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I've being doing several exercises and none was of this kind, which I can't figure out:

Let $V$ and $W$ be vector spaces with basis $B=\{\vec{v_1},\vec{v_2},\vec{v_3}\}$ and $B'=\{\vec{w_1},\vec{w_2},\vec{w_3},\vec{w_4}\}$, respectively.

Find the transformation matrix of the linear map, $f: V\longrightarrow W$, that verifies that: $$f(\vec{v_1}-\vec{v_2}+3\vec{v_3})=\vec{w_2}+2\vec{w_4}$$ $$f(\vec{v_1}-\vec{v_3})=\vec{w_1}+2\vec{w_2}+3\vec{w_3}+4\vec{w_4}$$ $$f(2\vec{v_1}-\vec{v_2}+\vec{v_3})=\vec{w_1}+3\vec{w_3}$$

I usually tend to approach this kind of problems by finding some auxiliar matrices that let me apply the transformation matrices formula (if thats how its called), that says: $$M_{B'B}(f)=M_{B'C'}\cdot M_{C'C}(f)\cdot M_{CB}$$ where $C,C'$ are the standard/canonical basis of $V,W$, respectively, hence them being easy to find.

Any ideas on how to approach this problem?

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Express $f(v_1),\ f(v_2),\ f(v_3)$ by the given equations and linearity of $f$, then put the coefficients of $w_i$ in the value of $f(v_j)$ into the $i$th row of $j$th column.

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  • $\begingroup$ Oh I see now, thanks for the help. Sorry for not having replied earlier but it was very late. $\endgroup$ – F.Webber Jan 22 '14 at 16:31

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