Exercise 146Linear algebra › Linear mapsL1Added 3 Nov 2026

Rank is invariant under invertible multiplications

#rank#invertible matrices#invariance

Statement

Let $A\in\mathcal{M}_{m,n}(\mathbb{R})$ , $P\in GL_m(\mathbb{R})$ , $Q\in GL_n(\mathbb{R})$ , and recall $\operatorname{rank}(M)=\dim\operatorname{Im} M$ . Show that
$\operatorname{rank}(PA)=\operatorname{rank}(A), \qquad \operatorname{rank}(AQ)=\operatorname{rank}(A), \qquad \operatorname{rank}(PAQ)=\operatorname{rank}(A).$

Hint

Multiplication by an invertible matrix is an isomorphism, hence preserves dimensions of subspaces.

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Solution

$PA$ . $\operatorname{Im}(PA)=P(\operatorname{Im} A)$ . Since $x\mapsto Px$ is an isomorphism of $\mathbb{R}^m$ , it maps the subspace $\operatorname{Im} A$ onto a subspace of the same dimension. Thus $\operatorname{rank}(PA)=\operatorname{rank}(A)$ .

$AQ$ . Since $Q$ is invertible, $Q(\mathbb{R}^n)=\mathbb{R}^n$ , so $\operatorname{Im}(AQ)=A(Q(\mathbb{R}^n))=A(\mathbb{R}^n)=\operatorname{Im} A$ . Thus $\operatorname{rank}(AQ)=\operatorname{rank}(A)$ .

$PAQ$ . Apply the first result to the matrix $AQ$ , then the second: $\operatorname{rank}(P(AQ))=\operatorname{rank}(AQ)=\operatorname{rank}(A)$ .

Conclusion. The rank is unchanged by left and right multiplication by invertible matrices.

Rank is invariant under invertible multiplications

Statement

Solution

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