= Solution
Here $v_3=2v_1-v_2$, so the <image of a linear map> is the <linear span> of $v_1,v_2$. These two <vectors> are <linearly independent>: if $av_1+bv_2=0$, the $e_1$ and $e_2$ coordinates give $a+b=0$ and $a-b=0$, hence $a=b=0$. Consequently the <rank of a linear map> is two. By the <rank-nullity theorem>, the <nullity> is $3-2=1$:
$$
\boxed{\operatorname{rank}\alpha=2,\qquad\operatorname{nullity}\alpha=1.}
$$
More explicitly, the relation $-2v_1+v_2+v_3=0$ gives the <kernel of a linear map>
$$
\ker\alpha=\operatorname{span}\{-2e_1+e_2+e_3\}.
$$
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