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Eigenvalues for an operator?

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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]##.
 

MathematicalPhysicist

Gold Member
4,052
122
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
1,050
337
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:
1,682
50
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
1,050
337
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?
 

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