= Solution
The <image of a linear map> and <kernel of a linear map> are
$$
\boxed{\operatorname{im}T=\{T(v):v\in V\}\subseteq W,\qquad\ker T=\{v\in V:T(v)=0\}\subseteq V.}
$$
A <linear map> sends zero to zero, since $T(0)=T(0+0)=T(0)+T(0)$. Thus both displayed <sets> contain zero. If $w_1=T(v_1)$ and $w_2=T(v_2)$ belong to the <image of a linear map>, then for any real <scalars> $a,b$,
$$
aw_1+bw_2=T(av_1+bv_2)\in\operatorname{im}T.
$$
If $v_1,v_2$ belong to the <kernel of a linear map>, then
$$
T(av_1+bv_2)=aT(v_1)+bT(v_2)=0,
$$
so $av_1+bv_2\in\ker T$. Containing zero and being closed under <linear combinations> is the <vector subspace> criterion. Hence the <image of a linear map> is a <vector subspace> of $W$, and the <kernel of a linear map> is a <vector subspace> of $V$.
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