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- Apr 14, 2013

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Let $\mathbb{K}$ be a field and let $V$ a $\mathbb{K}$-vector space. Let $\phi, \psi:V\rightarrow V$ be linear operators, such that $\phi\circ\psi=\psi\circ\phi$.

Show that:

- For $\lambda \in \text{spec}(\phi)$ it holds that $\text{Eig}(\phi, \lambda )\leq_{\psi}V$.
- Let $n=\dim_{\mathbb{K}}V$ and $n=|\text{spec}(\phi)|$. Then $\psi$ is diagonalizable.

At

**question 1**we have that $\lambda$ is an eigenvalue of $\phi$. We have that $\phi (v )=\lambda v$, for the respective eigenvector $v$.

Then we have that $\phi (v_1+v_2)=\lambda (v_1+v_2)=\lambda v_1+\lambda v_2=\phi (v_1)+\phi (v_2)$ and $\phi (cv_1)=\lambda (cv_1)=c\left (\lambda v_1\right )=c\phi (v_1)$.

Is everything correct so far? So it follows that $\text{Eig}(\phi, \lambda )\leq_{\psi}V$, right?

Could you give me a hint for

**question 2**?