Past exam of the linear algebra course of the University of Oxford / 2021 (A0, A15293W1)

Solutions to all three questions of the University of Oxford Part A linear algebra paper, sat on 15 June 2021. Source: released examination PDF, parent PDF pages 253–254; paper pages 1–2. All questions and parts are covered, including both alternatives in questions asking whether an assertion is true.

The quotient vector space $V/U$ identifies two vectors $v,w$ precisely when $v-w\in U$. Since $U$ is an invariant subspace, $T(U)\subseteq U$. ThusConsequently $T(v)+U=T(w)+U$: the output does not depend on the chosen representative. This proves that the induced linear map on a quotient vector spaceis well-defined. It is a linear map because, for scalars $a,b$,

No. Over any field $F$, take $V=F^2$, let $T=I$ be the identity operator, and choose the vector subspaceBoth $U$ and the quotient vector space $V/U$ have dimension one. A one-dimensional vector space has no nonzero proper vector subspace, so neither can have a nonzero proper invariant subspace. The induced linear map on a quotient vector space is again the identity operator.

However, $L=\mathrm{span}\{e_2\}$ is another nonzero proper vector subspace of $V$, and $T(L)=L$. HenceThe hypotheses do not prevent a direct sum $V=U\oplus L$ whose two summands are invariant subspaces.

Write $K_j=\ker T^j$, where $T^0=I$ and $K_0=\{0\}$. These kernels satisfy $K_j\subseteq K_{j+1}$. Also $T(K_{i+1})\subseteq K_i$ and $T(K_i)\subseteq K_{i-1}$: for example, $T^i(Tv)=T^{i+1}v=0$ when $v\in K_{i+1}$. Thus the proposed linear map between quotient vector spaces is well-defined and has the formulaIf $v+K_i$ belongs to its kernel, then $Tv\in K_{i-1}$. Therefore $T^{i}v=T^{i-1}(Tv)=0$, so $v\in K_i$ and $v+K_i$ is zero. We have provedThese are the injections between successive kernel quotients. In finite dimension, they imply $\dim K_{i+1}-\dim K_i\le \dim K_i-\dim K_{i-1}$.

Surjectivity need not hold. Take a nonzero vector space $V$ and the zero linear map $T=0$. At $i=1$, we have $K_2=K_1=V$ and $K_0=0$, so

Subtract the identity matrix and putDirect matrix multiplication givesFor the last equality, each column of $N^2$ is a multiple of $(1,-1,0{)}^T$, and $N(1,-1,0{)}^T=0$. Thus $N$ is a nilpotent matrix. The minimal polynomial of $A_{\lambda ,\mu }$ divides $(t-1{)}^3$ and equals $(t-1{)}^s$, where $s$ is the nilpotency index of $N$.

The first row of $N$ forces $\lambda =0$ if $N=0$, and its third row then forces $\mu =0$. Conversely $\lambda =\mu =0$ gives $N=0$. If exactly one parameter is nonzero, then $N\ne 0$ but $N^2=0$. If both are nonzero, then $N^2\ne 0$ while $N^3=0$. Consequently the complete answer isNo lower power can annihilate the matrix in any of these cases, as the displayed nonvanishing matrix powers show.

Let $F$ be the common field. The dual map is $S^{\prime }(f)=f\circ S$ for $f\in W^{\prime }$, where $W^{\prime }$ and $V^{\prime }$ are the dual spaces.

First assume $S^{\prime }$ is injective. If $S$ were not surjective, its image of a linear map $\mathrm{im}S$ would be a proper vector subspace of $W$. Choose a basis $w_1,\dots ,w_r$ of $\mathrm{im}S$ and use basis extension to obtain a basis $w_1,\dots ,w_m$ of $W$, with $r<m$. There is a linear functional $f\in W^{\prime }$ withIt is nonzero and vanishes on $\mathrm{im}S$, so $S^{\prime }(f)=f\circ S=0$, contradicting injectivity. Equivalently, the annihilator of a vector subspace $\mathrm{im}S$ would give a nonzero kernel for $S^{\prime }$. Therefore

Now assume $S^{\prime }$ is surjective. If $0\ne v\in \ker S$, basis extension gives a linear functional $h\in V^{\prime }$ with $h(v)=1$: include $v$ in a basis and take its coordinate functional. By surjectivity, $h=S^{\prime }(f)$ for some $f\in W^{\prime }$. Buta contradiction. Hence $\ker S=0$, which proves

In the monomial basis $\mathcal{B}=(1,x,x^2,x^3)$ of the complex vector space $V=\mathbb{C}[x{]}_{\le 3}$, the binomial theorem givesThe characteristic polynomial of this upper triangular matrix is the product of $t$ minus its diagonal entries, so ${\chi }_T(t)=(t-1{)}^4$.

