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Show that Characteristic polynomial = minimal polynomial

  1. Mar 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Let A = [itex]\begin{pmatrix}1 & 1 & 0 & 0\\-1 & -1 & 0 & 0\\-2 & -2 & 2 & 1\\ 1 & 1 & -1 & 0 \end{pmatrix}[/itex]

    The characteristic polynomial is [itex]f(x)=x^2(x-1)^2[/itex]. Show that f(x) is also the minimal polynomial of A.

    Method 1: Find v having degree 4.
    Method 2: Find a vector v of degree 2, whose minimal polynomial (on A of v) is [itex]x^2[/itex], and another, w, whose minimal polynomial is [itex](w-1)^2[/itex]. Or, just show that v and w exist.


    3. The attempt at a solution
    I'm confused as to how a vector can have a degree of more than 1. Isn't a vector simply:
    [itex]v= \begin{pmatrix}a\\b\\c\\...\\n\end{pmatrix}[/itex] in [itex]R^n[/itex]? I think I can get the question once I understand this. Thanks!
     
  2. jcsd
  3. Mar 20, 2012 #2

    Dick

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    I think what they mean by a vector v having degree 4 is that A^4(v)=0 but A^3(v) is not equal to zero.
     
  4. Mar 21, 2012 #3
    Oh, ok. Thanks!
     
  5. Mar 21, 2012 #4

    Dick

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    You're welcome but rereading that I'm not sure what the hint really means. Since f is the characteristic polynomial you know f(A)=A^2(A-1)^2=0. To show it's minimal you need to show A(A-1)^2 and A^2(A-1) are not zero. The 'degree of a vector' seems pretty unclear to me as well.
     
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