Past exam of the mathematical methods I course of the University of Oxford / 2014 resit / 6

For an idempotent matrix, $A^k=A$ for every integer $k\ge 1$. Thus any scalar polynomial reduces toor equivalently $aI+bA$. This expression describes all polynomials in $A$, although its two coefficients need not be unique if $A=0$ or $A=I$.

If $Av=\lambda v$ for a nonzero eigenvector, idempotence gives ${\lambda }^{2}v=A^{2}v=Av=\lambda v$. HenceThese are the only possible eigenvalues; either may be absent. In fact $v=Av+(I-A)v$ decomposes every vector into the image and kernel of $A$, so every idempotent matrix is diagonalizable even when it is not an orthogonal projection.