For the minimal polynomial, set $\Delta =T-I$, the forward difference with step one. For a nonconstant polynomial of polynomial degree $d$, the binomial theorem shows that $f(x+1)-f(x)$ has polynomial degree $d-1$. Constant polynomials have zero forward difference. Consequently ${\Delta }^4=0$ on $V$. To prove that the nilpotency index is exactly four, computeIt follows thatIf $T$ were diagonalizable, its only eigenvalue would be $1$, and its diagonal matrix in an eigenbasis would be $I$. This would make $T=I$, whereas $T(x)=x+1\ne x$. Thus $T$ is not diagonalizable. Equivalently, its minimal polynomial has a repeated root.

By Jordan normal form, each complex nilpotent matrix of size six can be put by matrix similarity into a block diagonal matrix whose Nilpotent Jordan blocks have sizesA Nilpotent Jordan block of size $s$ has matrix rank $s-1$ and minimal polynomial $t^s$. Therefore, for the whole matrix,Thus the given matrix rank determines the number $k$ of Jordan blocks, and the given minimal polynomial determines the largest size $s_1$. For size six these two data determine the entire integer partition. Here is the complete list, arranged by the largest size:To see that the list is exhaustive, choose the largest part $s_1$ between one and six and partition the remainder $6-s_1$ into parts at most $s_1$. The resulting rows all have distinct pairs $(s_1,k)$. Hence $A$ and $B$ have the same Jordan normal form $J$.

Explicitly, choose invertible matrices $X,Y$ such that $A=XJ{X}^{-1}$ and $B=YJ{Y}^{-1}$. ThenThis proves matrix similarity and illustrates the classification of six-dimensional nilpotent matrices by rank and nilpotency index.

We construct a cube root of a matrix on each Jordan block. Since $A$ is an invertible matrix over the complex numbers, its Jordan normal form isConsider one block $J_m(\lambda )=\lambda I+N$, where $N=J_m(0)$ is a Nilpotent Jordan block and $N^m=0$. Choose a complex cube root $\beta$ of $\lambda$, so ${\beta }^3=\lambda$.

We need a polynomial $q(z)$ with constant term one satisfyingThis can be constructed using only polynomial multiplication. Write $q(z)=1+{\sum }_{j=1}^{m-1}{c}_j{z}^j$. Once $c_1,\dots ,c_{j-1}$ have been chosen, the coefficient of $z^j$ in $q(z{)}^3$ is $3c_j$ plus an expression depending only on those earlier coefficients. Choose $c_j$ to make this coefficient one at $j=1$ and zero at $2\le j<m$. Division by three is possible over the complex numbers, so the recursion always succeeds. Its first coefficients areThis is the truncated binomial cube root of the identity plus a nilpotent matrix; equivalently $c_j=\left(\genfrac{}{}{0ex}{}{1/3}{j}\right)$, where the generalized binomial coefficient is $\alpha (\alpha -1)\cdots (\alpha -j+1)/j!$ and the empty product is one.

Substitute $z=N/\lambda$. The polynomial remainder is a multiple of $N^m$, hence zero. It follows thatFor $m=1$, take $q=1$, so the same construction covers single-entry blocks. Assemble the block diagonal matrix $C={⨁}_r{C}_{m_r}({\lambda }_r)$. Taking matrix powers respects these blocks, and conjugating back givesThus every invertible matrix over the complex numbers has at least one cube root of a matrix. The construction uses only finitely many matrix powers.

Take the Nilpotent Jordan blockThenSuppose that a complex matrix $B$ of size three satisfied $B^3=A$. We would have $B^9=A^3=0$, so $B$ would be a nilpotent matrix. Every eigenvalue of $B$ must then be zero: if $Bv=\theta v$ with $v\ne 0$, applying $B^9$ gives ${\theta }^{9}v=0$. Its characteristic polynomial is therefore $t^3$, and the Cayley-Hamilton theorem gives $B^3=0$. This contradicts $B^3=A\ne 0$. HenceThe obstruction is that a nilpotent matrix of size three cannot have nilpotency index greater than three.

