Difference map, cumulative right inverse and canonical decomposition
Define \(\Delta:F^6\to F^5\) by
\[
\Delta(x_1,\ldots,x_6)^T=(x_2-x_1,x_3-x_2,x_4-x_3,x_5-x_4,x_6-x_5)^T.
\]
Define \(H:F^5\to F^6\) by
\[
H(y_1,\ldots,y_5)^T=(0,y_1,y_1+y_2,\ldots,y_1+\cdots+y_5)^T.
\]
(a) Write the matrices of \(\Delta\) and \(H\). [4 marks]
(b) Find \(\ker\Delta\) and determine \(\operatorname{rk}\Delta\). [4 marks]
(c) Prove \(\Delta H=I_{F^5}\). Deduce that \(H\) is injective and that
\(\operatorname{im}H\cap\ker\Delta=\{0\}\). [5 marks]
(d) Prove \(F^6=\ker\Delta\oplus\operatorname{im}H\), and give the
decomposition of an arbitrary \(x\in F^6\) explicitly. [7 marks]
Worked solution and marking guidance
**(a)** The matrix of \(\Delta\) has \(-1,1\) in consecutive columns:
\[
[\Delta]=\begin{pmatrix}
-1&1&0&0&0&0\\0&-1&1&0&0&0\\0&0&-1&1&0&0\\
0&0&0&-1&1&0\\0&0&0&0&-1&1
\end{pmatrix}.
\]
The \(j\)-th column of \(H\) has zeros through coordinate \(j\) and ones
afterwards; equivalently
\[
[H]=\begin{pmatrix}
0&0&0&0&0\\1&0&0&0&0\\1&1&0&0&0\\
1&1&1&0&0\\1&1&1&1&0\\1&1&1&1&1
\end{pmatrix}.
\]
**(b)** \(\Delta x=0\) exactly when
\(x_1=\cdots=x_6\). Thus
\(\ker\Delta=\operatorname{span}\{(1,1,1,1,1,1)^T\}\).
Rank-nullity gives \(\operatorname{rk}\Delta=5\).
**(c)** Taking consecutive differences of the cumulative sums in \(H(y)\)
returns \(y_1,\ldots,y_5\), so \(\Delta H=I\). Therefore \(H(y)=0\)
implies \(y=\Delta H(y)=0\). If \(H(y)\in\ker\Delta\), then
\(0=\Delta H(y)=y\), so the intersection is zero.
**(d)** For \(x=(x_1,\ldots,x_6)^T\), put \(y=\Delta x\). Telescoping gives
\[
H(\Delta x)=(0,x_2-x_1,\ldots,x_6-x_1)^T.
\]
Hence
\[
x=x_1(1,1,1,1,1,1)^T+H(\Delta x),
\]
with the first term in the kernel and the second in the image. The zero
intersection from (c) makes this decomposition unique, proving the direct sum.
Polynomial-space involution and operator ranks (D)
Let \(V=F[x]_{\le 4}\), with ordered basis
\(\mathcal B=(1,x,\ldots,x^4)\). Define the involution by \((Jp)(x)=p(2-x)\), and let \(D\) be formal differentiation.
**Variant-specific requirement.** The reflection is about x=1, not x=0. Introduce y=x-1 and prove that the affine substitution becomes y↦-y. Translate the resulting bases back into polynomials in x when stating the final answer, and distinguish the characteristic-three loss in D(y^3) from the standing assumption that 2 is invertible.
