Let $V$ be a vector space over a field $\mathbb{K}$ and $\Phi$ an endomorphism of $V$.
I want to check if the following statements are correct or not.
- If $\Phi^2=\Phi$, then $\Phi$ has no other eigenvalues than $0$ and $1$.
This statement is correct.
Let $\lambda\in \mathbb{K}$ an eigenvaulue of $\Phi$, $v\in V\setminus\{0\}$ a corresponding eigenvector.
Then it holds that $\Phi (v)=\lambda \cdot v$ and $\Phi^2(v)=\Phi \left (\Phi (v)\right )=\Phi (\lambda v)=\lambda \cdot \left (\lambda \cdot v\right )=\lambda^2\cdot v$.
Since $\Phi^2=\Phi$, it follows that $\lambda^2\cdot v=\lambda \cdot v\Rightarrow \lambda^2\cdot v-\lambda \cdot v=0\Rightarrow \left (\lambda^2-\lambda \right )\cdot v=0$.
Since $v\neq 0$ it must be $\lambda^2-\lambda=0$. From that we get that $\lambda\left (\lambda-1\right )=0$. So it is either $\lambda=0$ or $\lambda=1$.
Is everything correct?
- If $\Phi^2$ has the eigenvalue $\lambda^2$, then $\Phi$ has the eigenvalue $\lambda$.
Could you give me a hint for this statement?
- If $\Phi$ is bijective and $\lambda$ is an eigenvalue of $\Phi$, then $\lambda^{-1}$ is an eigenvalue of $\Phi^{-1}$.
Let $\lambda\in \mathbb{K}$ an eigenvaulue of $\Phi$, $v\in V\setminus\{0\}$ a corresponding eigenvector.
Then it holds that $\Phi (v)=\lambda v$.
Since $\Phi$ is bijective, it s invertible, i.e. $\Phi^{-1}$ exists.
So we get $\Phi^{-1}\left (\Phi (v)\right )=\Phi^{-1}\left (\lambda v\right )\Rightarrow v=\Phi^{-1}\left (\lambda v\right )$.
Since $\Phi$ is a linear map, $\Phi^{-1}$ is also linear, or not?
So we get $v=\lambda \Phi^{-1}\left (v\right )\Rightarrow \Phi^{-1}\left (v\right )=\lambda^{-1}v$.
But how do we know that $\lambda$ is not equal to $0$ ?
From $\Phi^{-1}\left (v\right )=\lambda^{-1}v$ we get $\lambda^{-1}$ is an eigenvalue of $\Phi^{-1}$, or not?
Looks good to me.
Consider $1\times 1$ matrices and find a counterexample
$\lambda\neq0$ because $\Phi$ is injective. No injective endomorphism can have $0$ as an eigenvalue.