Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 7A (Vectors and Matrices)

An eigenvalue $\lambda$ is a scalar for which $\ker (A-\lambda I)$ contains a nonzero vector; that kernel is its eigenspace. The characteristic polynomial has degree $n$, so the fundamental theorem of algebra supplies at least one complex root and hence an eigenvalue. Its eigenspace dimension can be any integer from $1$ to $n$.

With ${\chi }_A(t)=\det (tI-A)$, direct expansion givesThus the eigenvalues are $3,1,-1,-2$. Kernel calculation gives eigenspace dimensions $1,2,1,1$, respectively.

The eigenvalue equation is $u_{i+1}=\lambda u_i$, soThis is a nonzero square-summable vector exactly when $c\ne 0$ and $|\lambda |<1$. Hence every point of the open unit disc is an eigenvalue, with the displayed one-dimensional eigenspace.

For the transpose shift, $(Cu{)}_1=0$ and $(Cu{)}_i=u_{i-1}$ for $i\ge 2$. If $Cu=\lambda u$, the first equation and the subsequent recurrence force every component to vanish, both for $\lambda =0$ and for $\lambda \ne 0$. Thus $C$ has no eigenvalues, illustrating that a bounded infinite-dimensional operator need not have one.