08 - Matrix Representation of a Composition of Linear Maps

Matrix Representation of a Composition of Linear Maps

Let $U,V,W$ be finite-dimensional vector spaces over $\mathbf{F}$ of dimensions $m,n,p$ with ordered basis $\mathcal{U},\mathcal{V},\mathcal{W}$ respectively
Let $S:U\to V$ and $T:v\to W$ be linear

Let

$$A={}_{\mathcal{V}}{S}_{\mathcal{U}}\text{ and }B={}_{\mathcal{W}}{T}_{\mathcal{V}}$$

Then

$$BA={}_{\mathcal{W}}T{S}_{\mathcal{U}}$$

Illegal way to think but the middle vector space ($\mathcal{W}$) “cancels out” ;(

Proof

Note that $A$ is a $n\times m$ matrix and $B$ is a $p\times n$ matrix hence $BA$ is a $p\times m$ matrix
Let $\mathcal{U}$ be $u_1,\cdots ,u_m$
Let $\mathcal{V}$ be $v_1,\cdots ,v_n$
Let $\mathcal{W}$ be $w_1,\cdots ,w_p$

Suppose that $A=(a_{ij})$ and $B=(b_{ij})$ then by definition

$$\begin{array}{rl}S(u_i)& =\sum _{j=1}^n{a}_{ji}{v}_j\text{ for }1\le i\le m\\ T(v_j)& =\sum _{k=1}^p{b}_{kj}{w}_k\text{ for }1\le j\le n\end{array}$$

Then for $1\le i\le m$ we have

$$\begin{array}{rl}(T\circ S)(u_i)=T(S(u_i))& =T\left(\sum _{j=1}^n{a}_{ji}{v}_j\right)\\ & =\sum _{j=1}^n{a}_{ji}T(v_j)\text{ as }T\text{ is linear}\\ & =\sum _{j=1}^n{a}_{ji}\sum _{k=1}^p{b}_{kj}{w}_k\\ & =\sum _{k=1}^p\left(\sum _{j=1}^n{b}_{kj}{a}_{ji}\right)w_k\\ & =\sum _{k=1}^p(BA{)}_{ki}{w}_k\end{array}$$

Hence by definition ${}_{\mathcal{W}}T{S}_{\mathcal{U}}=BA$