbf v$ such that $BA \mathbf y = B\mathbf v $

Answer 3

For any two matrices such that $AB$ makes sense, $$\DeclareMathOperator{\rk}{rk} \rk(AB)\le\rk(A) $$ If $B$ is invertible, then $$ \rk(A)=\rk(ABB^{-1})\le\rk(AB)\le\rk(A) $$