Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 2 / a (Dimensions and characteristic)

The rank-nullity theorem for a linear map $T:V\to W$, with $V$ a finite-dimensional vector space over a field $F$, isHere the nullity is the dimension of the kernel of a linear map, and the rank of a linear map is the dimension of its image of a linear map.

Define the linear functional $L:F^n\to F$ by $L(x)={\sum }_{i=1}^n{a}_i{x}_i$. The given hyperplane is $H=\ker L$. Choose $j$ with $a_j\ne 0$. For any $t\in F$, the vector whose only nonzero coordinate is $x_j=t/a_j$ satisfies $L(x)=t$. Hence $L$ is surjective and has rank of a linear map equal to ${\dim }_{F}F=1$.

Applying the rank-nullity theorem to this particular linear map givesThe nonzero-coefficient hypothesis is essential: if all $a_i$ were zero, the kernel of a linear map would instead be all of $F^n$.

Over the finite field $F={\mathbb{Z}}_p$, a basis of the hyperplane has $n-1$ vectors by ii. Each vector on the hyperplane has a unique expression as a linear combination of that basis, and each coefficient has $p$ possibilities. ThusEquivalently, choose freely all coordinates except $x_j$, where $a_j\ne 0$, and solve uniquely for $x_j$. For $n=1$ this counts just the zero vector, consistently with $p^0=1$.

The transpose is a linear map, so $T(A)=A+A^T$ is a linear map on the $n^2$-dimensional vector space $M_{n\times n}(F)$. Let $E_{ij}$ denote a matrix unit. The answer depends on the characteristic of a field.

If $\mathrm{char}F\ne 2$, then $T(A)=0$ means $A^T=-A$. Such skew-symmetric matrices have zero diagonal, and their entries above the diagonal can be chosen freely. A basis of the kernel of a linear map is $E_{ij}-E_{ji}$ for $i<j$, giving nullity $n(n-1)/2$. The image of a linear map consists of all symmetric matrices: $T(A)$ is symmetric, and every symmetric matrix $S$ is $T(S/2)$. Its basis comprises $E_{ii}$ and $E_{ij}+E_{ji}$ for $i<j$, giving rank of a linear map $n(n+1)/2$.

If $\mathrm{char}F=2$, then $-A=A$, so $\ker T$ consists of all symmetric matrices, with dimension $n(n+1)/2$. Every $T(A)$ is symmetric with zero diagonal, and every symmetric zero-diagonal matrix $S$ arises by taking $A_{ij}=S_{ij}$ for $i<j$ and all other entries zero. The image of a linear map therefore has basis $E_{ij}+E_{ji}$ for $i<j$, and dimension $n(n-1)/2$. This is the symmetrization map in characteristic two.

In both cases the rank-nullity theorem is verified explicitly: