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Minimal Polynomial

  1. Mar 25, 2009 #1
    Here is my problem:

    Let A be a complex n x n matrix with minimal polynomial q(x)=the sum from j=0 to m of [tex]\alpha_j x^j[/tex] where [tex]m\leq n[/tex] and [tex]\alpha_m[/tex] = 1.

    Show: If A is non-singular then [tex]\alpha_0[/tex] does not equal 0.

    So, I get that 0=q(A)=[tex]\alpha_0 I_n + \alpha_1 A + \alpha_2 A^2 +...+A^m[/tex], but I'm not sure what to do here. I assume we will have to use the fact that A is non-singular, but I'm not sure how. Does it maybe involve multiplying both sides by x on the right side? Any hints would be much appreciated! :)
     
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  3. Mar 25, 2009 #2

    Office_Shredder

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    Assume a0 is zero. See if you can find a way to reduce the degree of the minimal polynomial such that A is still a zero
     
  4. Mar 25, 2009 #3
    So, first, I've tried rewriting q(A) like this:
    0=[tex]\alpha_0 I_n+(\alpha_1+\alpha_2 A+...+A^{m-1})A[/tex]

    Then, if [tex]\alpha_0[/tex] is 0, then 0=[tex](\alpha_1+\alpha_2 A+...+A^{m-1})A[/tex].

    So, if we multiply both sides by a nonzero vector x, we have:
    0=[tex](\alpha_1+\alpha_2 A+...+A^{m-1})Ax[/tex].
    But since x is nonzero and A is nonsingular, this cannot equal 0. Therefore, [tex]\alpha_0[/tex] must equal 0. Is my reasoning correct? I'm still not sure. Thanks!
     
  5. Mar 25, 2009 #4

    Office_Shredder

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    You don't know that [tex] \alpha_1 + ... + A^{m-1}[/tex] is non-singular. Instead of applying both operators to a vector, try multiplying on the right by A-1 and see what you have
     
  6. Mar 25, 2009 #5
    Well, then [tex]\alpha_1 + \alpha_2 A+...+A^{m-1}[/tex]=0. But I don't see how that helps because if we are assuming that [tex]\alpha_0[/tex] is 0, and then [tex]\alpha_1 + \alpha_2 A+...+A^{m-1}[/tex]=0, q(A) still equals 0, so it would give us what we want. I must be missing something :confused:
     
  7. Mar 27, 2009 #6
    Hey guys, I'm still having problems with this. As Office_Shredder suggested, I assume [tex]\alpha_0[/tex] is 0, and then multiply by the inverse of A on the right side of each side which leaves me with:

    0=[tex]\alpha_1 + \alpha_2 A+...+A^{m-1}[/tex], but what to do with this I can't seem to see.
     
  8. Mar 27, 2009 #7
    Oh!!! Does this last equation set equal to 0 imply that the there would have to be a polynomial of lower degree (m-1)? Which would be impossible because we were talking about the minimal polynomial to begin with.
     
  9. Mar 27, 2009 #8
    Yea, that last statement sounds about right.
     
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