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

[mind0nmath]

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.

 

[mathwonk]

do you know what a characteristic polynomial is?

 

[Marco_84]

how can u prove that?
S T must sutysfy more conditions... that u didint typed?

wich ones??

 

[mind0nmath]

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.

 

[Marco_84]

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

 

[HallsofIvy]

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?

 

[HallsofIvy]

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]"!

 

[mind0nmath]

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.

 

[HallsofIvy]

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.

 

[mathwonk]

i seem to have constructed counter examples from the simplest non trivial 3x3 jordan forms.