 Solution

If $T^{\prime }=\lambda T$ with $\lambda \ne 0$, then $T^{\prime }(x)=0$ 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 $x\in V$,Therefore $T^{\prime }(x)=T(x)T^{\prime }(u)$. Moreover $T^{\prime }(u)\ne 0$, since $u\notin K$. Thus $\boxed{T^{\prime }=T^{\prime }(u)T}$, 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