Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2025 / Paper 2 / 18 / b

Take the dot product of the orthogonal basis expansion with ${\mathbf{u}}_j$. All terms except the $j$th vanish by orthogonality, leavingThereforeThe basis is not assumed normalized, so its squared lengths must remain in the denominators. Applying $S$ and using $S{\mathbf{u}}_i={\lambda }_i{\mathbf{u}}_i$ gives

Induction on $p$ gives $S^p{\mathbf{u}}_i={\lambda }_i^p{\mathbf{u}}_i$, including $p=0$. The absolutely convergent matrix exponential series may therefore be applied term by term to each eigenvector:ThusFor a repeated eigenvalue, every nonzero vector in its eigenspace remains an eigenvector. Because the real exponential function is injective, distinct real eigenvalues of $S$ do not merge under this transformation.

Since every eigenvalue is nonzero, $S$ is an invertible matrix. Multiplying $S{\mathbf{u}}_i={\lambda }_i{\mathbf{u}}_i$ by the inverse matrix givesHence the eigenvectors are unchanged and the corresponding eigenvalues are $1/{\lambda }_i$.

Choose the unnormalized eigenvectorsTheir linear combination giving the specified vector isApplying the reciprocal eigenvalues of the inverse matrix therefore yieldsIn Cartesian coordinates, this is $(-1/3,2/3{)}^T$, and direct matrix multiplication gives $S\mathbf{y}=(1,0{)}^T$.