Show that Characteristic polynomial = minimal polynomial

  • #1

Homework Statement


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.


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!
 

Answers and Replies

  • #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.
 
  • #3
Oh, ok. Thanks!
 
  • #4
Dick
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Oh, ok. Thanks!

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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