Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 18

Adding the diagonal entries gives $\mathrm{tr}A=4+4+2=10$. Expansion along the first row gives $\det A=4(8-4)+2(0-8)=0$. For the remaining trace, direct matrix multiplication givesThus $\mathrm{tr}(A^2)=20+20+12=52$. By multiplicativity of the determinant, $\det (A^2)=(\det A{)}^2=0$. Altogether,

The vector $(1,-1,0{)}^T$ is an eigenvector with eigenvalue $4$. The complementary subspace of vectors with first two components equal is preserved by $A$: a vector $(s,s,t{)}^T$ maps to $(4s+2t,4s+2t,4s+2t{)}^T$. There, the eigenvalue equations reduce to the $2\times 2$ matrix $\left(\begin{array}{cc}4& 2\\ 4& 2\end{array}\right)$, whose characteristic polynomial is $\lambda (\lambda -6)$. For $\lambda =6$, $t=s$; for $\lambda =0$, $t=-2s$.

Normalising by the Euclidean norm, a complete orthonormal basis of eigenvectors isSign changes of any of these eigenvectors are equally valid. The eigenvalues also check part (a): their sum is $10$ and the sum of their squares is $52$.

Expand the right-hand side in the orthonormal basis from part (b). Its coefficients areThus $\mathbf{b}=2\sqrt{2}{\mathbf{e}}_4+\sqrt{3}{\mathbf{e}}_6$. If $\mathbf{x}=c_0{\mathbf{e}}_0+c_4{\mathbf{e}}_4+c_6{\mathbf{e}}_6$, then matrix multiplication gives $A\mathbf{x}=4c_4{\mathbf{e}}_4+6c_6{\mathbf{e}}_6$. Matching coefficients fixes $c_4=\sqrt{2}/2$ and $c_6=\sqrt{3}/6$, while $c_0$ is arbitrary. Consequently all solutions areGeometrically, the solution set is an affine line in a vector space parallel to the kernel of $A$. The zero eigenvalue makes the component in that direction invisible to $A$; the other two components are uniquely determined. The displayed particular solution is perpendicular to the kernel, so it is also the point of this line closest to the origin.