= Solution
{c}
The <rank of a linear map> and <nullity of a linear map> are
$$
r(T)=\dim\operatorname{im}T,\qquad n(T)=\dim\ker T.
$$
The <rank-nullity theorem> states that
$$
\boxed{r(T)+n(T)=\dim V=n.}
$$
To prove it, choose a <basis> $u_1,\ldots,u_k$ of the <kernel of a linear map> and use <basis extension> to obtain a <basis>
$$
u_1,\ldots,u_k,v_1,\ldots,v_{n-k}
$$
of $V$. The <vectors> $T(v_1),\ldots,T(v_{n-k})$ <span> the <image of a linear map>: applying $T$ to an arbitrary <linear combination> of the displayed <basis> removes all the $u_i$ terms. They are also <linearly independent>. Indeed,
$$
\sum_j a_jT(v_j)=0\quad\Longrightarrow\quad\sum_j a_jv_j\in\ker T,
$$
so $\sum_j a_jv_j=\sum_i b_iu_i$ for some <scalars> $b_i$. The <linear independence> of the extended <basis> forces every $a_j$ and $b_i$ to vanish. Consequently $r(T)=n-k$ and $n(T)=k$, proving the identity. The same argument covers an empty <basis> when one of these <dimensions> is zero.
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