Past exam of the mathematics course of the University of Oxford / 2021 / Prelims

Source: released examination PDF, bundle pages 25–32.

Solutions cover all seven questions: Section A, Questions 1–4, and Section B, Questions 5–7, including every optional question. The module label is I for Mathematics and Mathematics and Philosophy, and M1 for Mathematics and Computer Science; the formal paper code is A10138W1. In Question 2(b), $n\ge 1$ is understood, as required by $V_{n-1}$.

Write $A=(a_{ij})$ as an $m\times r$ matrix and $B$ as an $r\times n$ matrix, with rows $b_1,\dots ,b_r$. The rule for matrix multiplication givesEvery row of $AB$ is therefore a linear combination of rows of $B$. Taking linear spans gives

When $A$ is an invertible matrix, $B=A^{-1}(AB)$. Apply i first to $AB$ and then to $A^{-1}(AB)$:Thus the two row spaces are equal. In particular, invertible left matrix multiplication preserves row rank.

For any matrix $Y$ and compatible invertible matrix $P$, the linear map $r\mapsto rP$ maps $\mathrm{rowspace}(Y)$ onto $\mathrm{rowspace}(YP)$: a linear combination of rows of $Y$ is sent to the same linear combination of rows of $YP$. Its inverse function is $s\mapsto s{P}^{-1}$, so these row spaces have equal dimensions. Hence right matrix multiplication by $P$ also preserves row rank.

If $X=P^{-1}YP$, ii now givesThis proves the required matrix similarity invariant without needing to assume equality of row rank and column rank.

For the given real matrix $M$, direct matrix multiplication gives the eigenvectorsPut these eigenvectors into the columns ofSince $\det P=-2\ne 0$, $P$ is an invertible matrix. The three column identities give $MP=PD$, soIn particular, the matrix similarity is established by an explicit change-of-basis matrix.

Write $C=(c_{ij})$ and $(d_1,d_2,d_3)=(0,2,4)$. Because $D$ is a diagonal matrix,Thus $CD=DC$ exactly when $(d_j-d_i)c_{ij}=0$ for every $i,j$. The three eigenvalues are distinct, so every off-diagonal entry of $C$ must be zero; diagonal entries are unrestricted. Consequently the centralizer of a diagonal matrix here isEvery displayed matrix indeed commutes with $D$, since diagonal matrices multiply entry by entry on the diagonal.

Matrix similarity transports commuting operators: with $M=PD{P}^{-1}$,By ii, all such matrices are thereforeFor completeness, matrix multiplication givesThe parameterization is injective, since matrix multiplication by $P^{-1}$ and $P$ recovers the parameter diagonal matrix. Thus there are infinitely many choices. Already the distinct scalar matrices $t{I}_3$, $t\in \mathbb{R}$, give an infinite family.

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.

Put $U=⟨u_1,\dots ,u_k⟩$. Taking all coefficients zero puts the zero vector in $U$. If $s={\sum }_i{\alpha }_i{u}_i$, $t={\sum }_i{\beta }_i{u}_i$ lie in $U$ and $a,b\in \mathbb{R}$, thenThus $U$ is closed under vector addition and scalar multiplication, contains the zero vector, and is therefore a linear subspace of $V$. This is precisely why the linear span is a linear subspace.

Suppose ${\sum }_{i=1}^{k+1}{\alpha }_i{u}_i=0$. If ${\alpha }_{k+1}\ne 0$, thencontrary to the hypothesis. Hence ${\alpha }_{k+1}=0$. The assumed linear independence of $u_1,\dots ,u_k$ then gives ${\alpha }_1=\cdots ={\alpha }_k=0$. Therefore $u_1,\dots ,u_{k+1}$ are linearly independent. Adding a vector outside the previous linear span creates no nontrivial relation.

Write $v={\sum }_{i=1}^{k+1}{\alpha }_i{u}_i$. Since $v\notin ⟨u_1,\dots ,u_k⟩$, necessarily ${\alpha }_{k+1}\ne 0$. Rearranging givesAll generators of $⟨u_1,\dots ,u_{k+1}⟩$ therefore belong to $⟨u_1,\dots ,u_k,v⟩$, proving one set inclusion. Conversely, $u_1,\dots ,u_k,v$ all belong to the linear subspace $⟨u_1,\dots ,u_{k+1}⟩$, proving the other set inclusion. ThusThis linear span exchange does not require linear independence of the original vectors.

For $\varphi (\alpha )={\sum }_{i=1}^k{\alpha }_i{u}_i$, its image of a linear map is exactly $⟨u_1,\dots ,u_k⟩$. Consequently $\varphi$ is surjective if and only if $u_1,\dots ,u_k$ are a spanning set for $V$.

If the vectors are linearly independent and $\varphi (\alpha )=\varphi (\beta )$, then ${\sum }_i({\alpha }_i-{\beta }_i)u_i=0$, so ${\alpha }_i={\beta }_i$ for all $i$. Hence $\varphi$ is injective. Conversely, if $\varphi$ is injective, any relation ${\sum }_i{\alpha }_i{u}_i=0$ says $\varphi (\alpha )=\varphi (0)$ and thus $\alpha =0$, proving linear independence.

Combining both equivalences givesA basis is therefore exactly a choice of vectors giving existence and uniqueness of coordinates, expressed by a bijection.

Take any finite selection of distinct indices $i_1,\dots ,i_r$ and a relationThe coordinate at $i_j$ on the left is exactly ${\alpha }_j$, while the zero vector has coordinate zero. Hence ${\alpha }_j=0$ for each $j$. Every finite subset of $E$ is therefore linearly independent, which is the definition of linear independence for an infinite set. Thus $E$ is linearly independent in the vector space of finitely supported sequences.

Every $v\in V$ is a finitely supported sequence, so for a finite set of indices $S$,Thus $E$ is a spanning set. If $F$ contained $v\notin E$, the displayed expression would give a nontrivial relationamong the distinct vectors of the finite subset $\{v\}\cup \{e^{(i)}:i\in S\}$ of $F$: the coefficient of $v$ is $1$. This contradicts the linear independence of $F$. It also covers $v=0$, whose singleton is not linearly independent. Hence $F=E$. In particular, $E$ is both a basis and a maximal linearly independent subset of $V$.

The hypothesis implies $v_1,v_2\ne 0$, since the zero vector is a scalar multiple of every vector. SetSince $v_1$ is not a scalar multiple of $v_2$, $w\ne 0$. Positive definiteness and bilinearity of the real inner product giveMultiplying by $\Vert v_2{\Vert }^2>0$ and taking square roots proves the strict Cauchy-Schwarz inequalitySince every real number $s$ satisfies $s\le |s|$, this proves, in particular,The argument derives the required inequality directly from the inner product axioms.

The two-vector Gram matrix isThe upper bound follows because the subtracted square is nonnegative. If either vector is zero, the determinant is zero. Otherwise apply the non-strict version of the calculation in i: $\Vert v_1-t{v}_2{\Vert }^2\ge 0$ with $t=⟨v_1,v_2⟩/\Vert v_2{\Vert }^2$ gives $⟨v_1,v_2{⟩}^2\le \Vert v_1{\Vert }^2\Vert v_2{\Vert }^2$. ThereforeEquality on the left holds exactly when the two vectors are linearly dependent; equality on the right holds exactly when they are orthogonal. The Gram determinant measures the squared area of their parallelogram: in the nonzero case, $\det A=\Vert v_2{\Vert }^2\Vert v_1-t{v}_2{\Vert }^2$, the square of base times perpendicular height.