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Math Help - Rank and characteristic/minimal Polynomial

  1. #1
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    Rank and characteristic/minimal Polynomial

    I am having trouble with this question and need some help.

    Let A be a square matrix of order n over F with a rank of 1.

    1. Show that the characteristic polynomial of A is P(t)=t^{n}-tr(A)t^{n-1} and find the minimal polynomial of A.
    2. Let T:F^{n}\rightarrow F^{n} be the linear map defined by Tv=Av for every v\in F^{n}. Show that for every 1\leq k\leq n there exists a subspace of F^{n} of dimension k that is T-reserved.

    I can show the first part of 1. If the rank is 1 then the nullity of A is of dimension n-1 and therfore the geometric multiplicity of the eignvalue 0 is n-1, so we get that the algebraic multiplicity of 0 is at least n-1. Therfore the characteristic polynomial is of the form t^{n-1}(t-a)=t^{n}-at^{n-1} (where a is the other eignvalue). We have shown before in a previous lin alg class that the coefficient of t^{n-1} of the characteristic polynomial of any square matrix of order n is -tr(A), so we get then that the characteristic polynomial is indeed t^{n}-tr(A)t^{n-1}.

    Now for the minimal polynomial I know that it has to be of the form t^{k}(t-tr(A)) where k\leq n-1, but I am not sure how to proceed.

    As for 2 I would appreciate some direction.


    Thanks,

    SK
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    We have

    \dim (\ker A)=n-\textrm{r}(A)=n-1.

    This implies that \lambda=0 is an eigenvalue of A.

    The algebraic multiplicity of \lambda=0 can't be n (why?).

    So, the characteristic polynomial is

    \chi(t)=t^{n-1}(t-a),\;\;(a\neq 0).

    On the other hand:

    \chi(t)=t^n-(\textrm{tr}A)t^{n-1}+\ldots\Rightarrow a=\textrm{tr}A

    As \lambda=a is simple then, \dim\ker(A-aI)=1 , so A is diagonalizable on F .

    Then, the minimal polynomial \mu(t) is:

    \mu(t)=t(t-\textrm{tr}A) .


    Fernando Revilla


    Edited: Sorry, now I see you solved the first part.
    Last edited by FernandoRevilla; January 18th 2011 at 10:21 AM.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by skyking View Post
    Show that for every 1\leq k\leq n there exists a subspace of F^{n} of dimension k that is T-reserved.
    What does T-reserved mean?.


    Fernando Revilla
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  4. #4
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    [QUOTE=FernandoRevilla;606818]What does T-reserved mean?.

    First thanks, for the reply.

    I wasn't sure I was translating this correctly, I wasn't sure what the term in english is. But if W is " T-reserved" it means that for every u\in W we get Tu\in W.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by skyking View Post
    I wasn't sure I was translating this correctly, I wasn't sure what the term in english is. But if W is " T-reserved" it means that for every u\in W we get Tu\in W.
    All right, then you meant T-invariant subspace .

    Hint :

    Use the fact that in a determined basis B of F^{n} the matrix of T is:

    D=\textrm{diag}(a,0,\ldots,0)


    Fernando Revilla
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