# 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.

• Use $\text{rank}(B^{-1}BA) \leq \text{rank}(BA) \leq \text{rank}(A)$
– Anon
Commented May 17, 2016 at 8:28
• How is rank defined if the textbook has not introduced dimension yet? Commented May 17, 2016 at 8:54

## 4 Answers

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.

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:

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

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.

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.