= Solution
{c}
Use <basis extension> to choose a <basis> $(v,v_2,\ldots,v_n)$ of the <vector space> $V$ over the <finite field> $\mathbb F_p=\mathbb Z_p$. The condition $T(v)=0$ fixes the first value of a <linear functional>; its remaining $n-1$ values are arbitrary. There are $p^{n-1}$ such <linear maps>, exactly one of which is zero. Hence the required number is
$$
\boxed{p^{n-1}-1.}
$$
For $n=1$ this is zero, as it should be: vanishing at a <basis vector> then forces the entire <linear functional> to vanish.
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