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?

which 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 \lambda is an eigenvalue for ST. That is, there is some non-zero vector v such that STv= \lambdav. Let u= Tv. Then Su= STv= \lambdav. What happens if you apply T to Su? That shows that \lambda 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.