Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics I / 2 (Rank-nullity and polynomial identities)

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:

The monomials $1,x,\dots ,x^n$ form a basis of $V_n$: they span every polynomial of degree of a polynomial at most $n$, and equality of polynomials forces equality of every coefficient, proving linear independence. ThereforeThe zero polynomial is included in $V_n$ even though its degree of a polynomial need not be assigned an ordinary nonnegative value.

Suppose, for a contradiction, that $p$ and $q$ have no common root of a polynomial. Since the polynomial ring $\mathbb{C}[x]$ has no zero divisors and $p,q\ne 0$, the relation $ap+bq=0$ with $a,b$ not both zero forces both $a$ and $b$ to be nonzero.

By the fundamental theorem of algebra, writewhere the ${\alpha }_j$ are distinct. For each $j$, $q({\alpha }_j)\ne 0$. The equality $bq=-ap$ implies that $(x-{\alpha }_j{)}^{m_j}$ divides $bq$. The multiplicity of a root in polynomial multiplication is the sum of its multiplicities in the factors: factor out the largest powers of $x-{\alpha }_j$ and note that the two remaining factors have nonzero values at ${\alpha }_j$. Since $q({\alpha }_j)\ne 0$, $b$ must contain $(x-{\alpha }_j{)}^{m_j}$.

Thus $b$ has at least ${\sum }_j{m}_j=n$ roots of a polynomial counted with multiplicity of a root, so $\mathrm{deg}b\ge n$. This contradicts $b\in V_{n-1}$. ConsequentlyCounting multiplicity of a root is necessary: counting only distinct roots of a polynomial would miss repeated factors.

Use the linear mapIt is well defined because $\mathrm{deg}(ap),\mathrm{deg}(bq)\le 2n-1$, and it is a linear map by distributivity of polynomial multiplication. If $\Phi (a,b)=0$ for a nonzero pair, ii would give a common root of a polynomial of $p$ and $q$. The hypothesis therefore implies $\ker \Phi =\{(0,0)\}$, so $\Phi$ is injective.

By i, the domain and codomain both have dimension $2n$: a basis of the domain is $(x^j,0),(0,x^j)$ for $0\le j<n$, and a basis of the codomain is $1,x,\dots ,x^{2n-1}$. The stated finite-dimensional result makes $\Phi$ surjective. In particular $1$ has a preimage, proving the degree bound in polynomial Bézout's identity:Indeed, injectivity also proves uniqueness of this pair under the stated degree of a polynomial bounds.