Solving For A Linear Operator I'm currently learning about linear operators, and the chapter in my book describing them only has examples with predefined linear operators.
One of the first questions asks:
Given L([1,2]) = [-2,3] and L([1,-1]) = [5,2] find the value of L([7,5])

Since all the examples in my book explicitly give you linear operators, I am completely confused as to how to find a linear operator L.
Thanks
 A: first write $(7,5)$ as a linear combination of $(1,2)$ and $(1,-1)$, or more generally
$$
(1,0)=\frac{2}{3}(1,-1)+\frac{1}{3}(1,2), (0,1)=-\frac{1}{3}(1,-1)+\frac{1}{3}(1,2).
$$
so $(7,5)=7(1,0)+5(0,1)=3(1,-1)+4(1,2)$.
now because $L$ is linear we have
$$
L[(7,5)]=L[3(1,-1)+4(1,2)]=3L[(1,-1)]+4L[(1,2)]=3(5,2)+4(-2,3)=(7,18).
$$
A: You're not required to find $L$ explicitly. Write $(7,5)$ as a linear combination of $(1,2)$ and $(1,-1)$. Use the same combination to compute the required value from the given values.
A: If you want to find L explicitly, the approach yoyo gave will take you there. As he says, $(1,0)=\frac{2}{3}(1,-1)+\frac{1}{3}(1,2), (0,1)=-\frac{1}{3}(1,-1)+\frac{1}{3}(1,2)$.Once you have the basis vectors $(1,0)$ and $(0,1)$ expressed in terms of what you are given, you can write
$L(a,b)=aL(1,0)+bL(0,1)=a(\frac{2}{3}L(1,-1)+\frac{1}{3}L(1,2))+\frac{b}{3}(-L(1,-1)+L(1,2))$
$=\frac{2a-b}{3}(5,2)+\frac{a+b}{3}(-2,3)=(\frac{8a-7b}{3},\frac{7a+b}{3})$
A: I think you are talking about the linear operator on the finite dimensional vector space. For the linear operator $L$ in your question, it's a linear transformation in ${\bf R}^2$. To define a linear operator on the vector space, in your word, to "find" a linear operator, one needs to define the the image of the basis of the vector space under the map. 
