Show that rank($AQ$) = rank($A$)

Question

Let $A$ be an $m × n$ matrix. If $P$ and $Q$ are invertible $m×m$ and $n×n$ matrices, respectively, then (a) rank($AQ$) = rank($A$), (b) rank($PA$) = rank($A$), and therefore, (c) rank($PAQ$) = rank($A$).

The rank of the matrix is defined to be the rank of the linear transformation $L_A: F_n \to F_m$, where $L_A(x) = Ax$ and $R$ is the range.

Proof. First observe that $R(L_{AQ})=R(L_AL_Q)=L_AL_Q(F_n)=L_A(L_Q(F_n)) = L_A(F_n)=R(L_A)$,

Can you explain to me why is the $L_Q(F_n) = F_n$? As I think about $L_Q(F_n)$ it must map to $F_m$ because for any vector $v \in F_n$ $L_Q(v) = Qv \in F_m$ by definition.

Answer 1

$Q$ is an invertible matrix, which means that it, for every $b\in F^n$, the equation $Qx=b$ has a solution (i.e., $x=Q^{-1} b$), which means that $L_Q$ must be surjective.

$L_Q$ is a mapping from $F_n$ to $F_n$, which means that $L_Q(F_n)$ is equal to the entire codomain of $L_Q$, which is $F_n$.

Answer 2

Think of the interpretation of matrices in terms of linear maps: $A$ is the matrix of a linear map $f_A:F^n\longrightarrow F^m$; $P$ the matrix of a linear map $u_P:F^m \longrightarrow F^m$ and $Q$ the matrix of a linear map $u_Q:F^n \longrightarrow F^n$.

The invertibility of $P$ and $Q$ means $u_P$ and $u_Q$ are isomorphisms, and the rank of a matrix is the dimension of the image of the associated linear map. Now, an isomorphism does not change the dimension of a subspace, so $$\operatorname{rank}(PA)=\dim\operatorname{Im}(u_P\circ f_A)=\dim\operatorname{Im}( f_A)=\operatorname{rank}(A).$$ Similar proof, mutatis mutandis, for $\operatorname{rank}(AQ)$.