{c}

If $Ev=\lambda v$ with $v\ne0$, applying $E^2=E$ gives $\lambda^2v=\lambda v$, 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 \b[$\{0\}$ if $E=0$, and $\{0,1\}$ otherwise]. Their algebraic multiplicities are $N-r$ and $r$ respectively.

Solution

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

 Solution (source code)