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There's linear map $L: \mathbb{R}^2 \to \mathbb{R}^2$ with reserves:

$L([1,2])=[1,1], \ \ \ \ L([2,2])=[2,1]$

a) calculate $L([1,0])$ and $L([0,1])$

b) determine $L([x,y])$

I don't know how to solve it. I tried first to determine transformation matrix of $L$, assuming that $[1,2]$ and $[2,2]$ are bases of $\mathbb{R}^2$ spaces (I'm not sure about that).

$L([1,2])=[1,1]=a[1,2]+b[2,2] \Rightarrow a=0, b= \frac{1}{2}$ $L([2,2])=[2,1]=c[1,2]+d[2,2] \Rightarrow c=-1, b= \frac{3}{2}$

$A_L=\begin{bmatrix} 0&-1\\ \frac{1}{2}& \frac{3}{2} \end{bmatrix} \Rightarrow L(x,y)=(-y, \frac{1}{2}x+ \frac{3}{2}y) $

Then $L([1,0])=(\frac{1}{2},0), \ \ \ \ \ L([0,1])=(-1,\frac{3}{2})$

Is that correct?

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First of all yes, $[1,2],[2,2]$ is indeed a basis of $\mathbb{R}^2$. To see why, it's enough to note thath $[1,0]=[2,2]-[1,2]$ (1) and $[0,1]=[1,2]-\frac{1}{2}[2,2]$ (2).

Now for the problem: Using (1),(2), you get

$L([1,0])=L([2,2])-L([1,2])=\\=[2,1]-[1,1]=[1,0]$,

and similarly for $[0,1]$.

To get $L([x,y])$, just consider $xL([1,0])+yL([0,1])$.

About your calculations: The matrix you compute is referred to $\mathbb{R}^2$ with basis $[1,2],[2,2]$. To get $L([x,y])$ you should express it, by means of a change of basis, as the matrix of $L$ with respect to the canonical basis of $\mathbb{R}^2$

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Since $(1,0)=-(1,2)+(2,2)$, $L(1,0)=-(1,1)+(2,1)=(1,0)$. And, since $(0,1)=(1,2)-\frac12(2,2)$, $L(0,1)=(1,1)-\frac12(2,1)=\left(0,\frac12\right)$. Can you take it from here?

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