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Math Help - trace, jordan carnonical form

  1. #1
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    trace, jordan carnonical form

    Can anyone help me showing that

    a)Let A is a realnumber matrix nxn. Show that tr (A) =sum of all eigenvalues of A.
    b)Let A is a realnumber matrix nxn. Show that det (exp (A)) = det (exp (J)) where J is
    the Jordan canonical form of A.
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: trace, jordan carnonical form

    We know that \mbox{Tr(A)}=\sum_{i=1}^{n}a_{ii}=a_{11}+a_{22}+..  .a_{nn}

    Now, have you any idea about the matrix of the linear map? (I would choose a base wherefore the matrix of the linear map is a diagonal matrix)
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  3. #3
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    Re: trace, jordan carnonical form

    I know TAT^-1 is D and D is the diagonal matrix whose diagonal entries are the eigenvalues of A. Hmm. I think I kind of getting it but I think I need a smoother answer please!
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Re: trace, jordan carnonical form

    Siron's answer is correct for diagonalizable matrices in \mathbb{R} . In general:

    Quote Originally Posted by mathsohard View Post
    a)Let A is a realnumber matrix nxn. Show that tr (A) =sum of all eigenvalues of A.
    Use that similar matrices have the same trace and A is similar to its canonical form of Jordan (triangular matrix whose elements in the diagonal are exactly the eigenvalues of A ) .

    b)Let A is a realnumber matrix nxn. Show that det (exp (A)) = det (exp (J)) where J isthe Jordan canonical form of A.
    Use that if P^{-1}AP=J then, e^A=Pe^JP^{-1} .
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  5. #5
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    Cool Re: trace, jordan carnonical form

    Quote Originally Posted by FernandoRevilla View Post
    Siron's answer is correct for diagonalizable matrices in \mathbb{R} . In general:



    Use that similar matrices have the same trace and A is similar to its canonical form of Jordan (triangular matrix whose elements in the diagonal are exactly the eigenvalues of A ) .

    Use that if P^{-1}AP=J then, e^A=Pe^JP^{-1} .
    Said well, support you !

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