Complementary subspaces with a characteristic-three exception
Let \(F\) be a field. Define \(D:F^5\to F^4\) and \(G:F^4\to F^5\) by
\[
D(a,b,c,d,e)^T=(a+b,\ b+c,\ c+d,\ d+e)^T
\]
and
\[
G(s,t,u,r)^T=(s,t,u,r,s-t+u-r)^T.
\]
(a) Write the matrices of \(D\) and \(G\) in the standard bases. [4 marks]
(b) Find a basis of \(\ker D\), and determine \(\operatorname{rk}D\). [4 marks]
(c) Prove that \(G\) is injective and determine \(\dim\operatorname{im}G\). [3 marks]
(d) Assume that \(3\) is invertible in \(F\). Prove that
\(\ker D\cap\operatorname{im}G=\{0\}\), and hence prove
\(F^5=\ker D\oplus\operatorname{im}G\). [6 marks]
(e) If \(3=0\) in \(F\), exhibit a non-zero element of
\(\ker D\cap\operatorname{im}G\). [3 marks]
Worked solution and marking guidance
**(a)** Reading coefficients gives
\[
[D]=\begin{pmatrix}1&1&0&0&0\\0&1&1&0&0\\0&0&1&1&0\\0&0&0&1&1\end{pmatrix},
\qquad
[G]=\begin{pmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\\1&-1&1&-1\end{pmatrix}.
\]
**(b)** The equations \(D(a,b,c,d,e)^T=0\) give
\(b=-a,c=a,d=-a,e=a\). Thus
\[
\ker D=\operatorname{span}\{k\},\qquad k=(1,-1,1,-1,1)^T.
\]
So \(\dim\ker D=1\), and rank-nullity gives \(\operatorname{rk}D=5-1=4\).
**(c)** If \(G(s,t,u,r)=0\), its first four coordinates give
\(s=t=u=r=0\). Hence \(G\) is injective and \(\dim\operatorname{im}G=4\).
**(d)** Suppose \(\lambda k=G(s,t,u,r)\). The first four coordinates force
\((s,t,u,r)=(\lambda,-\lambda,\lambda,-\lambda)\). The fifth coordinate
then gives \(\lambda=4\lambda\), so \(3\lambda=0\). Since \(3\) is
invertible, \(\lambda=0\). Hence the intersection is zero. Its two summands
have dimensions \(1\) and \(4\), so their direct sum has dimension \(5\)
inside \(F^5\), and therefore equals \(F^5\).
**(e)** In characteristic \(3\), \(4=1\). Taking
\((s,t,u,r)=(1,-1,1,-1)\) gives
\[
G(1,-1,1,-1)^T=(1,-1,1,-1,4)^T=k\ne0.
\]
Thus \(k\in\ker D\cap\operatorname{im}G\).
Polynomial-space involution and operator ranks (A)
Let \(V=F[x]_{\le 3}\), with ordered basis
\(\mathcal B=(1,x,\ldots,x^3)\). Define the involution by \((Jp)(x)=p(1-x)\), and let \(D\) be formal differentiation.
**Variant-specific requirement.** This is an affine reflection rather than reflection about the origin. You must first locate its fixed point and use the shifted coordinate y=x-1/2. The matrix in the printed monomial basis must still be obtained by expanding 1-x, (1-x)^2 and (1-x)^3; a diagonal matrix in the shifted basis alone does not answer part (a).
(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)** For \(\mathcal B=(1,x,x^2,x^3)\),
\[
[J]_{\mathcal B}=\begin{pmatrix}
1&1&1&1\\0&-1&-2&-3\\0&0&1&3\\0&0&0&-1
\end{pmatrix},
\]
because the columns are the coefficients of \(1,1-x,(1-x)^2,(1-x)^3\).
Substitution is linear and \(J^2p(x)=p(1-(1-x))=p(x)\).
**(b)** Put \(y=x-\tfrac12\). Then J sends y to -y, so
\[
V_+=\operatorname{span}\{1,y^2\},\qquad
V_-=\operatorname{span}\{y,y^3\}.
\]
The four displayed vectors are a basis and the two eigenspaces intersect
trivially because \(2\ne0\); hence \(V=V_+\oplus V_-\).
**(c)**
\[
[D]_{\mathcal B}=\begin{pmatrix}
0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0
\end{pmatrix}.
\]
For every p,
\[
(DJp)(x)=\frac d{dx}p(1-x)=-p'(1-x)=-(JDp)(x).
\]
Thus D sends \(V_+\) to \(V_-\) and \(V_-\) to \(V_+\).
**(d)** \(D(1)=0,D(y^2)=2y\), so
\(\operatorname{im}(D|_{V_+})=\operatorname{span}\{y\}\) and the rank is 1.
Also \(D(y)=1,D(y^3)=3y^2\). Hence
\(\operatorname{im}(D|_{V_-})=\operatorname{span}\{1,y^2\}\), rank 2, when
\(3\ne0\); in characteristic 3 the image is \(\operatorname{span}\{1\}\),
rank 1.
Spectral theorem and realification (A)
(a) Let
\[
A=\begin{pmatrix}3&1&1\\1&3&1\\1&1&3\end{pmatrix}\in M_3(\mathbb R).
\]
(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) For \(Z=X+iY\in M_n(\mathbb C)\), prove that the realification \(M=\begin{pmatrix}X&-Y\\Y&X\end{pmatrix}\) is orthogonal when Z is unitary, and prove \(MJ=JM\). 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 \((5,2,2)\). They may be obtained from the
characteristic polynomial or from the visible invariant directions.
**(a)(ii)** One valid ordered orthogonal eigenbasis gives
\[
Q=\begin{pmatrix}1/\sqrt3&1/\sqrt2&1/\sqrt6\\1/\sqrt3&-1/\sqrt2&1/\sqrt6\\1/\sqrt3&0&-2/\sqrt6\end{pmatrix},\qquad \Lambda=\operatorname{diag}(5,2,2).
\]
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 (A)
Let \(\sigma\in S_10\) be given in two-line notation by
\[
\sigma=\begin{pmatrix}
1&2&3&4&5&6&7&8&9&10\\
4&1&9&7&10&3&2&8&6&5
\end{pmatrix}.
\]
**Variant-specific requirement.** After finding the cycles, verify the decomposition by reconstructing the entire second row of the printed two-line notation. In the classification proof, explain explicitly why fixed points must be included as 1-cycles.
(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
\[
\sigma=(1 4 7 2)(3 9 6)(5 10).
\]
There are 1 fixed point(s), so the complete cycle type is
\([4, 3, 2, 1]\).
**(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 \([4, 3, 2, 1]\).
**Required variant check.** Reconstructing the second row from the cycles reproduces the printed mapping. Fixed points contribute 1-cycles, so omitting them would not give a partition of n.