= Solution
{c}
For a <linear map> $T:V\to V$, its <kernel of a linear map> and <image of a linear map> are
$$
\ker T=\{v\in V:T(v)=0\},\qquad\operatorname{im}T=\{T(v):v\in V\}.
$$
Its <nullity of a linear map> is $n(T)=\dim\ker T$, and its <rank of a linear map> is $r(T)=\dim\operatorname{im}T$.
Choose a <basis> $u_1,\ldots,u_k$ of $\ker T$ and extend it to a <basis> $u_1,\ldots,u_k,v_1,\ldots,v_m$ of $V$. The <vectors> $T(v_1),\ldots,T(v_m)$ span $\operatorname{im}T$, because applying $T$ to an arbitrary <linear combination> of the full <basis> eliminates the $u_i$ terms. They are also <linearly independent>: if $\sum_jb_jT(v_j)=0$, then $\sum_jb_jv_j\in\ker T$, so $\sum_jb_jv_j=\sum_i a_i u_i$ for some coefficients. Independence of the extended <basis> forces all $a_i,b_j$ to vanish.
Consequently $r(T)=m$, $n(T)=k$ and $\dim V=k+m$. This proves the <rank-nullity theorem>:
$$
\boxed{\dim V=r(T)+n(T).}
$$
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