Exercise 136Linear algebra › Linear mapsL1Added 3 Nov 2026

Composition and matrix product

#composition#matrix product#representation

Statement

Let $f:\mathbb{R}^n\to\mathbb{R}^m$ and $g:\mathbb{R}^m\to\mathbb{R}^p$ be linear maps with matrices $A$ (of size $m\times n$ ) and $B$ (of size $p\times m$ ) in the canonical bases. Show that the matrix of $g\circ f$ in the canonical bases is $BA$ .

Hint

The matrix of a linear map is the unique matrix $M$ with $\varphi(x)=Mx$ for all $x$ ; compute $g(f(x))$ with the matrix product.

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Solution

Uniqueness of the matrix. If $Mx=M'x$ for all $x$ , then taking $x=e_j$ (canonical basis vector) shows that the $j$ -th columns of $M$ and $M'$ coincide. So a linear map has exactly one matrix in the canonical bases.

Matrix action. For $x\in\mathbb{R}^n$ , $f(x)=Ax\in\mathbb{R}^m$ and $g(y)=By\in\mathbb{R}^p$ . By associativity of the matrix product,
$(g\circ f)(x)=g(f(x))=B(Ax)=(BA)x.$
Thus $BA$ (of size $p\times n$ ) satisfies the defining property of the matrix of $g\circ f$ .

Conclusion. Composition of linear maps corresponds to multiplication of matrices, in the same order: $\operatorname{Mat}(g\circ f)=\operatorname{Mat}(g)\operatorname{Mat}(f)$ .

Composition and matrix product

Statement

Solution

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