Therefore, p(A)X=p(c)X.This proves that p(c) is an eigenvalue of p(A).

[evilpostingmong]

Another proof...

Homework Statement

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).

Homework Equations

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

 

[CompuChip]

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?

 

[evilpostingmong]

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?

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

 

[Dick]

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.

 

[evilpostingmong]

Alright, got to finish up that way.

 

[Dick]

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.

 

[evilpostingmong]

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.

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.

 

[Office_Shredder]

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.

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?

 

[evilpostingmong]

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?

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

 

[Dick]

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.

 

[evilpostingmong]

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.

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.

 

[Dick]

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

 

[CompuChip]

Sure. If p(x)=2, P(A)=2I. Since A^0=I. I'm not exactly sure why Office_Shredder asked that.

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.

You are getting a lot better at this, evilpostingmong.

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.