Past exam of the linear algebra course of the University of Oxford / 2023 (A0, A15293W1)

Solutions to all three questions of the University of Oxford Part A linear algebra paper, sat on 20 June 2023. Source: released examination PDF, parent PDF pages 258–260; paper pages 1–3. Question labels follow the paper; the solutions supply their own reasoning. Part 2(c) includes the exceptional case of characteristic of a field three because that part does not specify its field separately.

The primary decomposition theorem for linear maps starts with a linear map $T:V\to V$ on a finite-dimensional vector space over a field $F$. Factor its minimal polynomial aswhere the $p_j$ are distinct monic polynomials that are irreducible polynomials and $a_j\ge 1$. Define $V_j=\ker p_j(T{)}^{a_j}$. ThenThus the direct sum separates the irreducible polynomial factors into uniquely determined invariant subspaces. The statement works even when the minimal polynomial does not split into linear polynomials over $F$.

All calculations here take place in the finite field $F={\mathbb{F}}_3$. For the given matrix $S$, its characteristic polynomial isThe values $q(0),q(1),q(2)$ are $2,2,1$, so this quadratic polynomial has no polynomial root in $F$ and is an irreducible polynomial. Direct calculation givesNeither $S$ nor $q(S)$ is zero. The Cayley-Hamilton theorem says that the minimal polynomial divides $xq(x)$; its proper monic polynomial divisors cannot annihilate $S$, so $m_S=xq$.

The two kernels give the required primary decomposition theorem for linear maps explicitly:The determinant of the matrix with these three basis vectors as columns is $1$ in $F$, proving $W=W_1\oplus W_2$. To check these are invariant subspaces, writing $u=(1,2,0{)}^T$ and $z=(0,0,1{)}^T$ gives $Su=u+2z$ and $Sz=2u$. Hence the matrix of the restriction to $W_2$ in the basis $(u,z)$ iswith minimal polynomial $q$; the nonzero vector space $W_1$ has zero restriction and minimal polynomial $x$.

The dual space is the vector space of all linear functionals on $V$:Addition and scalar multiplication are defined pointwise. The dual map sends an output measurement back along $T$:The function composition of these linear maps is linear, and $T^{\prime }$ is linear in $\phi$. This is the algebraic dual map, so its definition involves no complex conjugation or inner product.

For a vector subspace $U\subseteq V$, its annihilator of a vector subspace isLet $\dim U=k$ and $\dim V=n$. Extend a basis $u_1,\dots ,u_k$ of $U$ to a basis $u_1,\dots ,u_n$ of $V$, and let ${\epsilon }^1,\dots ,{\epsilon }^n$ be the dual basis. Every linear functional has the unique expression $\phi ={\sum }_{j=1}^n\phi (u_j){\epsilon }^j$. It belongs to $U^0$ precisely when its first $k$ coefficients vanish. ConsequentlyThe listed dual basis elements are linearly independent, so counting them proves the formula, including $U=0$ and $U=V$.

Suppose $U$ is an invariant subspace for $T$. For $\phi \in U^0$ and $u\in U$, the dual map satisfiesbecause $Tu\in U$ and the annihilator of a vector subspace consists of the linear functionals vanishing there. ThusThis is the correspondence between invariant subspaces and annihilators.

We prove by mathematical induction on $n=\dim V$ that every linear map on an $n$-dimensional complex vector space admits both triangular forms. For $n=0$ or $1$ this is immediate. When $n>1$, the fundamental theorem of algebra supplies an eigenvector $v$; its span $U=⟨v⟩$ is an invariant subspace.

