Product of Linear Maps¶

Definition¶

If $\boldsymbol{A}\in\mathcal{L}(V,W)$ and $\boldsymbol{B}\in\mathcal{L}(U,V)$$\mathcal{L}(U,V)$, the product $\boldsymbol{A}\boldsymbol{B}\in\mathcal{L}(U,W)$ is the composition defined by $(\boldsymbol{A}\boldsymbol{B})(\boldsymbol{u})=\boldsymbol{A}(\boldsymbol{B}\boldsymbol{u})$$\mathbf{A}(\mathbf{B}\mathbf{u})$.

Interpretation¶

The product $\boldsymbol{A}\boldsymbol{B}$ of two linear maps is their composition as functions: first apply the map on the right, $\boldsymbol{B}$, then the map on the left, $\boldsymbol{A}$. It is written as a product because, for matrices, composition is exactly matrix multiplication.

The figure reads from left to right: $\boldsymbol{B}$ takes $\boldsymbol{u}\in U$ to $\boldsymbol{B}\boldsymbol{u}\in V$$V$, and $\boldsymbol{A}$ takes that vector on to $\boldsymbol{A}(\boldsymbol{B}\boldsymbol{u})\in W$$W$. The curved arrow is the single map $\boldsymbol{A}\boldsymbol{B}\colon U\to W$ that does both steps at once. The middle space must match: $\boldsymbol{B}$ has to land in $V$, where $\boldsymbol{A}$ is defined, otherwise $\boldsymbol{A}(\boldsymbol{B}\boldsymbol{u})$ has no meaning.

The composition is again linear: $\boldsymbol{B}$ turns $\boldsymbol{u}+\boldsymbol{v}$ into $\boldsymbol{B}\boldsymbol{u}+\boldsymbol{B}\boldsymbol{v}$$\mathbf{B}\mathbf{v}$, and $\boldsymbol{A}$ turns that into $\boldsymbol{A}\boldsymbol{B}\boldsymbol{u}+\boldsymbol{A}\boldsymbol{B}\boldsymbol{v}$$\mathbf{A}\mathbf{B}\mathbf{v}$; scalar multiples pass through both maps in the same way. For matrices, $[\boldsymbol{A}\boldsymbol{B}]=[\boldsymbol{A}]\,[\boldsymbol{B}]$$[\mathbf{A}][\mathbf{B}]$, and the matching middle space becomes the size condition: $\boldsymbol{A}$ is $m\times n$ and $\boldsymbol{B}$ is $n\times k$$k$. Composition is associative, since $(\boldsymbol{A}\boldsymbol{B})\boldsymbol{C}$ and $\boldsymbol{A}(\boldsymbol{B}\boldsymbol{C})$ both mean "first $\boldsymbol{C}$, then $\boldsymbol{B}$, then $\boldsymbol{A}$".

A common slip is to apply $\boldsymbol{A}$ first because it is written first. The order matters: on $\mathbb{R}^2$, let $\boldsymbol{A}(x_1,x_2)=(x_1,0)$ and let $\boldsymbol{B}(x_1,x_2)=(x_2,x_1)$ swap the coordinates. Then $\boldsymbol{A}\boldsymbol{B}(x_1,x_2)=(x_2,0)$ but $\boldsymbol{B}\boldsymbol{A}(x_1,x_2)=(0,x_1)$$(0,x_1)$, so $\boldsymbol{A}\boldsymbol{B}\neq\boldsymbol{B}\boldsymbol{A}$$\mathbf{B}\mathbf{A}$, just as composition of functions is not commutative.

Key Results¶

• The product of compatible linear maps is linear.
• Matrix multiplication represents composition of linear maps.

Examples¶

• If $\boldsymbol{B}\colon\mathbb{R}\to\mathbb{R}^2$ and $\boldsymbol{A}\colon\mathbb{R}^2\to\mathbb{R}$$\mathbb{R}$, then $\boldsymbol{A}\boldsymbol{B}\colon\mathbb{R}\to\mathbb{R}$$\mathbb{R}$.
• For matrices, $\boldsymbol{A}\boldsymbol{B}\boldsymbol{x}=\boldsymbol{A}(\boldsymbol{B}\boldsymbol{x})$$\mathbf{A}(\mathbf{B}\mathbf{x})$.

Prerequisites¶

• Linear Map

Related Concepts¶

• Composition of Functions
• Matrix Product
• Linear Operator
• Identity Matrix

Sources¶

• Math II: Linear Algebra · Week 3

Aliases: composition of linear maps, composition of linear transformations Symbols: element of, real numbers

Last updated: 20 October 2026 03:02