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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.

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TheNekoMessiah
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4 Answers4

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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

  1. 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 $

  2. 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.

jean-luc thirion
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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:

$$\dim(b(a(\mathbb{K}^n)))=\dim(a(\mathbb{K}^n ))=\mathrm{rank}(A)$$

Kandinskij
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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.

modnar
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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.

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