By parts (ii) and (iii), $U^0$ has dimension $n-1$ and is an invariant subspace of the dual map $T^{\prime }$. Apply the induction hypothesis to $T^{\prime }{|}_{U^0}$ to choose a basis ${\phi }_1,\dots ,{\phi }_{n-1}$ with an upper triangular matrix. Extend it by a linear functional ${\phi }_n$ satisfying ${\phi }_n(v)=1$. Such a functional exists by basis extension, and cannot belong to $U^0$. In this basis of $V^{\prime }$, the matrix of $T^{\prime }$ has the formwhere $B$ is an upper triangular matrix; here the last diagonal entry follows by evaluating $T^{\prime }{\phi }_n$ at $v$, where $Tv=\lambda v$.

There is a basis $e_1,\dots ,e_n$ of $V$ with ${\phi }_i(e_j)={\delta }_{ij}$, since the dual basis construction gives a finite-dimensional bidual evaluation map isomorphism. In particular $e_n=v$. The matrix transpose relationship for a dual map givesa lower triangular matrix. Reversing this basis turns a lower triangular matrix into an upper triangular matrix. If $R$ is the permutation matrix reversing the order, we have provedThis completes the induction by triangularization using dual spaces.

Write $K_i=\ker T^i$, with $K_0=\{0\}$. The minimal polynomial $x^m$ means $T^m=0$ and $T^{m-1}\ne 0$. Choose $w$ with $T^{m-1}w\ne 0$. For each $0\le i<m$, set $w_i=T^{m-1-i}w$. ThenThus $w_i\in K_{i+1}\setminus K_i$. Also $K_i\subseteq K_{i+1}$ because applying $T$ to zero still gives zero. Therefore the kernel dimensions of powers of a nilpotent map increase strictly until the nilpotency index:The same reasoning covers $m=1$, when the nilpotent linear map is zero on a nonzero vector space.

Put $d_i=\dim K_i$, including $d_0=0$. The restriction of $T^i$ to $K_{i+1}$ has kernel $K_i$, so the rank-nullity theorem givesLikewise $T^{i-1}{|}_{K_i}$ has kernel $K_{i-1}$ and hence image dimension $d_i-d_{i-1}$. But $T(K_{i+1})\subseteq K_i$, which impliesComparing dimensions yields the required concavity of the kernel dimensions of powers of a nilpotent map:

The increments ${\Delta }_i=d_i-d_{i-1}$ are positive integers by part (a)(i), are nonincreasing by part (a)(ii), and sum to $d_m=4$. Thus they form an integer partition of $4$. The complete list is $(4)$, $(3,1)$, $(2,2)$, $(2,1,1)$, and $(1,1,1,1)$.

To prove all five occur over every field, let $J_s(0)$ be a Nilpotent Jordan block of size $s$. It sends each standard basis vector to its predecessor, with the first sent to zero, soFor a block diagonal matrix, the nullities add. The following explicit nilpotent matrices therefore realize every possible sequence:In each row the largest Jordan block size is exactly the nilpotency index $m$, so the displayed sequence stops at the correct place.

Direct matrix multiplication gives $A^2=0$. Denote the columns by $c_1,\dots ,c_4$; they satisfyThus the matrix rank is at most two. Over the real numbers or complex numbers, the upper-left two-by-two minor is $-3\ne 0$, so the matrix rank is exactly two. Every Jordan block has size at most two because $A^2=0$, and each size-two Nilpotent Jordan block contributes one to the matrix rank. ConsequentlyEquivalently the nullities are $(2,4)$ and the minimal polynomial is $x^2$.

There is a field qualification in the source: Question 2 initially allows an arbitrary field, and part (c) does not specify a new one. The preceding answer holds over every field whose characteristic is not three. In characteristic three, the first row vanishes and the other three rows all become $(2,1,0,2)$, giving matrix rank one while $A^2=0$. Under that reading the answer is instead

Throughout Question 3, the inner product is linear in its first argument and conjugate-linear in its second. Over the real numbers, conjugation has no effect. The symbol $T^{\ast }$ now denotes the adjoint operator for this inner product.

An adjoint operator of $T:V\to V$ is a linear map $T^{\ast }:V\to V$ satisfyingThis identity transfers $T$ from the first input of the inner product to the second.

