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How would I go about solving the following problem:

Suppose that $T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a linear transformation such that $T(1,-2) = (2,1)$ and $T(-1,3) = (3,0)$ Determine $T(x_1, x_2)$ for any $ (x_1, x_2)$ in $\mathbb{R}^n$

Am supposed to find $e_1$ and $e_2$ representation for $T(1,-2)$ and $T(-1,3)$?

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  • $\begingroup$ Can you use the given values of $T$ to work out $T(1,0)$ and $T(0,1)$? $\endgroup$ – Michael Biro Nov 1 '15 at 3:31
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HINT : $T(1,-2)=T(e_1)-2T(e_2)$ and $T(-1,3)=-T(e_1)+3T(e_2)$. Solve these two equations for $T(e_1)$ and $T(e_2)$ and substitute in $T(x_1,x_2)=x_1T(e_1)+x_2T(e_2)$

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  • $\begingroup$ Yes. Now substitute this in the other equation and you will get $T(e_2) =T(-1,3) +T(1,-2) =(5,1)$. So $T(e_1) =(12,3)$ $\endgroup$ – R_D Nov 1 '15 at 4:07
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$(1,-2)$ and $(-1,3)$ form a basis for $\mathbb{R}^2$, so any vector $v\in\mathbb{R}^2$ is a linear combination of them. So $v=(x_1,x_2)\to T(v)=T(c_1(1,-2)+c_2(-1,3))=c_1T(1,-2)+c_2T(-1,3)=c_1(2,1)+c_2(3,0).$ So we just need to determine $c_1$ and $c_2$.

That is you solve the system of equations: $x_1=c_12+c_23$ and $x_2=c_1+0$. So $c_1=x_2$ which implies $c_2=\frac{x_1-2x_2}{3}$.

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