# How to solve linear map with reserve?

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?

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 $$\mathbb{R}^2$$

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?