(a) Prove that the displayed involution is linear, show that its square is
the identity, and write its matrix in \(\mathcal B\). [4 marks]
(b) Assume \(2\ne0\) in \(F\). Find bases of its \(+1\) and \(-1\)
eigenspaces \(V_+\) and \(V_-\), and prove \(V=V_+\oplus V_-\). [6 marks]
(c) Write the matrix of the differential operator used in this question and
prove \(DJ=-JD\). Deduce the corresponding mapping relation between \(V_+\) and
\(V_-\). [5 marks]
(d) Determine the ranks of the restrictions to \(V_+\) and \(V_-\),
including every exceptional field characteristic caused by a coefficient
appearing in the derivative. [5 marks]
Worked solution and marking guidance
**(a)** The columns are the expansions of \(1,2-x,(2-x)^2,(2-x)^3,(2-x)^4\):
\[
[J]_{\mathcal B}=\begin{pmatrix}
1&2&4&8&16\\0&-1&-4&-12&-32\\0&0&1&6&24\\
0&0&0&-1&-8\\0&0&0&0&1
\end{pmatrix}.
\]
Substitution is linear and \(J^2p(x)=p(2-(2-x))=p(x)\).
**(b)** Put \(y=x-1\). Then
\[
\begin{aligned}
V_+&=\operatorname{span}\{1,y^2,y^4\}
=\operatorname{span}\{1,x^2-2x+1,x^4-4x^3+6x^2-4x+1\},\\
V_-&=\operatorname{span}\{y,y^3\}
=\operatorname{span}\{x-1,x^3-3x^2+3x-1\}.
\end{aligned}
\]
These form a direct-sum basis when \(2\ne0\).
**(c)**
\[
[D]_{\mathcal B}=\begin{pmatrix}
0&1&0&0&0\\0&0&2&0&0\\0&0&0&3&0\\0&0&0&0&4\\0&0&0&0&0
\end{pmatrix}.
\]
\((DJp)(x)=\frac d{dx}p(2-x)=-p'(2-x)=-(JDp)(x)\), so D interchanges
\(V_+\) and \(V_-\).
**(d)** The even-basis images are \(0,2y,4y^3\), giving rank 2.
The odd-basis images are \(1,3y^2\), giving rank 2 when \(3\ne0\) and
image \(\operatorname{span}\{1\}\), rank 1, in characteristic 3.
Spectral theorem and realification (D)
(a) Let
\[
A=\begin{pmatrix}2&1&0\\1&2&1\\0&1&2\end{pmatrix}\in M_3(\mathbb R).
\]
(i) Find all eigenvalues, with algebraic multiplicities. [3 marks]
(ii) Construct an orthogonal matrix \(Q\) and diagonal \(\Lambda\) such that
\(A=Q\Lambda Q^T\). The columns of Q must be stated explicitly. [4 marks]
(iii) Construct a real matrix B satisfying \(A=BB^T\), and justify why the
construction is valid. [3 marks]
(b) Starting from the two real identities equivalent to \(Z^*Z=I\), prove orthogonality of the realification. Then reverse every implication under the additional commutation condition with J. Give all block identities explicitly and prove both the
stated direction and any requested converse. [10 marks]
Worked solution and marking guidance
**(a)(i)** The eigenvalues are \((2+\sqrt2,2,2-\sqrt2)\). They may be obtained from the
characteristic polynomial or from the visible invariant directions.
**(a)(ii)** One valid ordered orthogonal eigenbasis gives
\[
Q=\begin{pmatrix}1/2&1/\sqrt2&1/2\\1/\sqrt2&0&-1/\sqrt2\\1/2&-1/\sqrt2&1/2\end{pmatrix},\qquad \Lambda=\operatorname{diag}(2+\sqrt2,2,2-\sqrt2).
\]
Direct multiplication verifies \(Q^TQ=I\) and \(AQ=Q\Lambda\), so
\(A=Q\Lambda Q^T\).
**(a)(iii)** All diagonal entries of \(\Lambda\) are non-negative. Hence
\[
B=Q\Lambda^{1/2}
\]
satisfies \(BB^T=Q\Lambda^{1/2}\Lambda^{1/2}Q^T=A\).
**(b)** Write \(Z=X+iY\). The unitary identity is equivalent to
\[
X^TX+Y^TY=I,\qquad X^TY=Y^TX.
