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

Every vector decomposes as $v=Ev+(v-Ev)$, where $Ev\in \mathrm{im}E$ and $E(v-Ev)=0$. If $w$ lies in both the image and kernel, write $w=Eu$ to obtain $w=E^{2}u=Ew=0$. Thusand $E$ is the projection onto its image along its kernel. It is the identity on its image and zero on its kernel. Let $N=\dim V$ and $r=\mathrm{rank}E$. A basis adapted to this direct sum gives matrix $\mathrm{diag}(I_r,0_{N-r})$. Since $E$ is not the identity, $N-r>0$, and therefore