Let $v_1,...,v_n$ be a basis of a vector space $V$ over a field $K$. Let $M(T)$ denote the matrix of a linear map $T:V \rightarrow V$ with respect to our basis.
Prove $$M(ST)=M(S)M(T)$$ for all $S,T:V \rightarrow V$

My Attempt
Let $n:= dim(V)$.
Now both $S,T$ have the same basis,and let $A=(\alpha_{ij})$ and $B=(\beta_{ij})$ be corresponding matrices.
And $AB$ be the $n\times n$ matrix $(\gamma_{ij})$. Then by definition of matrix multiplication, $$(\gamma_{ik})=\sum\limits_{j=1}^n \alpha_{ij}\beta_{ik}$$ for $1\leq i,k \leq n$.
Now to calculate the matrix $ST$ We have ; $$ST(v_k)=S(T(v_k))=S(\sum\limits_{j=1}^n \beta_{jk}v_j = \sum\limits_{j=1}^n \beta_{jk}S(v_j)= \sum\limits_{j=1}^n \beta_{jk} \sum\limits_{i=1}^n \alpha_{ij}v_i = \sum\limits_{i=1}^n(\sum\limits_{j=1}^n \alpha_{ij}\beta_{ik})v_i= \sum\limits_{i=1}^n (\gamma_{ij}) v_i$$
So the matrix of $ST$ is $(\gamma_{ij})=AB $ as claimed.

Is this proof correct?


1 Answer 1


While the idea is correct, there are a few errors with the subscripts in your proof, for example: $(\gamma_{ik})=\sum\limits_{j=1}^n \alpha_{ij}\beta_{ik}$ should be $(\gamma_{ik})=\sum\limits_{j=1}^n \alpha_{ij}\beta_{jk}$. And in the end you would like to have $ST(v_k)= \sum\limits_{i=1}^n (\gamma_{ik}) v_i$.

Here is another way, if you are allowed to make use of the fact that the map $\phi:\mathscr{L}(V,V)\rightarrow M_{n \times n}$ sending each operator to its matrix representation is a linear isomorphism, and $\psi:V\rightarrow \mathbb{F}^n$ sending each vector in $V$ to its coordinate vector with respect to any basis in $V$ is also a linear isomorphism, then we have the following for any vector $v \in V$: \begin{equation} M(ST)[v]=[ST(v)]=[S(T(v)]=M(S)[T(v)]=M(S)M(T)[v], \end{equation} where the square brackets indicate the coordinate vector with respect to the basis $\{v_1,v_2, \ldots,v_n\}$.


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