= Solution
{c}
If $T'=\lambda T$ with $\lambda\ne0$, then $T'(x)=0$ if and only if $T(x)=0$, so their <kernels> agree. Conversely, suppose both nonzero <linear functionals> have <kernel> $K$. Choose $u$ with $T(u)=1$, by scaling a <vector> on which $T$ is nonzero. For each $x\in V$,
$$
x=T(x)u+\bigl(x-T(x)u\bigr),\qquad x-T(x)u\in K.
$$
Therefore $T'(x)=T(x)T'(u)$. Moreover $T'(u)\ne0$, since $u\notin K$. Thus $\boxed{T'=T'(u)T}$, with a nonzero <scalar>.
Every nonzero <linear functional> $V\to\mathbb F_p$ is surjective, so its <kernel> has <dimension> $n-1$ by the <rank-nullity theorem>. Conversely, every <hyperplane> $K$ is such a <kernel>: extend a <basis> of $K$ by one <vector> $u$, and assign value zero on that <basis> and value one on $u$. Hence the <hyperplanes> containing $v$ are exactly the <kernels> of the $p^{n-1}-1$ nonzero <linear functionals> counted above. Each <kernel> occurs for exactly $p-1$ of them, its distinct nonzero <scalar multiples>. The required number is
$$
\boxed{\frac{p^{n-1}-1}{p-1}.}
$$
Back to article page
