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I would like to have a second opinion and assistance on the following:

Let there be a linear transformation going from $\mathbb{R}^3$ to $\mathbb{R}^2$, defined by $T(x,y,z)=(x+y,2z-x)$. Find the transformation matrix if base 1: $\langle (1,0,-1) , (0,1,1) ,(1,0,0)\rangle$ , base 2 : $\langle (0,1),(1,1)\rangle$ . An attempt at a solution included calculating the transformation on each of the bases in $\mathbb{R}^3$, (base 1) and then these vectors, in their column form, combined, serve as the transformation matrix, given the fact they indeed span all of $B_1$ in $B_2$.

Another point: if the basis for $\mathbb{R}^3$ and $\mathbb{R}^2$ are the standard basis for these spaces, the attempt at a solution is a correct answer.

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  • $\begingroup$ But you have to take the result vectors and express them with regard to the 2nd basis. $\endgroup$ – Gerry Myerson Aug 23 '13 at 12:59
  • $\begingroup$ and why is that? isn't it enough I had already put them through the transformation which puts them in R^2? consider this: the bases for R^3 and R^2 are the standard bases for these spaces...in that case the attempt at a solution works and gives the correct answer...how come? $\endgroup$ – Bak1139 Aug 23 '13 at 13:04
  • $\begingroup$ Because the definition of the transformation matrix is the matrix $A$ such that $Av=T(v)$ when $v$ is expressed in the basis of the domain and $Tv$ is expressed in the basis of the codomain. $\endgroup$ – Gerry Myerson Aug 23 '13 at 13:11
  • $\begingroup$ but they already are in that basis, aren't they? that of the image domain that is... $\endgroup$ – Bak1139 Aug 23 '13 at 13:24
  • $\begingroup$ Look: $T(1,0,-1)=(1,-3)$. But $(1,-3)=(-4)(0,1)+(1)(1,1)$, so what is $(1,-3)$ with regard to the standard basis is $(-4,1)$ with regard to the given basis, and if you want $Av=T(v)$ then you have to use that $(-4,1)$, not $(1,-3)$. $\endgroup$ – Gerry Myerson Aug 23 '13 at 13:31
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Let us call the basis of $\mathbb{R}^{3}$ $\{\lambda_{1},\lambda_{2},\lambda_{3}\}$ and the basis of $\mathbb{R}^{2}$ $\{\gamma_{1},\gamma_{2}\}$.

So all you need to do now is the following:

$T(\lambda_{1}) = \alpha_{11}\gamma_{1} +\alpha_{21}\gamma_{1} $

$T(\lambda_{2}) = \alpha_{12}\gamma_{1} +\alpha_{22}\gamma_{1} $

$T(\lambda_{3}) = \alpha_{13}\gamma_{1} +\alpha_{23}\gamma_{1} $

while $\forall \alpha_{i,j} \in \mathbb{R}$

And the matrix will look like:

$[T]=(\begin{matrix} \alpha_{11} & \alpha_{12} & \alpha_{13} \\ \alpha_{21} & \alpha_{22} & \alpha_{23} \\ \end{matrix})$

And this is how it is done

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