Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2014 / Paper 1 / 11

Expanding the determinant along the first row givesThe matrix trace is the sum of the diagonal entries, $\mathrm{tr}A=1+4+1=6$. Since the determinant is the product of the eigenvalues, at least one eigenvalue is zero. Thus $A$ is singular; the determinant alone does not specify the two other eigenvalues.

The symmetry between the first and third coordinates makes $(1,0,-1{)}^{\mathsf{T}}$ an eigenvector, with eigenvalue $-6$. In the remaining invariant plane, write a vector as $(s,t,s{)}^{\mathsf{T}}$. Its image is $(8s-4t,-8s+4t,8s-4t{)}^{\mathsf{T}}$, so the reduced eigenvalue equation has determinantThe corresponding directions have $t=2s$ for $\lambda =0$ and $t=-s$ for $\lambda =12$. Normalizing givesTheir squared unnormalized norms are $2,6,3$. The pairwise unnormalized dot products are $1-1=0$, $1-1=0$, and $1-2+1=0$, so these unit eigenvectors are mutually orthogonal. The eigenvalues sum to $6$ and have product zero, agreeing with part (a).

Expand an arbitrary vector in the orthonormal eigenbasis:Its image has no component along ${\mathbf{e}}_0$, soAlternatively, symmetry gives $(A\mathbf{r})\cdot {\mathbf{e}}_0=\mathbf{r}\cdot A{\mathbf{e}}_0=0$. This is image-kernel orthogonality for an adjoint, with $A^{\mathsf{T}}=A$.

Resolve the input along the three perpendicular eigendirections. The linear map removes the component along $(1,2,1)$, stretches the component along $(1,-1,1)$ by $12$, and reverses and stretches the component along $(1,0,-1)$ by $6$. Every output lies in the plane perpendicular to $(1,2,1)$. A nonzero input in the kernel of a linear map $A$ is sent to zero; other nonzero inputs remain in that output plane. Equivalently, take the orthogonal projection onto this plane, then scale by the two signed eigenvalues along its perpendicular eigendirections.