Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2010 / 1 / c / Solution

If $Ev=\lambda v$ with $v\ne 0$, applying $E^2=E$ gives ${\lambda }^{2}v=\lambda v$, hence $\lambda \in \{0,1\}$. The kernel is nonzero since $E\ne I$, so zero always occurs. One occurs exactly when the image is nonzero. Thus the eigenvalue set is $\{0\}$ if $E=0$, and $\{0,1\}$ otherwise. Their algebraic multiplicities are $N-r$ and $r$ respectively.