So we have a field K and a finite K-Vectorspace V.
$\phi: V \to V$ is a linear function.
How can I prove the question in the title? We've been told to use the following theorem:
A matrix A is diagonalizable if it's minimal polynomial decomposes into distinct linear factors.
I guess I need to somehow prove that the minimal polynomial of $\phi$ still decomposes into distinc linear factors for $\phi|_U$ but I don't know how to do this at all.
Denote $I=\{p:p(\phi|_U)=0\}$ the ideal generated by the minimal polynomial $m_{\phi|_U}$. We show that $m_\phi$, the minimal polynomial of $\phi$, is divisable by $m_{\phi|_U}$. It is enough to show that $m_\phi(\phi|_U)=0$. Indeed, since $m_\phi(\phi)(v)=0$ for all $v\in V$, we know that $$m_\phi(\phi)(u)=m_\phi(\phi|_U)(u)=0$$ for all $u\in U$ (the equality is correct since $\phi=\phi|_U$ on $U$. Therefore $m_\phi(\phi|_U)$ is the zero operator on $U$, and $m_\phi\in I$. As $m_{\phi|_U}$ is the generator, we have: $m_{\phi|_U}|m_\phi$. As $m_\phi$ decomposes to distinct linear factor, so does $m_{\phi|U}$.
