# Eigenspace is a subspace of V - ψ is diagonalizable

#### mathmari

##### Well-known member
MHB Site Helper
Hey!! Let $\mathbb{K}$ be a field and let $V$ a $\mathbb{K}$-vector space. Let $\phi, \psi:V\rightarrow V$ be linear operators, such that $\phi\circ\psi=\psi\circ\phi$.
Show that:
1. For $\lambda \in \text{spec}(\phi)$ it holds that $\text{Eig}(\phi, \lambda )\leq_{\psi}V$.
2. Let $n=\dim_{\mathbb{K}}V$ and $n=|\text{spec}(\phi)|$. Then $\psi$ is diagonalizable.

At question 1 we have that $\lambda$ is an eigenvalue of $\phi$. We have that $\phi (v )=\lambda v$, for the respective eigenvector $v$.
Then we have that $\phi (v_1+v_2)=\lambda (v_1+v_2)=\lambda v_1+\lambda v_2=\phi (v_1)+\phi (v_2)$ and $\phi (cv_1)=\lambda (cv_1)=c\left (\lambda v_1\right )=c\phi (v_1)$.

Is everything correct so far? So it follows that $\text{Eig}(\phi, \lambda )\leq_{\psi}V$, right? Could you give me a hint for question 2? • Klaas van Aarsen

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Let $\mathbb{K}$ be a field and let $V$ a $\mathbb{K}$-vector space. Let $\phi, \psi:V\rightarrow V$ be linear operators, such that $\phi\circ\psi=\psi\circ\phi$.
Show that:
1. For $\lambda \in \text{spec}(\phi)$ it holds that $\text{Eig}(\phi, \lambda )\leq_{\psi}V$.
Hey mathmari !!

What does $\leq_{\psi}$ mean? At question 1 we have that $\lambda$ is an eigenvalue of $\phi$. We have that $\phi (v )=\lambda v$, for the respective eigenvector $v$.
Then we have that $\phi (v_1+v_2)=\lambda (v_1+v_2)=\lambda v_1+\lambda v_2=\phi (v_1)+\phi (v_2)$ and $\phi (cv_1)=\lambda (cv_1)=c\left (\lambda v_1\right )=c\phi (v_1)$.

Is everything correct so far? So it follows that $\text{Eig}(\phi, \lambda )\leq_{\psi}V$, right?
Isn't all of that already implied by the fact that $\phi$ is linear? That is given so there is no need to prove it, is there?
And it doesn't involve $\psi$, so we can't say anything with respect to $\psi$ yet, can we? 2. Let $n=\dim_{\mathbb{K}}V$ and $n=|\text{spec}(\phi)|$. Then $\psi$ is diagonalizable.

Could you give me a hint for question 2?
$\text{spec}(\phi)$ is the set of eigenvalues isn't it?
So we have $n$ distinct eigenvalues, and therefore the corresponding eigenvectors must form a basis of $V$.
Can we find the eigenvalues and eigenvectors of $\psi$ using that $\phi\circ\psi=\psi\circ\phi$? • mathmari

#### mathmari

##### Well-known member
MHB Site Helper
Isn't all of that already implied by the fact that $\phi$ is linear? That is given so there is no need to prove it, is there?
And it doesn't involve $\psi$, so we can't say anything with respect to $\psi$ yet, can we? I am also confused with that notation. So does this not mean "subspace" ? Could it maybe be a typo and it should be $\phi$ instead of $\psi$ ? $\text{spec}(\phi)$ is the set of eigenvalues isn't it?
Yes!

So we have $n$ distinct eigenvalues, and therefore the corresponding eigenvectors must form a basis of $V$.
Why do the corresponding eigenvectors form a basis of $V$ ? Can we find the eigenvalues and eigenvectors of $\psi$ using that $\phi\circ\psi=\psi\circ\phi$? Let $\lambda_i$, $i\in \{1,2, \ldots , n\}$ be an eigenvalue of $\phi$ and $v_i$ the corresponding eigenvector.

Then we have that $\phi (v_i)=\lambda_i v_i$. We apply $\psi$ and we get $$\psi \left (\phi (v_i)\right )=\psi \left (\lambda_i v_i\right )\Rightarrow \left (\psi \circ \phi \right )(v_i)=\lambda_i \psi \left ( v_i\right )\Rightarrow \left (\phi\circ\psi \right )(v_i)=\lambda_i \psi \left ( v_i\right ) \Rightarrow \phi \left (\psi(v_i) \right )=\lambda_i \psi \left ( v_i\right )$$ Do we get from that that $\psi (v_i)$ is also an eigenvector for $\lambda_i$ ? #### Klaas van Aarsen

##### MHB Seeker
Staff member
I am also confused with that notation. So does this not mean "subspace" ? Could it maybe be a typo and it should be $\phi$ instead of $\psi$ ?
If it would merely be a subspace of $V$, then the proof would be trivial, wouldn't it?
And if $\psi$ were not involved, then there would be no reason to mention $\psi$ at all, nor that they commute. Why do the corresponding eigenvectors form a basis of $V$ ?
How many eigenvectors can we identify?
Can they be dependent? Let $\lambda_i$, $i\in \{1,2, \ldots , n\}$ be an eigenvalue of $\phi$ and $v_i$ the corresponding eigenvector.

