Past exam of the year 1 of the computer science course of the University of Oxford / 2005 / Calculus, linear algebra, discrete mathematics, logic and proof, and models of computation / 2 / c / Solution

Let $\lambda$ be the single eigenvalue. By diagonalizability, there is a basis $v_1,\dots ,v_n$ of eigenvectors, and every corresponding eigenvalue is $\lambda$. For any $v={\sum }_i{a}_i{v}_i$, linearity givesThus $\boxed{T=\lambda I}$, a scalar multiple of the identity map.

A counterexample to the conclusion without diagonalizability isIts characteristic polynomial is $(t-2{)}^2$, but $(J-2I)(a,b{)}^{\mathsf{T}}=(b,0{)}^{\mathsf{T}}$. The eigenspace for $2$ is therefore $\mathrm{span}\{(1,0{)}^{\mathsf{T}}\}$, of dimension one. There is no basis of eigenvectors in dimension two, so $J$ is not diagonalizable.