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I am given the characteristic polynomial $x^2(x^2-1)$ and am asked to find all possible jordan canonical forms. What I have so far is:

Possible elementary divisors are: 1) $x,x,(x+1),(x-1)$, 2) $x,x,(x+1)(x-1)$, 3) $x^2,(x+1)(x-1)$, and 4) $x^2(x+1)(x-1)$. I therefore got the possible Jordan forms as:

1)=2) \begin{matrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{matrix}

and

3)=4) \begin{matrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \end{matrix}

I'm really unsure if these are correct, so any insight would be greatly appreciated!

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  • $\begingroup$ The use of the term elementary divisors suggests you're looking for a way to relate JNF with smith normal form. Is this the case? $\endgroup$
    – Git Gud
    Mar 12, 2014 at 15:04

1 Answer 1

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They are correct. What the characteristic polynomial $$x^2(x-1)(x+1)$$tells you is that there are at least $3$ cages, of which the cages for eigenvalues $1$ and $-1$ have size $1$. This only leaves the eigenvalue $0$, for which you have $2$ options:

  1. There are $2$ linearly independent eigenvectors for the eigenvalue $0$. In that case, your matrix has $4$ eigenvectors and can be diagonalized.
  2. There is only $1$ linearly independent eigenvector for $0$. This means that the cage for $0$ is $2\times 2$ and the Jordan form is as your second matrix suggests.

Of course, all this is done up to permutation, meaning that you can change the order of the eigenvalues on the diagonal and still get a Jordan form.

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  • $\begingroup$ While basically correct, you should avoid saying "there are $2$ eigenvectors...". There are always infinitely many eigenvectors. $\endgroup$ Mar 12, 2014 at 15:10
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    $\begingroup$ How about "$2$ linearly independent eigenvectors"? $\endgroup$ Mar 12, 2014 at 15:13
  • $\begingroup$ Corrections noted. Thank you. $\endgroup$
    – 5xum
    Mar 12, 2014 at 15:38

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