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Math Help - Characteristic Polynomial

  1. #1
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    Characteristic Polynomial

    Given that the characteristic polynomial of A is t^4 + t + 1. How would I go about calculating the characteristic polynomial of A^2?

    I originally thought of using the fact that the eigenvalues of A^2 would simply be the square of the eigen values of A. But unfortunately I couldnt calculate the eigenvalues of A.

    Any help would be great, thanks!
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  2. #2
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    Quote Originally Posted by joe909 View Post
    Given that the characteristic polynomial of A is t^4 + t + 1. How would I go about calculating the characteristic polynomial of A^2?

    I originally thought of using the fact that the eigenvalues of A^2 would simply be the square of the eigen values of A. But unfortunately I couldnt calculate the eigenvalues of A.

    Any help would be great, thanks!


    Let \lambda_i\,,\,i=1,2,3,4 the eigenvalues of A\Longrightarrow \lambda_i^2 are the eigenvalues of A^2 . Now, applying Viete formulae we get:

    Coefficient of t^3=-\sum^4_{i=1}\lambda_i=0

    Coefficient of t^2=\sum^4_{i,j=1,\,i>j}\lambda_i\lambda_j=0

    Coefficient of t=-\!\!\!\!\!\sum^4_{i,j,k=1\,,\,i>j>k}\lambda_i\lamb  da_j\lambda_k=1

    Free coefficient =\prod^4_{i=1}\lambda_i=1 .

    Well, use the above relations for \lambda_i^2 and a little algebra to find out the coefficients of the char. pol. of A^2 . For example, it may interest you to use

    \sum^4_{i=1}\lambda_i^2=\left(\sum^4_{i=1}\lambda_  i\right)^2-2\sum^4_{i,j=1,\,i>j}\lambda_i\lambda_j ...
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  3. #3
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    I'm not sure if this is an acceptable answer, but it's slick! By the Cayley–Hamilton theorem, A^4+A+I=0. Therefore A = -(A^4+I) and so A^2 = (A^4+I)^2 = A^8+2A^4+I. Hence A^8 +2A^4-A^2+I = 0, which says that A^2 satisfies the polynomial equation t^4+2t^2-t+1=0. So I assume that t^4+2t^2-t+1 is the characteristic polynomial of A^2. But I don't know whether that conclusion is justified.
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    Quote Originally Posted by Opalg View Post
    I'm not sure if this is an acceptable answer, but it's slick! By the Cayley–Hamilton theorem, A^4+A+I=0. Therefore A = -(A^4+I) and so A^2 = (A^4+I)^2 = A^8+2A^4+I. Hence A^8 +2A^4-A^2+I = 0, which says that A^2 satisfies the polynomial equation t^4+2t^2-t+1=0. So I assume that t^4+2t^2-t+1 is the characteristic polynomial of A^2. But I don't know whether that conclusion is justified.

    It looks fine to me: it is of the correct degree, monic and vanishes at A^2...good!

    Tonio
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