Then we have that $\phi (v_i)=\lambda_i v_i$. We apply $\psi$ and we get $$\psi \left (\phi (v_i)\right )=\psi \left (\lambda_i v_i\right )\Rightarrow \left (\psi \circ \phi \right )(v_i)=\lambda_i \psi \left ( v_i\right )\Rightarrow \left (\phi\circ\psi \right )(v_i)=\lambda_i \psi \left ( v_i\right ) \Rightarrow \phi \left (\psi(v_i) \right )=\lambda_i \psi \left ( v_i\right )$$ Do we get from that that $\psi (v_i)$ is also an eigenvector for $\lambda_i$ ?
Yes - but only if $\psi (v_i)\ne 0$ because an eigenvector must be a nonzero vector. Either way, it means that $\psi (v_i)\in \text{Eig}(\phi, \lambda_i)$. • mathmari

#### mathmari

##### Well-known member
MHB Site Helper
How many eigenvectors can we identify?
Can they be dependent? We have $n$ eigenvalues. So do we have also $n$ eigenvectors? Yes - but only if $\psi (v_i)\ne 0$ because an eigenvector must be a nonzero vector. Either way, it means that $\psi (v_i)\in \text{Eig}(\phi, \lambda_i)$. We have that $\psi (v)=0 \iff v=0$ since $\psi$ is linear.

So we have that $v_i\in \text{Eig}(\phi, \lambda_i) \Rightarrow \psi (v_i) \in \text{Eig}(\phi, \lambda_i)$. This means that $\text{Eig}(\phi, \lambda_i)$ is $\psi$-invariant, right?
Is this meaybe related to $\leq_{\psi}$ ? #### Klaas van Aarsen

##### MHB Seeker
Staff member
We have $n$ eigenvalues. So do we have also $n$ eigenvectors?
Indeed. Every eigenvalue has at least 1 nonzero eigenvector associated with it.

We have that $\psi (v)=0 \iff v=0$ since $\psi$ is linear.
I don't think that is true. So we have that $v_i\in \text{Eig}(\phi, \lambda_i) \Rightarrow \psi (v_i) \in \text{Eig}(\phi, \lambda_i)$. This means that $\text{Eig}(\phi, \lambda_i)$ is $\psi$-invariant, right?
Invariant means that the image is the same as the original. But that is not the case here is it? Is this meaybe related to $\leq_{\psi}$ ?
I think so yes. #### mathmari

##### Well-known member
MHB Site Helper
Invariant means that the image is the same as the original. But that is not the case here is it? I think so yes. But how can we show that $\text{Eig}(\phi, \lambda_i)$ is $\psi$-invariant?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
But how can we show that $\text{Eig}(\phi, \lambda_i)$ is $\psi$-invariant?
I think that is something different.
It means that $\psi(\text{Eig}(\phi, \lambda_i))=\text{Eig}(\phi, \lambda_i)$, but that is not necessarily the case.
We can come up with a counter example if we want to. • mathmari

#### mathmari

##### Well-known member
MHB Site Helper
I think that is something different.
It means that $\psi(\text{Eig}(\phi, \lambda_i))=\text{Eig}(\phi, \lambda_i)$, but that is not necessarily the case.
We can come up with a counter example if we want to. Ahh ok, I thought we have to use this

• Klaas van Aarsen

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Ahh ok, I thought we have to use this
Interesting.

