# Eigenvalues for an operator?

#### joshmccraney

Problem Statement
Show no eigenvalues exist for the operator $Af(x) = xf(x)$ where $A:C[0,1]\to C[0,1]$.
Relevant Equations
Nothing comes to mind.
Eigenvalues $\lambda$ for some operator $A$ satisfy $A f(x) = \lambda f(x)$. Then

$$Af(x) = \lambda f(x) \implies\\ xf(x) = \lambda f(x)\implies\\ (\lambda-x)f(x) = 0.$$

How do I then show that no eigenvalues exist? Seems obvious one doesn't exists since $\lambda-x \neq 0$ for all $x\in [0,1]$.

Related Calculus and Beyond Homework News on Phys.org

#### MathematicalPhysicist

Gold Member
Well the argument is subtle.

Since $x$ ranges over every point in $[0,1]$ and the supposed eigenvalue is some constant you cannot have $\lambda=x$, since the function $x$ is not constant.
And obviously $f(x)\ne 0$, otherwise it wouldn't be an eigenvalues' problem.

#### Math_QED

Science Advisor
Homework Helper
You are in fact almost there. The condition
$(\lambda-x)f(x) =0$ implies that we have $f(x) = 0$ for all $x\neq \lambda$. By continuity however $f=0$. Thus, no eigenvector exists, as any possible eigenvalue leads to the trivial solution.

Last edited:
• joshmccraney

#### joshmccraney

You are in fact almost there. The condition
$(\lambda-x)f(x) =0$ implies that we have $f(x) = 0$ for all $x\neq \lambda$. By continuity however $f=0$. Thus, no eigenvector exists, as any possible eigenvalue leads to the trivial solution.
This just seems so simple. See, the HW was assigned in three parts, where part b) asked to show 0 is not an eigenvalue and part c) asked to prove no eigenvalues exist. But if this is part c), then why bother doing b)? Have I missed anything about $\lambda = 0$? It seems this logic holds for all $\lambda$.

#### Math_QED

Science Advisor
Homework Helper
This just seems so simple. See, the HW was assigned in three parts, where part b) asked to show 0 is not an eigenvalue and part c) asked to prove no eigenvalues exist. But if this is part c), then why bother doing b)? Have I missed anything about $\lambda = 0$? It seems this logic holds for all $\lambda$.
It also works for $\lambda = 0$. Then your equation leads to $xf(x) = 0 \forall x$ and for $x \neq 0$ we then have $f(x) = 0$. By continuity again also $f(0) = 0$, so $f=0$ everywhere.

So, to answer your question concretely, I don't think doing part (b) separately is necessary.

Note that eigenvalue $\lambda = 0$ can only happen when $A$ is non-injective. This map is injective however (continuity is key again here). Maybe that's what (b) wants you to write?

• joshmccraney

### Want to reply to this thread?

"Eigenvalues for an operator?"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving