I have this proof for Linear algebra, it is as follows:
Let A be an $m × n$ matrix. Prove that $rank(A) = r$ if and only if there is an invertible $m × m$ matrix $P$ and an invertible $n × n$ matrix $Q$ such that $PAQ= \begin{pmatrix} I_r & 0{_r}{_,}{_n}{_-}{_r}\\ 0{_m}{_-}{_r}{_,}{_r} & 0{_m}{_-}{_r}{_,}{_n}{_-}{_r}\\ \end{pmatrix}$
Honestly, I have no idea what this question is really asking. I understand what rank is, but I have no idea how to figure it out given the matrix. I don't really understand what the subscripts under the $0$s in the matrix really mean either. Can anyone provide some guidance? I'm pretty new when it comes to proofs in linear algebra.
The indices indicate the size of the submatrices, here $I_r$ is a $r\times r$ identity matrix and $0_{r,n-r}$ refers to a zero matrix of size $r\times n-r$.
What we have here is sometimes referred to as the normalform of the matrix determined by a linear map. As you know, a linear map $f:V\to W, x\to f(x)$ between finite dimensional linear spaces can always be rewritten in matrix form $A_f: V\to W, x \to A_f x$. Here the matrix $A_f$ depends on the choice of Bases for $V$ and $W$. Usually we choose the standard bases given by the unit vectors $e_1, e_2,\ldots,e_n$, but there are other choices. In particular for any matrix you can choose bases for $V$ and $W$ such that the matrix $A_f$ is in the normal form you are asked to prove.
Hint: Start out with a basis $(v_{n-r},\ldots,v_n)$ of $\ker f$ and extend it to a basis $(v_1,\ldots,v_n)$ of all of $V$. Then already you can say that $A_f$ looks like $\begin{pmatrix}*\cdots*&0\cdots0\\\vdots &\vdots\\ \underbrace{*\cdots*}_{r} &\underbrace{0\cdots0}_{n-r}\end{pmatrix}$
Now think about how to choose a basis of $W$ to make it look like the normal form. The matrices $P$ and $Q$ will then simply be the base change matrices from the standard base to the new bases.