Show that $f(E)=aI+bE$ where $E$ is a projection operator.

Question

$\mathbf{Question}:$Let $V$ be a vector space over a field $F$. If $E:V \to V$ be a projection operator and $f$ be a polynomial over the field $F$ then show that $f(E)=aI+bE$ for some $a,b \in F$; where $I$ is the identity operator on $V$. Find $a$ and $b$ in terms of coefficients of $f$.

$\mathbf{Approach}:$ Let $f(x)=a_0+a_1x+...+a_nx^n,$ where $a_i\in F$.

Now, $f(E)=a_0I+a_1E+...+a_nE^n$. But, the projection map being idempotent by definition, $E=E^2=...=E^n$. Hence, we end up with $f(E)=a_0I+(a_1+...+a_n)E$.

Thereby, $a=a_0$, and $b=a_1+...+a_n$.

Is this okay? The wording of the question is a bit too confusing for me to be sure of what I am doing here, so a verification would be nice.

Thank you.

Answer 1

Your solution is correct.

Alternatively, notice that $g(x)=x^2-x$ satisfies $g(E) = 0$. We can divide the polynomial $f$ by $g$ to obtain $$f(x) = g(x)q(x) + r(x)$$ for some polynomials $q,r$ with $\deg r \le 1$. If we write $r(x) = a+bx$ we have $$f(E) = \underbrace{g(E)}_{=0}q(E) + r(E) = r(E) = aI + bE.$$