We have found that every $v\in\text{Eig}(\phi,\lambda_i)$ is transformed by $\psi$ into a vector contained in $\text{Eig}(\phi,\lambda_i)$.
Therefore $\text{Eig}(\phi,\lambda_i)$ is an invariant subspace of $\psi$. That article also states: "An invariant subspace of T is also said to be T invariant."
So $\text{Eig}(\phi,\lambda_i)$ can be called $\psi$ invariant after all.
It's just that calling it $\psi$ invariant is a bit confusing, since it is not clear which property is supposed to be invariant. Last edited:
• mathmari

#### mathmari

##### Well-known member
MHB Site Helper
Interesting.

We have found that every $v\in\text{Eig}(\phi,\lambda_i)$ is transformed by $\psi$ into a vector contained in $\text{Eig}(\phi,\lambda_i)$.
Therefore $\text{Eig}(\phi,\lambda_i)$ is an invariant subspace of $\psi$. That article also states: "An invariant subspace of T is also said to be T invariant."
So $\text{Eig}(\phi,\lambda_i)$ can be called $\psi$ invariant after all.
It's just that calling it $\psi$ invariant is a bit confusing, since it is not clear which property is supposed to be invariant. The other definition is $\psi \left (\text{Eig}(\phi,\lambda_i)\right )\subset \text{Eig}(\phi,\lambda_i)$, or not?
Then doesn't this mean that $\psi (v)$ with $v\in \text{Eig}(\phi,\lambda_i)$, must be an element of $\text{Eig}(\phi,\lambda_i)$ ? #### Klaas van Aarsen

##### MHB Seeker
Staff member
The other definition is $\psi \left (\text{Eig}(\phi,\lambda_i)\right )\subset \text{Eig}(\phi,\lambda_i)$, or not?
Then doesn't this mean that $\psi (v)$ with $v\in \text{Eig}(\phi,\lambda_i)$, must be an element of $\text{Eig}(\phi,\lambda_i)$ ?
Yes - with the additional conditions that $\psi$ must be a linear mapping, which is given, and that $\text{Eig}(\phi,\lambda_i)$ must be a linear subspace of $V$, which is implied. Btw, I suggest to use $\subseteq$ instead of $\subset$ to avoid confusion with strict subsets. • mathmari

#### mathmari

##### Well-known member
MHB Site Helper
Yes - with the additional conditions that $\psi$ must be a linear mapping, which is given, and that $\text{Eig}(\phi,\lambda_i)$ must be a linear subspace of $V$, which is implied. Btw, I suggest to use $\subseteq$ instead of $\subset$ to avoid confusion with strict subsets. So for question 1 we have the following:

To show that $\text{Eig}(\phi, \lambda )\leq_{\psi}V$ we show that $\text{Eig}(\phi, \lambda )$ is $\ \psi$-invariant and a subspace of $V$.

Let $\lambda$ be an eigenvalue of $\phi$, and $v$ the respective eigenvector, so $v \in \text{Eig}(\phi, \lambda)$.
Then we have that $\phi (v)=\lambda v$.
We apply $\psi$ and get \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 the eigenvalue $\lambda$.
This means that $\psi (v)\in \text{Eig}(\phi, \lambda)$, i.e. $\psi \left (\text{Eig}(\phi, \lambda)\right )\subseteq \text{Eig}(\phi, \lambda)$.

This means that $\text{Eig}(\phi, \lambda)$ is $\psi$-invariant.

The fact that $\phi$ is linear is it trivial that $\text{Eig}(\phi,\lambda)$ is a linear subspace of $V$ ? #### Klaas van Aarsen

##### MHB Seeker
Staff member
This means that $\text{Eig}(\phi, \lambda)$ is $\psi$-invariant.

The fact that $\phi$ is linear is it trivial that $\text{Eig}(\phi,\lambda)$ is a linear subspace of $V$ ?
Yep. It's a known property of an eigenspace that it is a linear subspace.
We can verify it by observing that the sum of 2 eigenvectors in the eigenspace is also an eigenvector. And the scalar multiple of an eigenvector is an eigenvector as well. • mathmari

#### mathmari

##### Well-known member
MHB Site Helper
Yep. It's a known property of an eigenspace that it is a linear subspace.
We can verify it by observing that the sum of 2 eigenvectors in the eigenspace is also an eigenvector. And the scalar multiple of an eigenvector is an eigenvector as well. Ok, great!! At question 2 :

$\phi$ has $n$ eigenvalues and the dimension of $V$ is $n$.
Since $\phi$ has $n$ distinct eigenvalues there are $n$ linearly independent eigenvectors, right? Is this trivial or do we have to prove that?
Then we have shown that $\text{Eig}(\phi, \lambda)$ is $\psi$-invariant, i.e. $\psi \left (\text{Eig}(\phi, \lambda)\right )\subseteq \text{Eig}(\phi, \lambda)$. To show that $\psi$ has also $n$ linearly independent eigenvectors do we have to show that $\text{Eig}(\phi, \lambda)\subseteq \psi \left (\text{Eig}(\phi, \lambda)\right )$ ? #### Klaas van Aarsen

