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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!

    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


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    That's not correct.

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