Question about eigenvalues: eigenvalue $f^2 + f = -1 \rightarrow$ eigenvalue $f^3 = 1$

Question

I have to proof:

Let $f \in End(V)$. Show that if $f^2+f$ has eigenvalue $-1$ then $f^3$ has eigenvalue $1$.

My idea:

• If $-1$ is the eigenvalue of $f^2+f$ then there exists (per definition) a $v \in V, v \not= 0$ so that

• $(f^2+f)(v) = f^2(v) + f(v) = -1$ ??

• So $f^2(v) + f(v) + 1 = 0$

• Then $ 0 = f(f^2(v) + f(v) + 1) = f^3(v) + f^2(v) + f(v) = f^3(v) - 1$ or: $f^3 = 1$.

Is this true?

Answer 1

$f(v), f^2(v),\cdots$ are vectors. $$(f^2+f)(v)=f^2(v)+f(v)=−v$$ $$\cdots$$