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Eigenspace is a subspace of V - ψ is diagonalizable

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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?
Not really.
We have to find $n$ independent eigenvectors for $\psi$ don't we?
Is $v$ an eigenvector of $\psi$? 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
Not really.
We have to find $n$ independent eigenvectors for $\psi$ don't we?
Is $v$ an eigenvector of $\psi$? 🤔
I got stuck right now. How could we do that? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591

mathmari

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MHB Site Helper
Apr 14, 2013
4,713

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
For that we have to show that $\psi(v)=\lambda v$.
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)
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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. :eek:

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

mathmari

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MHB Site Helper
Apr 14, 2013
4,713

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
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:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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

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Apr 14, 2013
4,713
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 $n$ ?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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? 🤠