Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 1 / a / ii / Solution

Put $n=\dim V>0$ and $r=\mathrm{rank}T$. The direct sum in part (i) givesThe eigenvalues are the members of $\{0,1\}$ actually present: zero occurs exactly when $r<n$, and one exactly when $r>0$. Since $T$ is diagonalizable, its minimal polynomial contains each present eigenvalue once:In particular, a nonzero vector space does not force both eigenvalues to occur. The zero and identity maps are important special cases.