- Admin
- #26

- Mar 5, 2012

- 9,591

Not really.We have that $$\psi(v)=0 \Rightarrow \phi \left (\psi(v)\right )=\phi (0) \Rightarrow \left (\phi\circ\psi\right )(v)=0 \Rightarrow \left (\psi\circ\phi\right )(v)=0\Rightarrow \psi\left ( \phi(v)\right )=0$$ Does this help?

We have to find $n$ independent eigenvectors for $\psi$ don't we?

Is $v$ an eigenvector of $\psi$?