Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 1 / 18

The middle row is $-1/2$ times the sum of the first and third rows, so the rows are linearly dependent and the determinant is zero. The matrix trace is the sum of diagonal entries, $1+4+1=6$. A transpose preserves the determinant, while the determinant of a product is the product of its determinants. Consequently

The matrix is symmetric under interchange of the first and third coordinates. For an antisymmetric vector $(1,0,-1)$, direct multiplication gives $A(1,0,-1{)}^{\mathsf{T}}=-6(1,0,-1{)}^{\mathsf{T}}$. For a symmetric vector $(p,q,p)$, the linear transformation reduces toThe reduced characteristic polynomial is $\lambda (\lambda -12)$. At $\lambda =0$, its equations give $q=2p$; at $\lambda =12$, they give $q=-p$. Normalizing the three eigenvectors givesTheir norms are one. Their unnormalized pairwise dot products are $1-1=0$, $1-1=0$ and $1-2+1=0$, proving mutual orthogonality. The eigenvalues sum to six and have product zero, agreeing with the matrix trace and determinant from part (a).

The three eigenvectors form an orthonormal basis, so an arbitrary vector has expansion $\mathbf{r}=a{\mathbf{v}}_{-}+b{\mathbf{v}}_0+c{\mathbf{v}}_{+}$. Multiplication by $A$ removes the zero-eigenvalue component:Its dot product with ${\mathbf{v}}_0$ is zero for every $a,b,c$. Thus one convenient answer is the nonzero vectorEquivalently, $A^{\mathsf{T}}\mathbf{u}=A\mathbf{u}=0$, so $(A\mathbf{r})\cdot \mathbf{u}=\mathbf{r}\cdot (A^{\mathsf{T}}\mathbf{u})=0$. Any nonzero scalar multiple of this vector in the null space also works.