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I have the following linear map: $L: \mathbb{R}^2 \rightarrow \mathbb{R}^2$.

Now, suppose $L\left( \left[\begin{matrix} 1 \\ 1\end{matrix} \right]\right) = \left[\begin{matrix} 1 \\ 4\end{matrix} \right]$ and $L\left( \left[\begin{matrix} 2 \\ -1\end{matrix} \right]\right) = \left[\begin{matrix} -2 \\ 3\end{matrix} \right]$.

My question is, how can we find $L\left( \left[\begin{matrix} 3 \\ -1\end{matrix} \right]\right)$ from the information given above, and why?

I understood that a linear map associates (or transforms) a vector with another one, it is like a function, for example $f(x) := x^2$, for every input we have some output, and this mapping is related to the function itself.

Now, I suppose that I need to find what is this map anyway.

I have actually a solution here, but I don't understand it:

Assume $\left[\begin{matrix} 3 \\ -1\end{matrix} \right] = x \left[\begin{matrix} 1 \\ 1\end{matrix} \right] + y \left[\begin{matrix} 2 \\ -1\end{matrix} \right]$, then $x = \frac{1}{3}$, $y = \frac{4}{3}$. Therefore $L\left( \left[\begin{matrix} 3 \\ -1\end{matrix} \right]\right)$ = $L\left( x\left[\begin{matrix} 1 \\ 1\end{matrix} \right] + y\left[\begin{matrix} 2 \\ -1\end{matrix} \right]\right)$ = $\frac{1}{3}\left[\begin{matrix} 1 \\ 4\end{matrix} \right] + \frac{4}{3}\left[\begin{matrix} -2 \\ 3\end{matrix} \right]$ = $\left[\begin{matrix} \frac{-7}{3} \\ \frac{16}{3}\end{matrix} \right]$.

What I don't understand is:

  • why should the vector $\left[\begin{matrix} 3 \\ -1\end{matrix} \right]$ be represented as a linear combination of the other 2?

  • Why it uses this fact for applying then the linear map on $\left[\begin{matrix} 3 \\ -1\end{matrix} \right]$?

I cannot understand the connection between the whole parts of the problem sincerely.

How should one proceed in general in this kind of situations?

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    $\begingroup$ I have found this lecture by professor Strang that really explains what is behind this solutions. Check this out! $\endgroup$ – nbro Jun 2 '15 at 20:24
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You are given a linear transformation $L:\mathbb{R}^2 \to \mathbb{R}^2$. To define a linear transformation, it is sufficient (and necessairy) to define it on a base of your domain. In this case, the base is $B=\{(1,1);(2,-1)\}$. It is easy to see that $B$ is a base for $\mathbb{R}^2$ since the vectors are linearly independent.

Now, since $B$ is a base for $\mathbb{R}^2$, it means that any vector $v \in \mathbb{R}^2$ can be written as a linear combination of elements of the base $B$. In particular, your vector $(3,-1)=1/3(1,1) + 4/3(2,-1)$.

Now we know that $L$ is a linear transformation, which means that given $v,w \in \mathbb{R}^2$ and $\lambda, \mu \in \mathbb{R}$ we have that $L(\mu v+\lambda w)=\mu L(v) + \lambda L(w)$

Therefore, to calculate $L(3,-1)$, we first express our vector $(3,-1)$ as a linear combination of $(1,1)$ and $(2,-1)$ as $(3,-1)=1/3(1,1) + 4/3(2,-1)$ and apply our knowledge on linear transformations!

\begin{align} L(3,-1)&=L(1/3(1,1) + 4/3(2,-1)) \\ &=L(1/3(1,1))+L(4/3(2,-1)) \\ &=1/3L(1,1)+4/3L(2,-1) \\ &=1/3(1,4) + 4/3(-2,3) = (-7/3,16/3) \end{align}

And we are done!

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