If \(E\) is a projection and \(f\) is a polynomial, then \(f(E) = aI + bE\). What are \(a\) and \(b\) in terms of the coefficients of \(f\)?
Solution
Write \[ f(t)=c_0+c_1t+\cdots+c_mt^m. \] Since \(E\) is a projection, \(E^2=E\), and therefore \[ E^k=E\qquad(k\ge1) \] by induction. Hence \[ \begin{aligned} f(E) &=c_0I+c_1E+\cdots+c_mE^m\\ &=c_0I+(c_1+\cdots+c_m)E. \end{aligned} \] Thus one may take \[ \boxed{a=c_0=f(0),\qquad b=c_1+\cdots+c_m=f(1)-f(0).} \] This proves the claimed form [@HK71].