MHB Eigenspace is a subspace of V - ψ is diagonalizable

[mathmari]

Let's not call it $\lambda$ to avoid confusion with the $\lambda$ we already have. Let's call the eigenvalue $\mu$.
So we have to show that $\psi(v)=\mu v$... (Sweating)

We have that $\phi (v)=\lambda v \Rightarrow v=\phi (v)\lambda^{-1}$. Then we get $\psi (v)=\psi \left (\phi (v)\lambda^{-1}\right )\Rightarrow \psi (v)=\lambda^{-1} \phi \left (\psi (v)\right )$.

That doesn't help us, does it? :unsure:

 

[I like Serena]

We have that $\phi (v)=\lambda v \Rightarrow v=\phi (v)\lambda^{-1}$. Then we get $\psi (v)=\psi \left (\phi (v)\lambda^{-1}\right )\Rightarrow \psi (v)=\lambda^{-1} \phi \left (\psi (v)\right )$.

That doesn't help us, does it?

That's not necessarily true is it? Suppose $\lambda=0$, then $\lambda^{-1}$ is not defined.

We have that $\psi(v)=0$.
Can we find a scalar $\mu$ such that $\psi(v)=\mu v$?

 

[mathmari]

We have that $\psi(v)=0$.
Can we find a scalar $\mu$ such that $\psi(v)=\mu v$?

It holds for $\mu=0$.

 

[I like Serena]

It holds for $\mu=0$.

Indeed...

 

[mathmari]

Indeed...

What does this mean? :unsure:

 

[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?