Put $b(v,w)=\phi (w)(v)$. Each $\phi (w)$ belongs to the dual space $V^{\prime }$, so it is a linear functional. For real numbers $a,b_0$ and vectors $v_1,v_2,w$, this givesThe linearity of $\phi$ gives linearity in the second argument as well:Thus $b$ is a bilinear form.

Its right radical of a bilinear form is exactly $\ker \phi$, becauseIf $b$ is a nondegenerate bilinear form, this kernel is zero. Since $V$ is a finite-dimensional vector space and $\dim V^{\prime }=\dim V$, the rank-nullity theorem makes $\phi$ surjective.

Conversely, if $\phi$ is surjective, the same dimension equality makes it injective, so the right radical is zero. If $v$ lies in the left radical, then $f(v)=0$ for every $f\in V^{\prime }$, because every $f$ is some $\phi (w)$. A nonzero $v$ can be included in a basis and detected by its coordinate functional, so this forces $v=0$. Both radicals vanish, proving

For the given nondegenerate bilinear form, define the linear mapBy part (a), $\Phi$ is a linear isomorphism. The paper's bilinear orthogonal complement uses $U$ in the first argument; with this convention,Here $U^0$ is the annihilator of a vector subspace $U$. Since $\Phi$ is a linear isomorphism, its restriction gives a linear isomorphism $U^{\perp }\to U^0$.

For completeness, choose a basis $u_1,\dots ,u_k$ of $U$ and extend it to a basis $u_1,\dots ,u_n$ of $V$. In its dual basis $u_1^{\ast },\dots ,u_n^{\ast }$, a linear functional vanishes on $U$ exactly when its first $k$ coefficients are zero. ThereforeConsequentlyThe argument includes $U=0$ and $U=V$, and requires no symmetry assumption on the bilinear form.

No. On the real vector space $V={\mathbb{R}}^2$, use the symmetric bilinear formIts matrix of a bilinear form is $\mathrm{diag}(1,-1)$, whose determinant is $-1\ne 0$, so it is a nondegenerate bilinear form. LetA vector $(s,t)$ belongs to the bilinear orthogonal complement $U^{\perp }$ exactly when $s-t=0$. HenceIn particular, the claimed direct sum fails. The isotropic vector $(1,1)$ has zero self-pairing, although it pairs nontrivially with other vectors of $V$.

The dimension identity in part (i) counts both vector subspaces but does not force their intersection of vector subspaces to be zero. For a symmetric bilinear form, $U\cap U^{\perp }$ is the radical of a bilinear form restricted to $U$. Thus the precise criterion isIndeed, trivial intersection of vector subspaces and part (i) give a sum of vector subspaces of dimension $\dim V$. A positive-definite bilinear form would guarantee this for every $U$; symmetry alone does not.

Writeon $V=W\times W^{\prime }$, where $W^{\prime }$ is the dual space. The dual pairing is linear separately in the vector and the linear functional. For example, for real numbers $a,b$,Thus $\omega$ is a bilinear form. It is also an alternating bilinear form, since

To prove that this is a nondegenerate bilinear form, suppose $(v,f)$ pairs to zero with every $(w,g)$. Setting $g=0$ gives $f(w)=0$ for every $w\in W$, hence $f=0$. Setting $w=0$ gives $g(v)=0$ for every $g\in W^{\prime }$. If $v\ne 0$, basis extension supplies a coordinate functional with $g(v)=1$, a contradiction. Thus $v=0$. The left radical is zero, and the displayed change of sign implies the right radical is zero as well. Hence $\omega$ is a nondegenerate bilinear form, making $V$ a symplectic vector space.

For the required example take the coordinate subspaceUsing the definition of the bilinear orthogonal complement in part (b),ThereforeThis is a Lagrangian subspace. Its dimension is $\dim W$, exactly half of $\dim V=2\dim W$, so part (b)(i) is satisfied.

There is no contradiction with part (b)(ii): that part asks whether a direct sum is guaranteed, and the answer is already no. Moreover, when $W\ne 0$, the present alternating bilinear form is not a symmetric bilinear form. Choose $v\ne 0$ and $f\in W^{\prime }$ with $f(v)=1$; thenIt therefore does not satisfy the symmetry hypothesis of part (b)(ii). If $W=0$, then $V=U=0$, and both $U=U^{\perp }$ and $V=U\oplus U^{\perp }$ hold trivially.