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Linear Algebra Proof

  • #1

Homework Statement


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.


Homework Equations



none

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:

Answers and Replies

  • #2
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5,287

Homework Statement


Suppose S,T ∈L(V) and S is invertible. Prove that if p∈P(F) is a polynomial, then p(STS-1)=Sp(T)S-1.
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.

Homework Equations



none

The Attempt at a Solution


Suppose by C!
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?
that p(STS-1)≠Sp(T)S-1. Since p=1∈P(F),
Why is p = 1? All you're given is that p is polynomial over some field F.
1(STS-1)≠S1(T)S-1

Multiplying through by S and S-1, we see that T≠T.

Does this work?
 
  • #4
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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
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
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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
Fredrik
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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
Fredrik
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Sp(T), which might be (incorrectly) interpreted as the span of T.
Or the trace of T. That's what I thought this meant at first. I think the Germans write Sp instead of Tr.
 
  • #9
STEMucator
Homework Helper
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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
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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 OP also posted this problem in the Math technical section, under Linear Algebra. I have locked that thread.
 
  • #11
Fredrik
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The OP also posted this problem in the Math technical section, under Linear Algebra. I have locked that thread.
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
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micromass came along after I did, and deleted the thread.
 
  • #13
Thanks Fredric that's all I wanted to know!
 

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