0
$\begingroup$

Consider maps $S: U \mapsto V$ and $T:V\mapsto W.$ The composition $TS$ is a linear map from $U \mapsto W.$

Let $ \mathcal{M(T)} = \begin{pmatrix} a_{1,1}& \ldots & a_{1,n}\\ \vdots & & \vdots \\ a_{m,1}& \ldots & a_{m,n} \end{pmatrix}$ and $ \mathcal{M(S)} = \begin{pmatrix} b_{1,1}& \ldots & b_{1,p}\\ \vdots & & \vdots \\ b_{n,1}& \ldots & b_{n,p} \end{pmatrix}$

$\begin{align} \mathcal{TS}u_k =& \mathcal{T} \left(\sum_{r=1}^n b_{r,k}v_r \right) \\=& \sum_{r=1}^n b_{r,k} \sum_{j=1}^m a_{j,r} w_j \\=&\sum_{j=1}^m \left(\sum_{r=1}^n a_{j,r} b_{r,k} \right)w_j\end{align}$

My main question is, why can the matrix of a linear map be expressed as a sum?

$\endgroup$
1
$\begingroup$

If I've understood the spirit of your question: "Because a linear transformation is uniquely determined by its values on a basis of the domain (and we define the matrix of a transformation to encode this fact)."

In your notation, if $B = \{u_{1}, \dots, u_{p}\}$ is a basis of $U$, $B' = \{v_{1}, \dots, v_{n}\}$ is a basis of $V$, and $S:U \to V$ is linear, there exist unique scalars $b_{r,k}$ with $r = 1, \dots, n$ and $k=1, \dots, p$, such that $$ S(u_{k}) = \sum_{r=1}^{n} b_{r,k} v_{r}. $$ The "matrix of $S$ with respect to the bases $B$ and $B'$" is, by definition, the rectangular array obtained by putting $b_{r,k}$ into the $r$th row and $k$th column. This array completely specifies $S$: Every element $u$ of $U$ can be written uniquely as a linear combination $u = \sum_k c_{k} u_{k}$ for some scalars $c_{k}$, and by linearity of $S$, $$ S(u) = \sum_{k=1}^{p} c_{k} S(u_{k}) = \sum_{k=1}^{p} c_{k} \sum_{r=1}^{n} b_{r,k} v_{r} = \sum_{r=1}^{n} \left(\sum_{k=1}^{p} b_{r,k} c_{k} \right) v_{r}. $$ (This equation also explains where the seemingly-peculiar definition of matrix multiplication comes from. The expression in parentheses is the $r$th coordinate of $S(u)$, a.k.a. the $r$th row of the product of the matrix of $S$ with the coordinate vector of $u$ with respect to $B$.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.