MATH40003 2026 Exam Standard Full Paper C

**(a)** The equations are \(x_1=x_3\), \(x_2=x_4\), and \(x_2=x_5\). Hence \[ \ker D=\{(a,b,a,b,b)^T:a,b\in F\} =\operatorname{span}\{(1,0,1,0,0)^T,(0,1,0,1,1)^T\}. \] It has dimension \(2\); rank-nullity gives \(\operatorname{rk}D=3\). **(b)** The columns of \[ [G]=\begin{pmatrix}1&0&0\\0&1&0\\-1&0&0\\0&-1&0\\0&0&1\end{pmatrix} \] are independent, since the first, second and fifth coordinates of \(G(s,t,u)\) are \(s,t,u\). Thus \(G\) is injective and those three columns form a basis of its image. **(c)** If \(G(s,t,u)=(a,b,a,b,b)\), comparison gives \(s=a=-s\), \(t=b=-t\), and \(u=b=t\). Hence \(2s=2t=0\). If \(2\) is invertible, \(s=t=u=0\); the intersection is zero. The dimensions are \(2\) and \(3\), so their direct sum has dimension \(5\) and therefore equals \(F^5\). **(d)** In characteristic \(2\), \(-1=1\), so \[ G(1,0,0)^T=(1,0,1,0,0)^T\in\ker D \] is non-zero. Thus the sum is not direct.

**(a)** For \(\mathcal B=(1,x,x^2,x^3,x^4,x^5)\), \[ [J]_{\mathcal B}=\operatorname{diag}(1,-1,1,-1,1,-1). \] Substitution is linear and \(J^2=I\). **(b)** When \(2\ne0\), \[ V_+=\operatorname{span}\{1,x^2,x^4\},\qquad V_-=\operatorname{span}\{x,x^3,x^5\}, \] and these complementary coordinate subspaces give \(V=V_+\oplus V_-\). **(c)** \[ [D]_{\mathcal B}=\begin{pmatrix} 0&1&0&0&0&0\\0&0&2&0&0&0\\0&0&0&3&0&0\\ 0&0&0&0&4&0\\0&0&0&0&0&5\\0&0&0&0&0&0 \end{pmatrix}. \] \((DJp)(x)=-p'(-x)=-(JDp)(x)\), so D interchanges \(V_+\) and \(V_-\). **(d)** On \(V_+\), the images are \(0,2x,4x^3\), so the image is \(\operatorname{span}\{x,x^3\}\) and the rank is 2. On \(V_-\), the images are \(1,3x^2,5x^4\). Thus the image is \(\operatorname{span}\{1,x^2,x^4\}\), rank 3, when \(3,5\ne0\); \(\operatorname{span}\{1,x^4\}\), rank 2, in characteristic 3; and \(\operatorname{span}\{1,x^2\}\), rank 2, in characteristic 5.

**(a)(i)** The eigenvalues are \((6,1,1)\). They may be obtained from the characteristic polynomial or from the visible invariant directions. **(a)(ii)** One valid ordered orthogonal eigenbasis gives \[ Q=\begin{pmatrix}1/\sqrt5&-2/\sqrt5&0\\2/\sqrt5&1/\sqrt5&0\\0&0&1\end{pmatrix},\qquad \Lambda=\operatorname{diag}(6,1,1). \] 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.

**(a)(i)** Following each orbit gives \[ \sigma=(1 2 3 4 5 6)(7 8 9)(10 11). \] There are 1 fixed point(s), so the complete cycle type is \([6, 3, 2, 1]\). **(a)(ii)** Use \((a_1\ldots a_k)=(a_1a_k)\cdots(a_1a_2)\) on every non-trivial cycle. This uses \(8\) 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 \([6, 3, 2, 1]\). **Required variant check.** Advancing two positions around each cycle gives \(\sigma^2=(1\ 3\ 5)(2\ 4\ 6)(7\ 9\ 8)(10)(11)(12)\). Thus the 6-cycle splits into two 3-cycles, the 3-cycle remains a 3-cycle, and the transposition splits into fixed points. In general a k-cycle splits into \(\gcd(k,2)\) cycles of length \(k/\gcd(k,2)\).