Past exam of the mathematics course of the University of Oxford / 2023 / Prelims / Mathematics I / 2 (Vector spaces over finite fields)

A vector space over a field $\mathbb{F}$ is a set $V$ with vector addition $V\times V\to V$ and scalar multiplication $\mathbb{F}\times V\to V$ satisfying the following vector space axioms. Under vector addition, $V$ is an abelian group: addition is associative and commutative, there is a zero vector $0_V$, and each $v$ has an additive inverse $-v$. For all $a,b\in \mathbb{F}$ and $u,v\in V$,The operations are defined for every indicated pair and take their values in $V$.

A map $T:V\to W$ between vector spaces over the same field $\mathbb{F}$ is a linear map precisely when it preserves vector addition and scalar multiplication:Equivalently, $T(au+bv)=aT(u)+bT(v)$ for all $u,v\in V$ and $a,b\in \mathbb{F}$.

Every $v\in V$ has a unique basis expansion $v={\sum }_{i=1}^n{a}_i{v}_i$. Define the linear functionalUniqueness of the coordinates makes this definition well defined. The coordinates of a sum or a scalar multiple are the corresponding sums or scalar multiples of the coordinates, so this formula defines a linear map. It has $T(v_i)={\mu }_i$. Conversely, any linear map with those values must satisfy the same formula by linearity, proving uniqueness.

If the finite field $\mathbb{F}$ has $q$ elements, each of the $n$ values ${\mu }_i$ can be chosen independently in $q$ ways. The number of linear maps $V\to \mathbb{F}$ is therefore $\boxed{q^n}$, including the zero linear map.

Use basis extension to choose a basis $(v,v_2,\dots ,v_n)$ 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 isFor $n=1$ this is zero, as it should be: vanishing at a basis vector then forces the entire linear functional to vanish.

Put $L=\mathrm{span}\{v\}$. Every two-dimensional vector subspace containing $v$ is $W=\mathrm{span}\{v,w\}$ for some $w\notin L$. There are $p^n-p$ choices for $w$ in $V$. Within any one such $W$, exactly $p^2-p$ choices of $w\in W\setminus L$ yield that same $W$: each forms a linearly independent pair with $v$, and that pair is a basis of $W$.

These sets of choices partition $V\setminus L$, because $w$ determines $\mathrm{span}\{v,w\}$ uniquely. Thus the number isEquivalently these vector subspaces correspond to the one-dimensional vector subspaces of the quotient vector space $V/L$. For $n=1$ there are none and the formula gives zero.

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