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Good day!

Given a characteristic polynomial $P$ of matrix $A$ I need to show that the characteristic polynomial $O$ of $A^2$ can't have more different real roots than $P$.

I know that the characteristic polynomial for both cases can be calculated like this:
$P = |A - \lambda I| = 0$
$O = |A^2 - \lambda^2 I| = 0$
But in a general case with $n*n$ matrices they become way too complicated.

Can anyone guide me in the right direction?
Thanks!

  • 2
    That's wrong, isn't it? $A = \begin{pmatrix} 0 & -1 \\\ 1 & 0 \end{pmatrix}$ has charakteristic polynomial $P(t) = t^2 + 1$ and $A^2 = -\mathrm{Id}$, which has $Q(t) =(t+1)^2$?2012-05-10
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    Shouldn't it be $Q(t) = (t^2 + 1)^2$? In this case both $P(t)$and $Q(t)$ have 0 real roots.2012-05-10
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    Because $Q(t) = A^2 - t^2I = \begin{vmatrix} -1 - t^2 & 0\\ 0 & -1 - t^2 \end{vmatrix} = (1+t^2)^2$2012-05-10
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    In your equation for the characteristic polynomial of $A^2$ you shouldn't have $\lambda^2$, you should just have $\lambda$. It *is true* though that the square of any eigenvalue for $A$ will automatically be an eigenvalue of $A^2$.2012-05-10
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    It's plain to see that $A^2=\begin{pmatrix} -1 & 0 \\\ 0 & -1 \end{pmatrix}$ has $-1$ as an eigenvalue. You can't go wrong: *every* vector is an eigenvector for that particular $A^2$!2012-05-10
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    Yes, thanks you're right2012-05-10

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