[Math] Prove $\mathrm{rank}(BAC)=\mathrm{rank}(BA)=\mathrm{rank}(AC)=\mathrm{rank}(A)$

linear algebramatricesmatrix-rank

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.

Best Answer

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.

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