Reverse construction of a projection from a complement
Let \(T:F^5\to F^3\) be
\[
T(a,b,c,d,e)^T=(a+b+c,\ b+c+d,\ c+d+e)^T
\]
and let \(W=\{(u,v,w,0,0)^T:u,v,w\in F\}\).
(a) Write the matrix of \(T\), find a basis of \(\ker T\), and determine its
dimension. [7 marks]
(b) Prove directly that the restriction \(T|_W:W\to F^3\) is an
isomorphism. [4 marks]
(c) Deduce that \(F^5=\ker T\oplus W\). [3 marks]
(d) For \(x=(a,b,c,d,e)^T\), find explicitly the projection \(P(x)\) onto
\(\ker T\) along \(W\), and write the matrix of \(P\). [6 marks]
Worked solution and marking guidance
**(a)** The matrix is
\[
[T]=\begin{pmatrix}1&1&1&0&0\\0&1&1&1&0\\0&0&1&1&1\end{pmatrix}.
\]
Solving \(Tx=0\) gives
\(c=-d-e,\ b=e,\ a=d\). Therefore
\[
\ker T=\{(d,e,-d-e,d,e)^T:d,e\in F\}
=\operatorname{span}\{(1,0,-1,1,0)^T,(0,1,-1,0,1)^T\},
\]
so \(\dim\ker T=2\).
**(b)** On \(W\),
\[
T(u,v,w,0,0)^T=(u+v+w,v+w,w)^T.
\]
Its matrix is upper triangular with diagonal entries \(1,1,1\), hence
invertible. Explicitly, for \(y=(y_1,y_2,y_3)\),
\[
w=y_3,\quad v=y_2-y_3,\quad u=y_1-y_2.
\]
Thus \(T|_W\) is bijective.
**(c)** If \(z\in\ker T\cap W\), then \(T|_W(z)=0\), so \(z=0\). Moreover
\(\dim\ker T+\dim W=2+3=5\), hence \(F^5=\ker T\oplus W\).
**(d)** Choose \(w\in W\) with \(T(w)=T(x)\). Back-substitution gives
\[
w=(a-d,\ b-e,\ c+d+e,\ 0,\ 0)^T.
\]
Therefore
\[
P(x)=x-w=(d,e,-d-e,d,e)^T,
\quad
[P]=\begin{pmatrix}
0&0&0&1&0\\0&0&0&0&1\\0&0&0&-1&-1\\
0&0&0&1&0\\0&0&0&0&1
\end{pmatrix}.
\]
One may check \(P^2=P\), \(\operatorname{im}P=\ker T\), and
\(\ker P=W\).
Polynomial-space involution and operator ranks (B)
Let \(V=F[x]_{\le 4}\), with ordered basis
\(\mathcal B=(1,x,\ldots,x^4)\). Define \(R\) by coefficient reversal and define \(L=xD-2I\), where \(D\) is formal differentiation.
**Variant-specific requirement.** Here the involution reverses coefficients; it is not substitution x↦-x. Establish the palindromic and anti-palindromic coefficient conditions. The operator L=xD-2I has been chosen so that its diagonal weights reverse sign under coefficient reversal, and that observation should replace any attempted chain-rule proof.
(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 \(LR=-RL\). 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)** In \(\mathcal B=(1,x,x^2,x^3,x^4)\), coefficient reversal has
\[
[R]_{\mathcal B}=\begin{pmatrix}
0&0&0&0&1\\0&0&0&1&0\\0&0&1&0&0\\0&1&0&0&0\\1&0&0&0&0
\end{pmatrix}.
\]
It is linear and reversing twice gives \(R^2=I\).
**(b)** Comparing paired coefficients gives
\[
V_+=\operatorname{span}\{1+x^4,x+x^3,x^2\},\quad
V_-=\operatorname{span}\{1-x^4,x-x^3\}.
\]
When \(2\ne0\), these five independent vectors give
\(V=V_+\oplus V_-\).
**(c)** Since \(L=xD-2I\),
\[
[L]_{\mathcal B}=\operatorname{diag}(-2,-1,0,1,2).
\]
For a monomial \(x^k\), \(Lx^k=(k-2)x^k\), whereas reversal changes k to
\(4-k\), whose weight is \(2-k=-(k-2)\). Thus \(LR=-RL\), so L interchanges
the two eigenspaces.
**(d)**
\[
\begin{aligned}
L(1+x^4)&=-2(1-x^4),&L(x+x^3)&=-(x-x^3),&L(x^2)&=0,\\
L(1-x^4)&=-2(1+x^4),&L(x-x^3)&=-(x+x^3).
\end{aligned}
\]
As \(2\ne0\), both restrictions have rank 2. The displayed images also give
the required bases and show directly why characteristic 2 was excluded.
Spectral theorem and realification (B)
(a) Let
\[
A=\begin{pmatrix}2&0&0\\0&3&-1\\0&-1&3\end{pmatrix}\in M_3(\mathbb R).
\]
(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) Assume a real orthogonal \(2n\times2n\) matrix Q commutes with \(J=\begin{pmatrix}0&I\\-I&0\end{pmatrix}\). Derive its block form and construct the unique unitary complex matrix whose realification is Q. 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 \((2,2,4)\). They may be obtained from the
characteristic polynomial or from the visible invariant directions.
**(a)(ii)** One valid ordered orthogonal eigenbasis gives
\[
Q=\begin{pmatrix}1&0&0\\0&1/\sqrt2&1/\sqrt2\\0&1/\sqrt2&-1/\sqrt2\end{pmatrix},\qquad \Lambda=\operatorname{diag}(2,2,4).
\]
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 (B)
Let \(\sigma\in S_9\) be given in two-line notation by
\[
\sigma=\begin{pmatrix}
1&2&3&4&5&6&7&8&9\\
2&3&4&1&6&7&5&9&8
\end{pmatrix}.
\]
**Variant-specific requirement.** As a corollary of the classification theorem, prove that every permutation is conjugate to its inverse: reverse each non-trivial cycle and compare its length. State why this conclusion is stronger than merely saying the two permutations have the same sign.
(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
\[
\sigma=(1 2 3 4)(5 6 7)(8 9).
\]
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.** The inverse of a k-cycle has the same k entries in reverse order. Hence σ and σ^{-1} have the same complete cycle type and are conjugate; equal sign alone would not imply this.