Products of Linear Maps

Often it makes no sense to multiply together two elements of a vector space, but for some pairs of linear maps a useful product exists.

Product of linear maps

If $T\in \mathcal{L}(U,V)$ and $S\in \mathcal{L}(V,W)$, then the product $ST\in \mathcal{L}(U,W)$ is defined by

$$(ST)(u)=S(Tu)$$

for all $u\in U$.

Thus $ST$ is just the usual composition $S\circ T$ of two functions, but when both functions are linear, we usually write $ST$ instead of $S\circ T$. The product notation $ST$ makes the distributive properties (see below) seem natural.

Note that $ST$ is defined only when $T$ maps into the domain of $S$. We can verify that $ST$ is indeed a linear map from $U$ to $W$ whenever $T\in \mathcal{L}(U,V)$ and $S\in \mathcal{L}(V,W)$.

Algebraic properties of products of linear maps

• Associativity: $(T_1{T}_2)T_3=T_1(T_2{T}_3)$ whenever $T_1$, $T_2$, and $T_3$ are linear maps such that the products make sense (meaning $T_3$ maps into the domain of $T_2$, and $T_2$ maps into the domain of $T_1$).
• Identity: $TI=IT=T$ whenever $T\in \mathcal{L}(V,W)$; here, the first $I$ is the identity operator on $V$, and the second $I$ is the identity operator on $W$.
• Distributive properties: $(S_1+S_2)T=S_{1}T+S_{2}T$ and $S(T_1+T_2)=S{T}_1+S{T}_2$ whenever $T,T_1,T_2\in \mathcal{L}(U,W)$ and $S,S_1,S_2\in \mathcal{L}(V,W)$.

Multiplication of linear maps is not commutative. In other words, it is not necessarily true that $ST=TS$, even if both sides of the equation makes sense.

Example: Two noncommuting linear maps from $\mathcal{P}(\mathbb{R})$ to $\mathcal{P}(\mathbb{R})$.

Suppose $D\in \mathcal{L}(\mathcal{P}(\mathbb{R}))$ is the differentiation map (as we saw in here), and $T\in \mathcal{L}(\mathcal{P}(\mathbb{R}))$ is the multiplication by $x^2$ map we saw. Let’s consider applying these maps to some polynomial $p(x)$.

Then:

$$((TD)p)(x)=x^2{p}^{\prime }(x)$$

but

$$((DT)p)(x)=x^2{p}^{\prime }(x)=2xp(x)$$

Thus, $TD\ne DT$ – differentiating and then multiplying by $x^2$ is not the same as multiplying by $x^2$ and then differentiating.