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

• mind0nmath
In summary: When ST is applied to u, STv still equals \lambdav. This shows that ST has an eigenvalue for u. But what about when S is applied to u? That doesn't change anything. u still equals \lambdav. This shows that S has an eigenvalue for u. So it seems that ST and S have the same eigenvalue for any vector u. But what if u is zero?I seem to have constructed counter examples from the simplest non trivial 3x3 jordan forms.When ST is applied to u, STv still equals \lambdav. This shows that ST has an eigenvalue for u. But what about when S is applied
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.

do you know what a characteristic polynomial is?

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

wich 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 $\lambda$ is an eigenvalue for ST. That is, there is some non-zero vector v such that STv= $\lambda$v. Let u= Tv. Then Su= STv= $\lambda$v. 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.

Last edited by a moderator:
i seem to have constructed counter examples from the simplest non trivial 3x3 jordan forms.

## 1. What is the meaning of "eigenvalues"?

Eigenvalues are a set of numbers associated with a square matrix that represent the scalar values by which a particular vector is scaled when the matrix is applied to it. They are important in understanding the behavior and properties of linear transformations.

## 2. What does it mean for two matrices to have the same eigenvalues?

If two matrices, S and T, have the same eigenvalues, it means that when these matrices are applied to a vector, they will produce the same set of scaled vectors. In other words, the eigenvectors of S and T are the same.

## 3. What is the importance of proving that ST and TS have the same eigenvalues?

Proving that ST and TS have the same eigenvalues is important because it helps us understand the relationship between the two matrices. It also allows us to simplify computations and make predictions about the behavior of these matrices.

## 4. Can you provide an example to demonstrate that ST and TS have the same eigenvalues?

Yes, for example, let S be a 2x2 matrix with eigenvalues 2 and 3, and T be a 2x2 matrix with eigenvalues 5 and 6. We can see that when we multiply these matrices in either order, the resulting matrix will have the same eigenvalues of 10 and 18. This demonstrates that ST and TS have the same eigenvalues.

## 5. Are there any exceptions to the rule that ST and TS have the same eigenvalues?

Yes, there are exceptions to this rule. It is possible for two matrices to have the same eigenvalues, but when multiplied, the resulting matrix does not have the same eigenvalues. This can occur when the matrices do not commute, meaning that the order in which they are multiplied matters. However, it is always true that if two matrices commute, then they will have the same eigenvalues.

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