Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / 18

The first coordinate direction is already an eigenvector of $A$ with eigenvalue $4$. On the last two coordinates, the matrix is $\left(\begin{array}{cc}2& 4\\ 4& 2\end{array}\right)$, with characteristic polynomialThus the three eigenvalues are $4,6,-2$. For $\lambda =6$, the last two components are equal; for $\lambda =-2$, they are opposite. Normalising by the Euclidean norm gives the complete listEach sign-reversed vector is an equally valid choice. The three displayed eigenvectors are mutually orthogonal, as expected for a real symmetric matrix with distinct eigenvalues.

Subtracting the second row from the third gives a third row $(0,2,0)$, so expansion along it shows $\det B=-2\ne 0$. Thus the matrix inverse exists and every right-hand side has a unique solution.

To calculate it, temporarily solve $B\mathbf{x}=\mathbf{b}$ with general $\mathbf{b}=(b_1,b_2,b_3{)}^T$. Subtracting the second equation from the third and from the first givesBack-substitution into the second equation givesThereforeFor the given right-hand side, matrix multiplication yields the unique answer

The three coordinate basis vectors are eigenvectors, so $M$ is diagonal in those coordinates. Exactly one coordinate is multiplied by $-1$, while the other two are fixed. The three possibilities areThey are respectively reflections in the coordinate planes $x=0$, $y=0$ and $z=0$. Each is a Householder reflection with unit normal equal to the coordinate eigenvector carrying the eigenvalue $-1$.

For each possibility in part (i), squaring the diagonal entries gives $M^2=I$. Equivalently, the linear transformation fixes a vector space basis after being applied twice. ThereforeGeometrically, reflecting twice in the same plane returns every vector to itself.