If $R$ and $S$ both satisfy the defining identity of an adjoint operator, thenfor every $u,w\in V$. Choosing $u=(R-S)w$ gives $\Vert (R-S)w{\Vert }^2=0$. The positive definiteness of the inner product forces $(R-S)w=0$ for every $w$, proving

For the normal operator $T$, the adjoint operator identity and $T^{\ast }T=T{T}^{\ast }$ giveFor the last equality, $⟨T{T}^{\ast }v,v⟩=⟨T^{\ast }v,T^{\ast }v⟩$ and complex conjugation allow reversal of the two arguments. Taking the nonnegative square roots provesIn particular the two kernels coincide.

Let $w=Tv$. If $T^{2}v=0$, then $Tw=0$, and part (i) gives $\Vert T^{\ast }w\Vert =\Vert Tw\Vert =0$. Thus $T^{\ast }w=0$ andThe positive definiteness of the inner product gives $w=0$. Since the reverse kernel inclusion always holds, we obtainThis is kernel stabilization for a normal operator.

For a polynomial $f(x)={\sum }_{j=0}^d{a}_j{x}^j$, the defining identity and uniqueness of the adjoint operator giveIndeed, the adjoint operator of a sum is the sum of the adjoint operators, the adjoint of $aT$ is $\bar{a}{T}^{\ast }$, and the adjoint of a product $AB$ is $B^{\ast }{A}^{\ast }$. Since the normal operator $T$ commutes with $T^{\ast }$, each power of $T$ commutes with each power of $T^{\ast }$. Multiplying the two finite sums therefore givesHence polynomials of a normal operator are normal, including constant polynomials.

Suppose some exponent $q_j\ge 2$. DefinePart (iii) makes $N$ a normal operator. For every $v\in V$, put $w=h(T)v$. The minimal polynomial identity givesPart (ii), applied to $N$, forces $Nw=0$. As this holds for every $v$,This contradicts minimality: $m_T/f_j$ is a monic polynomial of smaller degree annihilating $T$. Therefore the minimal polynomial of a normal operator is a square-free polynomial over either of the given fields:This argument does not require the irreducible polynomials to be linear, so it also covers the real case.

Take $N=T-\lambda I$. It is a polynomial in the normal operator $T$, so part (b)(iii) makes $N$ a normal operator. Also $N^{\ast }=T^{\ast }-\bar{\lambda }I$. If $Tv=\lambda v$, part (b)(i), applied to $N$, givesThe norm vanishes only at the zero vector, soThis is the adjoint eigenvector identity for a normal matrix expressed independently of coordinates.

Let $u\in \ker (T-\lambda I)$ and $w\in \ker (T-\mu I)$, with $\lambda \ne \mu$. By part (i) and the adjoint operator identity,The last step uses conjugate-linearity in the second argument of the inner product. Therefore $(\lambda -\mu )⟨u,w⟩=0$, proving the orthogonality of the two eigenspaces:

Over the complex numbers, the fundamental theorem of algebra makes every irreducible polynomial factor $f_j(x)$ equal to $x-{\lambda }_j$. Part (b)(iv) giveswith distinct eigenvalues. The primary decomposition theorem for linear maps consequently expresses $V$ as the direct sum of the actual eigenspaces $E_{{\lambda }_j}=\ker (T-{\lambda }_{j}I)$, with no higher generalized eigenspaces needed. Part (ii) makes these summands mutually orthogonal.

Apply the Gram-Schmidt process to a basis within each eigenspace. Its linear combinations stay within that eigenspace, so all resulting vectors are still eigenvectors. Their union is orthonormal by part (ii) and spans $V$ by the primary decomposition theorem for linear maps. Thusfor an orthonormal basis $\mathcal{B}$ of eigenvectors. This proves the finite-dimensional complex spectral theorem for normal operators by the requested decomposition argument.