Distinct Eigenvalues and Eigenvectors in Matrix Multiplication

[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 don't know how to Conclude that X is also an eigenvector of B. Does anyone know how to do it? Thanks!

 

[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.

 

[JerryKelly]

why you cannot help me?

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.

 

[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.

 

[Muzza]

But that was what JerryKelly had already shown...

 

[HallsofIvy]

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.

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.

 

[Hurkyl]

How many eigenvectors does A have for any particular eigenvalue?

 

[matt grime]

But that was what JerryKelly had already shown...

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.

 

[HallsofIvy]

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.

 

[Hurkyl]

Hey, no doing the whole problem for the OP!

 

[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.

 

[JerryKelly]

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

 

[HallsofIvy]

Hey, no doing the whole problem for the OP!

Mea Culpa, Mea Culpa, Mea Maxima Culpa.