Given $A\in \mathbb{M}_{m\times n}^\mathbb{R},\ B\in \mathbb{M}_{n\times m}^\mathbb{R}$ such that $AB=I_m$.
(a) Find $\rho(A),\rho(B)$ (the ranks of $A,B$) (SOLVED BY ME, $\rho(A)=\rho(B)=m$)
(b) Show the solution set of $(BA)\underline{x}=\underline{0}$ is the same as the solution set of $A\underline{x}=\underline{0}$, and find $\rho(BA)$.
My attempt at (b):
I want to show they have the same dimension and one is contained within the other.
I was able to show Null$(A) \subseteq$ Null$(BA)$, but not $\dim $Null$(A)=$ $\dim $Null$(BA)$. This is where I need help! I know that from the equality of the solution sets I can conclude $\rho(BA)=m$, so no need to explain that.
Thanks in advance.
So you have to check that:
$$\{x:Ax=0\}=\{x: BAx=0\}.$$
The inclusion $\subseteq$ is clear, because if $Ax=0$ then $BAx=B0=0.$ Now: suppose that $x$ is such that $BAx=0$; then
$$ABAx=A0=0 \Longrightarrow I_mAx=0,$$
since the identity matrix is invertible then $Ax=0$ so $x \in \{x:Ax=0\}$.