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A (a matrix) = 1 3 0 
               2 1 -1

A: R^3 ----> R^2 linear
v -----> Av

Let B = {(1,0,2),(0,1,-1),(2,1,3)} basis of R^3, C= {(2,1),(3,2)} basis of R^2

Compute the matrix A ith respect to B of R^3 and C of R^2

I have the solution but I don't really uunderstand the workings.

Sol: A(1,0,2) = (1,0) = a11 (2,1) + a21(3,2) = 2(2,1) - (3,2)

A(0,1,-1) = (3,2) = 0(2,1) + 1(3,2)

A(2,1,3) = (5,2) = 4(2,1) - 1(3,2)

this gives the matrix:

2 0 4
-1 1 -1

The main bit I don't understand is why (1,0) and (3,2) and (5,2) are chosen??

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The vectors $(1,0)$, $(3,2)$ and (5,2) are not chosen.

They are the values of the elements of your fixed basis $B$ under the map $A$. For instance, $$\begin{pmatrix}1&3&0\\2&1&-1\end{pmatrix}\begin{pmatrix}1\\0\\2\end{pmatrix}=\begin{pmatrix}1\cdot 1+3\cdot 0+0\cdot 2\\2\cdot 1+1\cdot 0+(-1)\cdot 2\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}.$$ This procedure is called matrix multiplication.

The vectors $(1,0)$, $(3,2)$ and (5,2) can be written in a unique way as linear combinations of elements of the fixed basis of the codomain $\mathbb R^2$. This gives you the entries of the matrix representing your linear map in the bases $B$ and $C$.

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But how do you know these are the values of the elements? how do you work them out? – user9656 Apr 16 '11 at 13:35
@Jenna: I expanded my answer. – Rasmus Apr 16 '11 at 13:43
Cheers. I understand a lot better now! – user9656 Apr 16 '11 at 13:47
@Jenna: Nice to here. Did you know that you can accept an answer, if it satisfies you? Then other people can see that your question has been answered. – Rasmus Apr 16 '11 at 18:29

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