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Characteristic polynomial

  1. May 11, 2012 #1
    "Given operators σ,τ on a finite-dimensional space V, show that στ=i, and that σ=p(τ) for some polynomial p in F[x]."

    The first part was no problem. As for the second, I have a strong suspicion that p is the characteristic polynomial, mostly because I believe I heard of that fact before (that a matrix inserted into its own characteristic polynomial it its inverse). However, I can't seem to find anything about that, and furthermore, the characteristic polynomial has not yet been mentioned in the text I'm using.

    Any idea how the proof should proceed?
     
  2. jcsd
  3. May 11, 2012 #2

    HallsofIvy

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    What is "i", the identity operator? Unless there are conditions on σ and τ you haven't mentioned, that simply isn't true.
     
  4. May 11, 2012 #3
    Yes, i is the identity. Here is the proof that I used:

    We know that
    dim V = rk i ≤ min { rk σ, rk τ },
    So both σ and τ have rank dim V, and so both are isomorphisms.

    Where did I go wrong here?

    It's from Steven Roman's Advanced Linear Algebra, I believe 1st edition. I'm told there are many many errors, and can't seem to find any errata online. In any case, I'm much more interested in the second part.
     
  5. May 11, 2012 #4


    You cannot prove something wrong unless you cheat. As you wrote your OP the conclusion is simply false. Check and correct this.


    DonAntonio
     
  6. May 11, 2012 #5
    Oh my, that wa a stupid mistake. I must not have been paying attention as i was typing. It was supposed to be:

    (sigma composed with tau equals i) implies (sigma and tau are both invertible).

    Apologies for the mixup. In either case DonAntonio, what you said about proofs is not at all correct: every sentence has a negation.
     
  7. May 12, 2012 #6


    So?? It still is true you cannot prove something wrong unless you cheat. Of course, this must be understood under the usual, standard

    assumption: that one has a consistent axiomatic system, with the usual logical rules and etc.

    DonAntonio
     
  8. May 12, 2012 #7


    So you want to prove [itex]\,\,A,B\in GL(V)\,\,,\,\,\dim V<\infty\,\,,\,\,AB=I\Longrightarrow B=p(A)\,\,,\,\,p(x)\in\mathbb{F}[x]\,\,,\,\,\mathbb{F}=\,[/itex] the field of definition of V.

    Well, you were then on the right track: as A (And also B, of course) is non-singular, its charac. polynomial[itex]\,\,p_A(x)\,\,[/itex] has non-zero free coefficient,

    so if we write [itex]\,\,p_A(x)=x^n+a_{n-1}x^{n-1}+...+a_1x+a_0\,\,,\,\,a_0\neq 0\Longrightarrow \,\,[/itex] , by the Cayley-Hamilton Theorem we

    have the operator (matricial) equality [itex]\,\,p_A(A)=0\,\,[/itex], we'll get that [itex]\,\,I=-\frac{1}{a_0}A^n-...-\frac{a_1}{a_0}A\Longrightarrow\,\,[/itex] ...finish the argument

    DonAntonio
     
  9. May 12, 2012 #8
    My point is that every time one proves "p," one disproves "not p." In fact, it's only when one has an INconsistent system that this fails, not the other way around. In any event, do you have any idea about the polynomia question?
     
  10. May 12, 2012 #9

    1) Well, it looks like you continue to misunderstand what I meant, so let's leave it

    2) You rushed so much to answer the above that you didn't see my answer to your OP.

    DonAntonio
     
  11. May 12, 2012 #10
    I'd really like to know what you meant by that, if you wouldn't mind giving further explanation.

    And apologies for missing the response, I was replying through e-mail since I wasn't at my computer, and it would only display the first intervening post. However, I don't think that's the proof I'm seeking: the book hasn't yet made any mention of the characteristic polynomial or the Cayley-Hamilton theorem, so I'm guessing there's a more subtle, elementary proof somewhere. Or else it may have just been another error; like I said, I've spotted more than a few in my edition of the text.

    Thanks anyways!
     
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