Linear Algebra ,rank, space

Question

My math-english is not very good so please be patient.

$A$ and $B$ are given matrices; $A$ is $m\times n$ , $B$ is $n\times m$, and $AB=I$, the identity of order $m$.

I need to prove that linear space of $BAx=0$ is the same as the linear space of $Ax=0$.

And I need to find the rank of $BA$.

Thank you!

Answer 1

Clearly, if $A\mathbf x = \mathbf 0$ then $B(A\mathbf x) = \mathbf 0$

So, you need to show that if $A\mathbf x\ne \mathbf 0,$ then $BA\mathbf x \ne \mathbf 0$

What else to you know? $AB = I.$ In which case $AB\mathbf x = \mathbf x$ and for all $\mathbf x \ne \mathbf 0, B\mathbf x \ne \mathbf 0.$

And so the kernel of $BA$ = kernel $A$

And the Rank of $BA$ = Rank $AB = m.$

Answer 2

Here's another proof. As Doug says, we need to show that if $Ax \neq 0$, then $BAx \neq 0$. So, suppose that $Ax \neq 0$. Then we know that $$ Ax = AB(Ax) = A(BA x) \neq 0 $$ Since $A(BA x) \neq 0$, we can conclude that $BAx \neq 0$, as desired.