= Solution
{c}
For the <linear map> $T:V\to W$, its <image of a linear map> and <kernel of a linear map> are
$$
\boxed{\operatorname{im}T=\{T(v):v\in V\},\qquad\ker T=\{v\in V:T(v)=0\}.}
$$
Both contain the <zero vector>, since $T(0)=0$. If $y_1=T(v_1)$ and $y_2=T(v_2)$ belong to the <image of a linear map>, then for any <real numbers> $a,b$,
$$
ay_1+by_2=T(av_1+bv_2)\in\operatorname{im}T.
$$
Thus $\operatorname{im}T$ is a <vector subspace> of $W$. If $u_1,u_2\in\ker T$, then
$$
T(au_1+bu_2)=aT(u_1)+bT(u_2)=0,
$$
so $\ker T$ is a <vector subspace> of $V$.
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