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 $-1$.