Given S, T, prove that ST and TS have the same eigenvalues.

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Discussion Overview

The discussion revolves around proving that the products of two linear operators, ST and TS, have the same eigenvalues. Participants explore the conditions under which this holds, the definitions involved, and the implications of the proof in the context of linear algebra and quantum mechanics.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant requests assistance in starting the proof regarding the eigenvalues of ST and TS.
  • Another participant inquires about the concept of a characteristic polynomial, suggesting it may be relevant to the proof.
  • Concerns are raised about the conditions that S and T must satisfy for the proof to hold, with a request for clarification on these conditions.
  • A participant proposes a method involving arbitrary vectors and eigenvalues to demonstrate the relationship between ST and TS, but this is challenged on the grounds that it assumes the existence of eigenvalues for S and T.
  • There is a mention of the commutator [S, T] and its relevance, with one participant asserting that it is useful in quantum mechanics.
  • Another participant clarifies that linear operators on complex vector spaces always have eigenvalues, while those on real vector spaces may not, which introduces a potential limitation to the proof.
  • A participant outlines a reasoning process to show that if λ is an eigenvalue for ST, it can be shown that λ is also an eigenvalue for TS, provided certain conditions about the vectors involved are met.
  • One participant claims to have constructed counterexamples using Jordan forms, indicating that the discussion may not reach a consensus on the proof's validity.

Areas of Agreement / Disagreement

Participants express differing views on the existence of eigenvalues for linear operators, particularly in relation to the underlying field (real vs. complex numbers). There is no consensus on the proof's validity, with some proposing methods while others raise challenges and counterexamples.

Contextual Notes

Limitations include the dependence on the definitions of eigenvalues and the conditions under which S and T operate. The discussion also highlights the potential for counterexamples in specific cases, particularly with non-trivial Jordan forms.

mind0nmath
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I need help starting/doing this proof.
Suppose S,T are Linear Operators on a Finite Dimensional Vector Space V. Prove that ST and TS have the same eigenvalues.
A linear operator is a linear map from a vector space to itself.
thanks.
 
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do you know what a characteristic polynomial is?
 
how can u prove that?
S T must sutysfy more conditions... that u didint typed?

which ones??
 
thanks but I think I have the answer. If you pick an arbitrary vector in V and define T(u) = au and S(u) = bu, where a,b are eigenvalues for u, then applying S to T(u) and T to S(u) will give the wanted results.
 
mind0nmath said:
thanks but I think I have the answer. If you pick an arbitrary vector in V and define T(u) = au and S(u) = bu, where a,b are eigenvalues for u, then applying S to T(u) and T to S(u) will give the wanted results.

yes and what is the answer?

[S,T]=0

think abou it... very usefull in QM
ciao
 
mind0nmath said:
thanks but I think I have the answer. If you pick an arbitrary vector in V and define T(u) = au and S(u) = bu, where a,b are eigenvalues for u, then applying S to T(u) and T to S(u) will give the wanted results.
But you can't "define" that because you want to show it is true for any T and S. What if T or S don't have eigenvalues?
 
Marco_84 said:
yes and what is the answer?

[S,T]=0

think abou it... very usefull in QM
ciao
And completely meaningless since you haven't bothered to define "[S, T]"!
 
isn't it true that all linear operators on a finite dimensional vector space have eigenvalues? I don't really know what [S, T] is though.
 
No, it is not. Linear operators on a vector space over the complex numbers always have eigenvalues. Linear operators on a vector space over the real numbers may not have (real) eigenvalues.

I took Marco84 to task for not defining it [S, T]. The standard definition is [S, T]= ST- TS but I really don't see how it will help here. ST and TS always have the same eigenvalues but not the same eigenvectors!

Suppose [itex]\lambda[/itex] is an eigenvalue for ST. That is, there is some non-zero vector v such that STv= [itex]\lambda[/itex]v. Let u= Tv. Then Su= STv= [itex]\lambda[/itex]v. What happens if you apply T to Su? That shows that [itex]\lambda[/itex] is an eigenvalue for TS as long as u is not 0. Now suppose u= 0.
 
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  • #10
i seem to have constructed counter examples from the simplest non trivial 3x3 jordan forms.
 

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