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The minimal polynomial

  1. Jun 16, 2009 #1
    I'm just learning a bit about the "minimal polynomial" today but there was a section from the book which I didn't understand. This is the section, and i've circled the bit i'm having trouble with.

    http://img15.imageshack.us/img15/1825/97503873.jpg [Broken] (sorry, it won't let me post an image for some reason??)

    Firstly it's a bit unclear to me what they mean by p(T)(v). Would this mean that you take the linear transformation T (or equivalently its matrix), stick it in the polynomial p to obtain a new linear transformation p(T), then perform this transformation on v?

    Okay, assuming that's correct I can understand that p(T)=0 <=> p(T)(v)=0. But then how does this imply that the minimal polynomial is the least common multiple of all those other ones?! They say it like it's completely obvious!

    Thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jun 16, 2009 #2
    Your interpretation of p(T) is correct. Take note of this, since it's a rather useful technique. As for the least common multiple thing, how obvious it is depends on how much algebra you've been doing lately. Your text probably should have supplied a proof, unless they already have results to that effect.

    I'll adapt the notation given in your text: let bi be a basis for V and ui be corresponding minimal polynomials killing each bi (bear with me, I don't know how to type this properly! I'll always use "i" as an index). We need to prove two directions. First, assume ui divides p for all i, say (fixing i) p=f*gi. Then p(T)(v)=f(T)*gi(T)(v)=f(T)(0)=0, so p(T) kills all of V. Conversely, say bi does not divide p, but p(bi)=0. Then p=f*ui+r, for some f and r with degree less than that of ui. But then p(T)(bi)=r(bi)=0, contradicting the minimality of ui.
     
    Last edited: Jun 16, 2009
  4. Jun 16, 2009 #3
    These posts are written in a hurry. If anything needs clarification, please ask. Above, I used two facts implicitly: the result given as an excercise immediately below the red box in the scan, and the fact that (f*g)(T)=f(T)*g(T). These are both easy.
     
  5. Jun 17, 2009 #4

    Hurkyl

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    That's not correct.

    p(T)=0 <=> { p(T)(v) = 0 for all v }
     
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