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Thread: Characteristic Polynomial

  1. #1
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    Characteristic Polynomial

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

    I originally thought of using the fact that the eigenvalues of $\displaystyle A^2$ would simply be the square of the eigen values of $\displaystyle A$. 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 $\displaystyle A$ is $\displaystyle t^4 + t + 1$. How would I go about calculating the characteristic polynomial of $\displaystyle A^2$?

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

    Any help would be great, thanks!


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

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

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

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

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

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

    $\displaystyle \sum^4_{i=1}\lambda_i^2=\left(\sum^4_{i=1}\lambda_ i\right)^2-2\sum^4_{i,j=1,\,i>j}\lambda_i\lambda_j$ ...
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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, $\displaystyle A^4+A+I=0$. Therefore $\displaystyle A = -(A^4+I)$ and so $\displaystyle A^2 = (A^4+I)^2 = A^8+2A^4+I$. Hence $\displaystyle A^8 +2A^4-A^2+I = 0$, which says that $\displaystyle A^2$ satisfies the polynomial equation $\displaystyle t^4+2t^2-t+1=0$. So I assume that $\displaystyle t^4+2t^2-t+1$ is the characteristic polynomial of $\displaystyle A^2$. 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, $\displaystyle A^4+A+I=0$. Therefore $\displaystyle A = -(A^4+I)$ and so $\displaystyle A^2 = (A^4+I)^2 = A^8+2A^4+I$. Hence $\displaystyle A^8 +2A^4-A^2+I = 0$, which says that $\displaystyle A^2$ satisfies the polynomial equation $\displaystyle t^4+2t^2-t+1=0$. So I assume that $\displaystyle t^4+2t^2-t+1$ is the characteristic polynomial of $\displaystyle A^2$. 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 $\displaystyle A^2$...good!

    Tonio
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