Eigenspace is a subspace of V - ψ is diagonalizable

[I like Serena]

What does this mean?

It means that $v$ is an eigenvector of $\psi$ with eigenvalue $0$ instead of eigenvalue $\lambda$.

 

[mathmari]

It means that $v$ is an eigenvector of $\psi$ with eigenvalue $0$ instead of eigenvalue $\lambda$.

For $\phi$ there are $n$ distinct eigenvalues $\lambda_i$'s and the corresponding eigenvectors $v_i$'s.
We want to show that each $v_i$ is also an eigenvector of $\psi$ for some eigenvalue $\mu$.
We shown that
\begin{equation*}\psi \left (\phi (v)\right )=\psi \left (\lambda v\right )\Rightarrow \left (\psi \circ \phi \right )(v)=\lambda \psi \left ( v\right )\Rightarrow \left (\phi\circ\psi \right )(v)=\lambda \psi \left ( v\right ) \Rightarrow \phi \left (\psi(v) \right )=\lambda \psi \left ( v\right )\end{equation*}
If $\psi (v)\ne 0$ then $\psi (v)$ is also an eigenvector for $\lambda$, i.e. $\psi (v)\in \text{Eig}(\phi, \lambda)$.
If $\psi (v)=0$ then the corresponding eigenvalue is $0$.

Is everything correct so far? Or am I thinking in a wrong way for that question? :unsure:

 

[I like Serena]

If $\psi (v)\ne 0$ then $\psi (v)$ is also an eigenvector for $\lambda$, i.e. $\psi (v)\in \text{Eig}(\phi, \lambda)$.
If $\psi (v)=0$ then the corresponding eigenvalue is $0$.

Is everything correct so far? Or am I thinking in a wrong way for that question?

All correct. (Nod)

Btw, $\psi (v)\in \text{Eig}(\phi, \lambda)$ holds in both cases, i.e. it does not distinguish the first case from the second.

 

[mathmari]

All correct. (Nod)

Btw, $\psi (v)\in \text{Eig}(\phi, \lambda)$ holds in both cases, i.e. it does not distinguish the first case from the second.

But how does it follow from that that the numbers of distinct eigenvectors is $n$ ?

 

[I like Serena]

But how does it follow from that that the numbers of distinct eigenvectors is $n$ ?

$\phi$ has $n$ distinct eigenvalues $\lambda_i$ with corresponding eigenvectors $v_i$.
The set of those eigenvectors is consequently an independent set of $n$ vectors.
Therefore the $v_i$ form a basis of $V$.

The same eigenvectors are also eigenvectors of $\psi$ aren't they?
They just don't necessarily have the same eigenvalues, and there are not necessarily $n$ distinct eigenvalues any more.
That is, $0$ can be an eigenvalue with multiplicity greater than 1.
Either way, they still form a basis of $V$, don't they?