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

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

In the adapted basis from part (a), the matrix has $r$ diagonal entries equal to one and $N-r$ equal to zero. With the characteristic polynomial convention ${\chi }_E(X)=\det (XI-E)$,The characteristic polynomial is independent of the choice of basis.

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.

The polynomial $X(X-1)$ annihilates $E$ by idempotence, so the minimal polynomial divides it. If $E=0$, its monic polynomial of smallest positive degree is $X$. If $E\ne 0$, both eigenvalues zero and one occur, so both factors must divide every annihilating polynomial. Therefore