# Another proof

1. Jun 11, 2009

### evilpostingmong

Another proof....

1. The problem statement, all variables and given/known data

Suppose c is an eigenvalue of a square matrix A with eigenvector X=/=0.
Show that p(c) is an eigenvalue of p(A) for any nonzero polynomial p(x).

2. Relevant equations

3. The attempt at a solution
Knowing that c is an eigenvalue of A, it is true that AX=cX.

p(A)X=a0(X)+a1AX+...+anA^nX
And p(c)X=a0(X)+a1cX+...+anc^nX.
Since AX=cX, A^kX=c^kX.
So if AX-c^kX=0, A^kX-c^kX=0.
Now to prove p(A)X-p(c)X=0,
(AX-cX)a1+...+(A^nX-c^nX)an=(0)*a1+...+(0)*an=0

2. Jun 11, 2009

### CompuChip

Re: Another proof....

Can't you just take
p(A)X=a0(X)+a1AX+...+anA^nX
and recursively replace Ax by cx (and pull all the c's to the front) to get
p(A)X=a0 X + a1 c X + ... + an c^n X = p(c) X?

3. Jun 11, 2009

### evilpostingmong

Re: Another proof....

Yeah, I killed a mouse with an A-bomb, instead of a knife, so to speak.

4. Jun 11, 2009

### Dick

Re: Another proof....

That works. You might think about writing it like p(A)X=a0*(X)+a1*(AX)+...+an*(A^nX)=a0(X)+a1*c*X+...an*c^n*X=(a0+a1*c+...+an*c^n)X=p(c)X, showing p(A)X=p(c)X directly rather than showing p(A)X-p(c)X=0. But that's a just a question of taste. Your proof is fine. But the last line should be, "Since p(A)X=p(c)X we see X is a eigenvector of p(A) with eigenvalue p(c)." Since that's what they actually want you to prove.

5. Jun 11, 2009

### evilpostingmong

Re: Another proof....

Alright, gotta finish up that way.

6. Jun 11, 2009

### Dick

Re: Another proof....

Right. It always helps to explicitly say why all of the messing around you just did proves what they want you to prove. Soon you will be writing EXCELLENT proofs.

7. Jun 11, 2009

### evilpostingmong

Re: Another proof....

Thank you Dick, I really think that looking at someone's answer won't help nearly
as much as taking an active role, and yeah, God forbid I write a textbook like that.
Gotta get my point across besides using some algorithmic approach.
I hate books that go like this
PROOF:
$$\sum$$$$\sum$$f(x)$$\otimes\ominus$$$$\subset$$A
clearly A$$\subseteq$$G thus A=G which is clear. But f(x)=/=$$\Gamma$$
$$\Delta$$ thus f(x)=$$\Psi$$ as given. Then I forget what the author was
proving in the first place.

8. Jun 11, 2009

### Office_Shredder

Staff Emeritus
Re: Another proof....

That is easily the best parody proof I've ever read. Good work.

Since you've got the answer to the thread, here's a tangent question that should be easy, but sometimes gets skipped over without consideration: In the OP question, what is p(A) if p(x) = 2?

9. Jun 11, 2009

### evilpostingmong

Re: Another proof....

2 is the a0 here, since the proof says that p(A)X=a0X+a1AX+...+anA^nX,
p(A)=2I

10. Jun 11, 2009

### Dick

Re: Another proof....

Sure. If p(x)=2, P(A)=2I. Since A^0=I. I'm not exactly sure why Office_Shredder asked that. You are getting a lot better at this, evilpostingmong.

11. Jun 11, 2009

### evilpostingmong

Re: Another proof....

Thank you! I want to get better at proofs, I think they're fun. I'm not sure why he asked
it either, to be honest.

12. Jun 11, 2009

### Dick

Re: Another proof....

They are fun. That's why we do them.

13. Jun 12, 2009

### CompuChip

Re: Another proof....

It was a bit trivial. Maybe the point was to stress that p(x) is an "ordinary" polynomial in the sense that you plug in a number and get a number, as you are used to (apart from some subtleties, of course ); but when you apply it to a matrix like p(A) you always get a matrix. Even when p(x) is a constant function -- i.e., proportional to the identity number 1 -- p(A) gives a matrix -- i.e., proportional to the identity matrix I.

Definitely. As I said in an earlier thread: doing a lot of proofs and asking others to be very critical is the best way to quickly make improvements.