# (distinct eigenvalues) help!

1. Dec 5, 2005

### JerryKelly

Let A be an nxn mx with n distinct eigenvalues and let B be an nxn mx with AB=BA. if X is an eigenvector of A, show that BX is zero or is an eigenvector of A with the same eigenvalue. Conclude that X is also an eigenvector of B.

I could show BX is zero or is an eigenvector of A with the same eigenvalue, but i dont know how to Conclude that X is also an eigenvector of B. Does anyone know how to do it? Thanks!

2. Dec 5, 2005

### JasonRox

Why don't we ever get neat questions like that?

I have an idea how to tackle this, but I can't help you at this moment.

3. Dec 5, 2005

### JerryKelly

why you cannot help me?

4. Dec 6, 2005

### matt grime

The answer follows from all the definitions given in one line:

If BX=0 we're done, if not to show BX is an eigenvalue of A we consider ABX and use the information in the question.

5. Dec 6, 2005

6. Dec 6, 2005

### HallsofIvy

Staff Emeritus
Yes, that part he said he could do. The remaining problem is to show that X is in fact an eigenvector of B.

You haven't used the fact that A has n distince eigenvalues. If that is true then there exist a basis for the vector space consisting of eigenvectors of A.

7. Dec 6, 2005

### Hurkyl

Staff Emeritus
How many eigenvectors does A have for any particular eigenvalue?

8. Dec 6, 2005

### matt grime

But it also proves the rest of the question (admittedly it is a trivial observation since any question that is doable is doable because of the information in the question, but here it is a case of follow your nose). We have proved B preserves (generalized) eigenspaces (of A) which are all 1-d according to the information in the question, thus answering the last part.

9. Dec 6, 2005

### HallsofIvy

Staff Emeritus
Since matt grime is about to burst trying to give clues without giving the whole thing, I'm going to give up and spell it out in "donkey steps".

Given A and B are n by n matrices with AB= BA.
1. If x is an eigenvector of A with eigenvalue $\lambda$ then Bx is also also an eigenvector of A with eigenvalue $\lambda$.
(A(Bx))= (AB)x= (BA)x= B(Ax)= B($\lambda$x)= $\lambda$(Bx))

2. If lambda is an eigenvalue of A then the set of all vectors x such that Ax= $\lambda$x forms a subspace (often called the "eigenspace" of $\lambda$). Further the eigenspaces of two distinct eigenvalues have only 0 in common.

3. Since A is an n by n matrix it is on an n dimensional vector space.

4. Since A has n distinct eigenvalues, each eigenspace has dimension 1.

5. If two vectors are in the same one-dimensional subspace then one is a multiple of the other.

6. Since x and Bx are in the same one-dimensional eigenspace, Bx is a multiple of x.

Last edited: Dec 7, 2005
10. Dec 6, 2005

### Hurkyl

Staff Emeritus
Hey, no doing the whole problem for the OP!

11. Dec 6, 2005

### JerryKelly

Thanks for help,HallsofIvy! It made perfert sence! for the first step, it is so similar what i was doing. I was doing ABx=BAx=B$\lambda$x=$\lambda$Bx.
since A(Bx)=$\lambda$(Bx)
Bx=0 or Bx is eigenvector.

12. Dec 6, 2005

### JerryKelly

your way is seeing better than my way. Thanks,agian! It is very helpful.

13. Dec 7, 2005

### HallsofIvy

Staff Emeritus
Mea Culpa, Mea Culpa, Mea Maxima Culpa.