Prove $\mathrm{rank}(BAC)=\mathrm{rank}(BA)=\mathrm{rank}(AC)=\mathrm{rank}(A)$ Would anyone know how to prove the following?
It is stated as a theorem in the textbook without further explanations.

Let $A$ be an $m \times n$ matrix, $B$ an $m\times m$ matrix, and $C$ a $n\times n$ matrix. Then if $B$ and $C$ are nonsingular matrices, it follows that:
$$ \mathrm{rank}(BAC)=\mathrm{rank}(BA)=\mathrm{rank}(AC)=\mathrm{rank}(A)$$

I have searched for other similar questions but the proofs seem to all rely on some notion of fields, and $\mathrm{dim}()$, but the textbook has yet to touch on such concepts at this point.
Your help would be greatly appreciated.
 A: I suppose you defined the rank as $rank(M) = dim(colspace(M)) = dim(rowspace(M))$
Also the picture of the 4 fundamental subspaces of a matrix picture makes the proof quite easy to follow.
Simple observations


*

*for a squared n x n matrix B we have $Nullspace(B) = 0 \iff Rowspace(B)= R^n \iff colspace(B)= R^n \iff B\ is\ invertible $

*if A and B are n x n, $dim(Nullspace(A))=dim(Nullspace(B)) \iff dim(rowspace(A))=dim(rowspace(B)) \iff rank(A)=rank(B)$
Proof that multiplying a matrix M on the left by an invertible matrix B results in $rank(BM)=rank(M)$
By definition: 
$$ x \in Nullspace(BM) \iff BMx = 0 \iff B(Mx) = 0 \\\iff Mx \in Nullspace(B) \iff Mx = 0 \ (because\ B\ has\ zero\ nullspace) \\ \iff x \in Nullspace(M)$$
So $Nullspace(B)=Nullspace(BM)$ so from observation 2, their rank are the same
Proof that multiplying a matrix M on the right by an invertible matrix A results in $rank(MA)=rank(M)$
By definition:
$$Colspace(MA) = \{MAx | \forall x \in R^n\} = \{M(Ax) | \forall x \in R^n \} = \{My,\ y=Ax | \forall x \} = \{My | \forall y \in Colspace(A) \} $$
 and because A is invertible $Colspace(A)=R^n$ from observation 1. Finally:  $$Colspace(MA) =  \{My | \forall y \in R^n\} = Colspace(M) $$ and hence ranks are equal from observation 2. 
A: The rank of a matrix is the dimension of the image of the associated linear application. Let $a,b,c$ be the linear application associated to $A,B,C$. So we have:
$$c: \mathbb{K}^n \to \mathbb{K}^n $$
$$a: \mathbb{K}^n \to \mathbb{K}^m $$
$$b: \mathbb{K}^m \to \mathbb{K}^m $$
We can write:
$$\mathrm{rank}(BAC)= \mathrm{rank}(b\circ a \circ c)$$
By the definition of rank of a linear application:
$$\mathrm{rank}(b\circ a \circ c)=\dim(b(a(c(\mathbb{K}^n))))$$
Since $C$ is non singular , $c$ is an isomorphism and 
$c(\mathbb{K}^n)= \mathbb{K}^n $:
$$\dim(b(a(\mathbb{K}^n)))$$
$b$ is an isomorphism so it preserves dimension of subspaces:
$$\dim(b(a(\mathbb{K}^n)))=\dim(a(\mathbb{K}^n ))=\mathrm{rank}(A)$$
A: This seems difficult, since you did not say how the rank of a matrix is defined and explicitly requrested not to rely on the concepts of field and dimension. I would like to give it try, though.
Let us start with the following definition of $\text{rank}(X)$ for an $m\times n$ matrix $X$:
Definition. If $X$ is the zero matrix, then $\text{rank}(X)=0$; Otherwise, $\text{rank}(X)$ is the smallest positive integer $r$ such that 
$$X=LR$$
holds for an $m\times r$-matrix $L$ and an $r\times n$ matrix $R$.
Observation 1. If $X$ is an $\ell\times m$ matrix and $Y$ is an $m\times n$ matrix, then 
$$\text{rank}(XY)\leq \min(\text{rank}(X),\text{rank}(Y)).$$
Proof.
The observation holds trivially if $X$ or $Y$ is a zero matrix. It remains to consider the case that $X$ and $Y$ are both nonzero matrices.
Let $r=\text{rank}(X)$. 
Let $L$ (respectively, $R$) be an $\ell\times r$ (respectively, $r\times m$) matrix with
$X=LR$.
We have 
$$\text{rank}(XY)\leq r=\text{rank}(X),$$ 
since $XY=L(RY)$ holds for the
$\ell\times r$ matrix $L$ and the $r\times m$ matrix $RY$. One can prove
$\text{rank}(XY)\leq \text{rank}(Y)$ similarly. $\Box$
Since $C$ is nonsingular, its inverse $C^{-1}$ exists. By Observation 1,
$$\text{rank}(A)=\text{rank}(ACC^{-1})\leq \text{rank}(AC)\leq\text{rank}(A),$$ 
implying $\text{rank}(A)=\text{rank}(AC)$. The other equalities can be proved similarly.
A: HINT
It is a known fact: $\text{Rank}(AB)\leq \min(\text{Rank}(A), \text{Rank}(B))\tag 1$
Using (1) you can prove $\text{Rank}(AB)= \text{Rank}(A)$ if $B$ nonsingular.
See this and this for proofs.