##### MHB Seeker
Staff member
At question 2 :

$\phi$ has $n$ eigenvalues and the dimension of $V$ is $n$.
Since $\phi$ has $n$ distinct eigenvalues there are $n$ linearly independent eigenvectors, right? Is this trivial or do we have to prove that?
It's a known property.
If you don't know that property then formally you should either look it up and quote it, or prove it. Then we have shown that $\text{Eig}(\phi, \lambda)$ is $\psi$-invariant, i.e. $\psi \left (\text{Eig}(\phi, \lambda)\right )\subseteq \text{Eig}(\phi, \lambda)$. To show that $\psi$ has also $n$ linearly independent eigenvectors do we have to show that $\text{Eig}(\phi, \lambda)\subseteq \psi \left (\text{Eig}(\phi, \lambda)\right )$ ?
That doesn't seem to help does it?
Hint: what is the dimension of $\text{Eig}(\phi, \lambda)$ for some $\lambda$? • mathmari

#### mathmari

##### Well-known member
MHB Site Helper
Hint: what is the dimension of $\text{Eig}(\phi, \lambda)$ for some $\lambda$? It is the dimension of the basis, i.e. the number of linearly independent eigenvectors, i.e. $n$, right?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
It is the dimension of the basis, i.e. the number of linearly independent eigenvectors, i.e. $n$, right?
Nope. What is $\text{Eig}(\phi,\lambda)$ again? • mathmari

#### mathmari

##### Well-known member
MHB Site Helper
What is $\text{Eig}(\phi,\lambda)$ again? Ahhh since we have $n$ distinct eigenvalues and the dimension of $V$ is $n$, each $\text{Eig}(\phi,\lambda)$ has the dimension $1$, i.e. each eigenvector has one eigenvector.

Is that correct? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Ahhh since we have $n$ distinct eigenvalues and the dimension of $V$ is $n$, each $\text{Eig}(\phi,\lambda)$ has the dimension $1$, i.e. each eigenvector has one eigenvector.

Is that correct?
Yep. More correctly: each eigenspace is the span of 1 eigenvector. • mathmari

#### mathmari

##### Well-known member
MHB Site Helper
Yep. More correctly: each eigenspace is the span of 1 eigenvector. Ok! So we have that $\psi \left (\text{Eig}(\phi, \lambda)\right )\subseteq \text{Eig}(\phi, \lambda)$ and the dimensions are then $\dim \left (\psi \left (\text{Eig}(\phi, \lambda)\right )\right )\leq \dim\text{Eig}(\phi, \lambda) \Rightarrow \dim \left (\psi \left (\text{Eig}(\phi, \lambda)\right )\right )\leq 1$.

Is it then $\dim \left (\psi \left (\text{Eig}(\phi, \lambda)\right )\right )= 1$ ? #### Klaas van Aarsen

##### MHB Seeker
Staff member
Ok! So we have that $\psi \left (\text{Eig}(\phi, \lambda)\right )\subseteq \text{Eig}(\phi, \lambda)$ and the dimensions are then $\dim \left (\psi \left (\text{Eig}(\phi, \lambda)\right )\right )\leq \dim\text{Eig}(\phi, \lambda) \Rightarrow \dim \left (\psi \left (\text{Eig}(\phi, \lambda)\right )\right )\leq 1$.

Is it then $\dim \left (\psi \left (\text{Eig}(\phi, \lambda)\right )\right )= 1$ ?
Nope. After all, $\psi(v)$ can be $0$, can't it? In that case we have $\dim \left (\psi \left (\text{Eig}(\phi, \lambda)\right )\right )= 0$. • mathmari

#### mathmari

##### Well-known member
MHB Site Helper
Nope. After all, $\psi(v)$ can be $0$, can't it? In that case we have $\dim \left (\psi \left (\text{Eig}(\phi, \lambda)\right )\right )= 0$. To show that $\psi$ is diagonalisable we have to show that there are $n$ eigenvectors, or not?

Could you give me a hint for that? #### Klaas van Aarsen

##### MHB Seeker
Staff member
To show that $\psi$ is diagonalisable we have to show that there are $n$ eigenvectors, or not?

Could you give me a hint for that?
Let $v$ be an eigenvector of $\phi$ for some eigenvalue $\lambda$.
Then we have a "problem" if $\psi(v)=0$ yes? Because then $v$ is not an eigenvector of $\psi$ with eigenvalue $\lambda$.
What can we say about such a $v$ with respect to $\psi$? • mathmari

#### mathmari

##### Well-known member
MHB Site Helper
Let $v$ be an eigenvector of $\phi$ for some eigenvalue $\lambda$.
Then we have a "problem" if $\psi(v)=0$ yes? Because then $v$ is not an eigenvector of $\psi$ with eigenvalue $\lambda$.
What can we say about such a $v$ with respect to $\psi$? 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? 