# Linear Algebra Proof

1. Mar 12, 2013

### blueberryfive

1. The problem statement, all variables and given/known data
Suppose S,T ∈ L(V) and S is invertible. Prove that if p ∈ P(F) is a polynomial, then p(S*T*S-1)=S*p(T)*S-1.

2. Relevant equations

none

3. The attempt at a solution

Suppose by contradiction that for any p ∈ P(F),

p(S*T*S-1)≠S*p(T)*S-1 for any p ∈ P(F).

Since this is true for any p∈ P(F), let p=1x ∈ P(F). Then

1*(S*T*S-1)≠S*(1)*(T)*S-1

This implies that T≠T, a contradiction. Therefore if p ∈ P(F) is a polynomial, then p(S*T*S-1)=S*p(T)*S-1.

Last edited: Mar 12, 2013
2. Mar 12, 2013

### Staff: Mentor

This is difficult to read due to its terseness. It's difficult to tell that Sp(T)S-1 means S * p(T) * S-1, as opposed to Sp(T), which might be (incorrectly) interpreted as the span of T.
Don't make us have to work so hard to translate what you are writing. I'm reasonably sure that C! is supposed to be an abbreviation for contradiction, but why should we have to work so hard?
Why is p = 1? All you're given is that p is polynomial over some field F.

3. Mar 12, 2013

### blueberryfive

I've redone it

4. Mar 12, 2013

### Staff: Mentor

That's a lot better - thanks!

If p = 1, then isn't p(STS-1) = 1? IOW, it wouldn't be 1 * STS-1, as you have.

I'm still struggling to make sense of this.

You have p $\in$ P(F) is a polynomial. I assume F is some field. How does it make sense to write p(STS-1)? Are we talking about polynomials in powers of S or T or the like?

Also, and this might seem like a dumb question, but what are S and T? Are they linear operators? Square matrices?

5. Mar 12, 2013

### blueberryfive

I'm sorry, you're absolutely correct. I was not thinking (x)=1, but p(x)=1x. Then p(STS-1)=1*p(STS-1).

S and T are both in L(V), the set of all linear transformations from V to V (V is a finite-dimensional vector space). F denotes ℝ or ℂ.

I'm not sure how to formally answer that question, but say p(x)=1+x+x2. Then p(T)=I +T+T2, where I is the identity transformation. Hence, p(T)v = Iv + Tv + T2v.

6. Mar 12, 2013

### Staff: Mentor

I would go at this directly, rather than by contradiction. p is any polynomial, so it has some representation, say p(X) = A0 + A1X + A2X2 + ... + AnXn.

Now expand p(STS-1) and show that it can be rewritten as S*p(T)*S-1. The trick is to show that, for example, A3(STS-1)3 can be simplified to SA3T3S-1. You might have to do a simple induction proof along the way.

7. Mar 12, 2013

### Fredrik

Staff Emeritus
This is really weird. I'm sure I wrote the first reply to this post yesterday. Was there multiple copies of this thread and one got deleted? Maybe I just wrote the reply and never submitted it.

The main point of what I wrote was that you seem to think that the negation of
For all p, we have $p(STS^{-1})=Sp(T)S^{-1}$.​
is
For all p, we have $p(STS^{-1})\neq Sp(T)S^{-1}$.​
It's not. The correct negation is
There exists a p such that $p(STS^{-1})\neq Sp(T)S^{-1}$.​
Because of this, you can't just choose p=1.

8. Mar 12, 2013

### Fredrik

Staff Emeritus
Or the trace of T. That's what I thought this meant at first. I think the Germans write Sp instead of Tr.

9. Mar 12, 2013

### Zondrina

Suppose that your polynomial looks like this :

$p(z) = a_0 + a_1z + a_2z^2 + ... + a_nz^n$.

You also have that S and T are linear transformations ( homomorphisms in fact ) and that S is invertible ( even stronger since S is an isomorphism now as well ).

You can't multiply the transformations together... you can compose them through the regular function composition though.

So $p(STS^{-1}) = ... = Sp(T)S^{-1}$

10. Mar 13, 2013

### Staff: Mentor

The OP also posted this problem in the Math technical section, under Linear Algebra. I have locked that thread.

11. Mar 13, 2013

### Fredrik

Staff Emeritus
I don't see it under linear algebra and it doesn't show up in a search. You must have deleted it when you meant to lock it, or someone else deleted it.

12. Mar 13, 2013

### Staff: Mentor

micromass came along after I did, and deleted the thread.

13. Mar 19, 2013

### blueberryfive

Thanks Fredric that's all I wanted to know!

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