For a linear map $T(a_1v_1+...+a_nv_n) = a_1v_1$, why does this automatically imply $Tv_1 = v_1$?

In my book, they have a basis $v_1,...,v_n$ of a finite dimensional vector space $V$ and they say they can construct the following linear map for $T \in L(V)$:

$T(a_1v_1+...+a_nv_n) = a_1v_1$.

Why can they do this linear map? It seems they just pulled it out of thin air.

Also, why does this linear map imply that $Tv_1 = v_1$?

Thank you!!

Answer

$x=a_1v_1+a_2v_2+\dots+a_nv_n\Rightarrow T(x)=a_1v_1$

$y=b_1v_1+b_2v_2+\dots+b_nv_n\Rightarrow T(y)=b_1v_1$

Then $T(x+y)=T(a_1v_1+a_2v_2+\dots+a_nv_n+b_1v_1+b_2v_2+\dots+b_nv_n)=T\{(a_1+b_1)v_1+\dots+(a_n+b_n)v_n)\}=(a_1+b_1)v_1=a_1v_1+b_1v_1=T(x)+T(y)$

This shows that $T$ is linear map.

Now If $a_1=1$ we get the desired result.