\]
For \(M=\begin{pmatrix}X&-Y\\Y&X\end{pmatrix}\), block multiplication
therefore gives \(M^TM=I\). With
\(J=\begin{pmatrix}0&I\\-I&0\end{pmatrix}\), direct block multiplication
gives \(MJ=JM\) (with the equivalent sign convention both sides change
consistently). Conversely, writing \(Q=\begin{pmatrix}A&B\\C&D\end{pmatrix}\)
and comparing \(QJ\) with \(JQ\) gives \(D=A\) and \(C=-B\), up to the same
fixed sign convention. Thus Q is the realification of \(A-iB\).
Its orthogonality equations are exactly the real and imaginary parts of the
unitary equation. Realification also respects multiplication because the
block product reproduces complex multiplication, and consequently respects
inverses. For \(n=1\), \(e^{i\theta}\) becomes the usual rotation matrix.
Permutations, sign and conjugacy (D)
Let \(\sigma\in S_9\) be given in two-line notation by
\[
\sigma=\begin{pmatrix}
1&2&3&4&5&6&7&8&9\\
4&1&5&2&6&7&3&9&8
\end{pmatrix}.
\]
**Variant-specific requirement.** Deduce that sign is constant on conjugacy classes, and then give a specific pair in a symmetric group which have the same sign but different cycle types. This must be used to explain why sign alone cannot classify conjugacy.
(a) (i) Write \(\sigma\) as a product of disjoint cycles, including the
fixed points when recording its cycle type. [4 marks]
(ii) Write it as a product of transpositions and compute
\(\operatorname{sgn}(\sigma)\). [4 marks]
(b) Let G be a group. Prove that conjugacy is an equivalence relation, naming
the conjugating element in the reflexive, symmetric and transitive steps.
Also prove \(gx^mg^{-1}=(gxg^{-1})^m\) for every \(m\ge1\). [6 marks]
(c) Prove that two permutations of \(S_n\) are conjugate if and only if they
have the same cycle type. Your proof must construct a conjugating permutation
in the reverse direction. Apply the result to state the cycle type of the
conjugacy class containing \(\sigma\). [6 marks]
Worked solution and marking guidance
**(a)(i)** Following each orbit gives
\[
\sigma=(1 4 2)(3 5 6 7)(8 9).
\]
There are 0 fixed point(s), so the complete cycle type is
\([4, 3, 2]\).
**(a)(ii)** Use
\((a_1\ldots a_k)=(a_1a_k)\cdots(a_1a_2)\) on every non-trivial cycle.
This uses \(6\) transpositions in total; hence
\(\operatorname{sgn}(\sigma)=+1\). Any equivalent transposition
decomposition earns full credit.
**(b)** Reflexivity uses \(e x e^{-1}=x\). If \(gxg^{-1}=y\), then
\(g^{-1}yg=x\), proving symmetry. If \(gxg^{-1}=y\) and
\(hyh^{-1}=z\), then \((hg)x(hg)^{-1}=z\), proving transitivity.
For powers, expand the product: adjacent factors \(g^{-1}g\) cancel,
or use induction, giving \(gx^mg^{-1}=(gxg^{-1})^m\).
**(c)** Conjugating a cycle replaces every entry a by g(a), so cycle
lengths, including fixed points, are preserved. Conversely, if σ and τ have
the same cycle type, pair cycles of equal length and define g to send the
j-th entry of each σ-cycle to the j-th entry of the paired τ-cycle.
This defines a bijection and direct evaluation gives \(g\sigma g^{-1}=\tau\).
Thus the class containing the printed σ is exactly the set of permutations
with cycle type \([4, 3, 2]\).
**Required variant check.** Since sign is multiplicative, sgn(gσg^{-1})=sgn(g)sgn(σ)sgn(g)^{-1}=sgn(σ). However the identity and a 3-cycle are both even and have different cycle types, so they are not conjugate.