Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2008 / 2 / i / Solution

A projection is a linear operator $E$ satisfying $E^2=E$; it need not be an orthogonal projection without an inner product. Pairwise annihilation givesThus their sum is a projection.

The identities $E_{i}P=E_i$ imply $T(I-P)=0$. Since $T$ is nonsingular, $P=I$. Alsoso $\mathrm{im}{E}_i\subseteq V_i$. Conversely, if $Tx={\lambda }_{i}x$, applying $E_j$ gives $({\lambda }_j-{\lambda }_i)E_{j}x=0$. For $j\ne i$ the eigenvalues are distinct, hence $E_{j}x=0$. Since $P=I$, $x=E_{i}x$, proving $\mathrm{im}{E}_i=V_i$.

Every vector is ${\sum }_i{E}_{i}x$. If ${\sum }_i{v}_i=0$ with $v_i\in V_i$, applying $E_j$ gives $v_j=0$. ThusThese operators are the spectral projections of a diagonalizable matrix, expressed without choosing an eigenbasis.