1. ## Linear map question

The question:

If ${v_1, v_2}$ are linearly independent in a real vector space V and $v_3 = 2v_1 + v_2$, is there a linear map $T:W -> \mathbb{R}^2$ where $W = span(v_1, v_2)$ such that

$T(v_1) = $\left( {\begin{array}{c} 1 \\ 2 \\ \end{array} } \right)$$
, $T(v_2) = $\left( {\begin{array}{c} -3 \\ 2 \\ \end{array} } \right)$$
, $T(v_3) = $\left( {\begin{array}{c} -1 \\ 3 \\ \end{array} } \right)$$
?

I'm not sure how to do this. I'm thinking it has something to do with the fact that a linear map needs to satisfy:

$T(\lambda v_1 + ... + \lambda_n v_n) = \lambda_1 T(v_1) + ... + \lambda_n T(v_n)$

Any assistance would be great.

2. Originally Posted by Glitch
The question:

If ${v_1, v_2}$ are linearly independent in a real vector space V and $v_3 = 2v_1 + v_2$, is there a linear map $T:W -> \mathbb{R}^2$ where $W = span(v_1, v_2)$ such that

$T(v_1) = $\left( {\begin{array}{c} 1 \\ 2 \\ \end{array} } \right)$$
, $T(v_2) = $\left( {\begin{array}{c} -3 \\ 2 \\ \end{array} } \right)$$
, $T(v_3) = $\left( {\begin{array}{c} -1 \\ 3 \\ \end{array} } \right)$$
?

I'm not sure how to do this. I'm thinking it has something to do with the fact that a linear map needs to satisfy:

$T(\lambda v_1 + ... + \lambda_n v_n) = \lambda_1 T(v_1) + ... + \lambda_n T(v_n)$

Any assistance would be great.
Is $T\left(2v_1+v_2\right)=2T(v_1)+T(v_2)$ equal to $T(v_3)$?

3. Nope. Thanks.