Eigenvalue of $f \circ f = id_V$

Question

An endomorphism $f$ in the vector space $V$ with the popperty $f \circ f = id_V$ is given. Is my assumption correct that the only Eigenvalues of $f$ are $\lambda_1=1$ and $\lambda_2 =-1$?

Reasoning

We have the defenition of the Eigenvalue:

$$f(v)=\lambda v \tag{1}$$

$$\Rightarrow_{\circ f} f(f(v))= \lambda f(v)$$

$$v=\lambda f(v) \tag{2}$$

Now we can put $(1)$ into $(2)$:

$$v=\lambda \lambda v$$

The only numbers that satisfy this equation for $\lambda$ are $+1$ and $-1$.

Answer 1

Although, @Doe's answers your question, the following method is a more general method for finding eigenvalues of a given map. Note that, by doing some extra computations, we can directly find the eigenvalues of $f$ for sure with this method.

First observe that the map $f$ is a zero of the the polynomial $$p(\lambda) = \lambda^2 - 1 = 0,$$ i.e $$p(f) = 0_v.$$

Hence, the minimal polynomial of $f$ has to divide $p$, but $this implies$, either $$m_f(\lambda) = \lambda -1,$$ or $$m_f(\lambda) = \lambda +1,$$ or $$m_f(\lambda) = \lambda^2 -1.$$

Since characteristic polynomial and the minimal polynomial always have the same roots, we can conclude that $f$ can only have the eigenvalues $\pm 1$.

Answer 2

Indeed, this is correct. You have $$v=\lambda^2 v\implies (1-\lambda^2)v=0$$ Either $\lambda=\pm1$ or $v=0$. But since $v$ is an eigenvector, it must be non-zero. Hence we're done.