Prove $\operatorname{rank} (BA) = \operatorname{rank} (A) = \operatorname{rank} (AC)$, for $B$ and $C$ invertible matrices

Question

This is my question:

Let $A\in F^{m \times n}$, $B\in F^{m \times m}$ and $C\in F^{n \times n}$ be invertible matrices. Prove that $$\operatorname{rank} (BA) = \operatorname{rank} (A) = \operatorname{rank} (AC).$$

How to proceed with this question? A complete answer will help my proper understanding. :)

Answer 1

To show that $\operatorname{rank} (A) = \operatorname{rank} (AC)$, column space of $AC$ is a subspace of column space of $A$. Now, see that $A = (AC)C^{-1}$, which says that column space of $A$ is a subspace of column space of $AC$. For the first part you may use row rank of $A$ equals the column rank of $A$.

Answer 2

The key is to note that because $A$ and $C$ are invertible, they are isomorphisms of vector spaces. Thus, if $U$ is a subspace of $V$, we have $\dim(B(U)) = \dim(U)$.

For the first part: note that $B(\Bbb R^n) = R(B)$ is a subspace of $\Bbb R^m$. It follows that $$ \dim(R(AB)) = \dim(AB(\Bbb R^n)) = \dim(A[B(\Bbb R^n)]) = \dim(A[R(B)]) = \dim(R(B)) $$ we conclude that $AB$ has the same rank as $B$.

For the second: note that $\dim(R(C)) = n$, from which it follows that $R(C) = \Bbb R^n$. So, we have $$ R(BC) = BC(\Bbb R^n) = B[C(\Bbb R^n)] = B[\Bbb R^n] = R(B) $$ it follows that $BC$ has the same rank as $B$.