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I need some clarification on this problem; my class notes and my current thought process are conflicting.

I have a linear transformation $$T(a,b) = (a+2b, 3a-b)$$ and I'm asked to find $[T]_B$ where $B = \{(1,1), (1,0)\}$.

I would answer the question by applying trhe transformation to both vectors of the matrix $B$, ofr example: $ T(1,1) = (3,2)$ and $ T(1,0) = (1,3)$ so $[T]_B$ would be $$\begin{pmatrix} 3 & 1 \\ 2 & 3\end{pmatrix}$$

In my notes for some reason, I took it a step further and converted $(3,2)$ into $2(1, 1) + 1(1,0)$ and converted $(1,3)$ into $3(1, 1) -2(1,0)$ for a final result of $$[T]_B = \begin{pmatrix} 2 & 3 \\ 1 & -2\end{pmatrix}$$ Which is the correct response for $[T]_B$?

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The second one, the columns of the matrix represent the images of the basis elements as you correctly say, but you need to express the images in terms of your basis vectors. The point is, $(3,2) = T(1,1)$ with respect to the usual basis, but $(2,1) = T(1,1)$ with respect to the basis $B$ so the latter matrix is the correct one.

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But, in the first one, I used the basis $B$ to come up with the those vectors. –  Imray Dec 9 '12 at 21:04
    
You used them to find out what they were, yes, but you expressed the co-ordinates of the vectors in terms of the standard basis and not $B$, like you do in the second part. –  Tom Oldfield Dec 9 '12 at 21:06
    
At what point did I use the standard basis - or that just automatic? So another way to solve it would have been to first convert $(1,0)$ into the basis $B$ and then apply the transformation? –  Imray Dec 9 '12 at 21:10
    
@Imray The transformation is only given in terms of the standard basis. You used the standard basis to say $T(1,1) = (3,2)$. We can write the same thing in terms of basis $B$, where all vectors are written with respect to $B$ by saying $T(1,0) = (2,1)$. This is the same transformation on the same vector- with the same result, just written with respect to the basis $B$. –  Tom Oldfield Dec 9 '12 at 21:14
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