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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!

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    Yes, thanks you're right2012-05-10

2 Answers 2

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This is false, for somewhat trivial reasons. Let $A$ be a matrix with eigen values $\lambda_1,\dots,\lambda_n$ then note that the eigenvalues of $A^2$ are $\lambda_1^2,\dots,\lambda_n^2$ since if $v_i$ is an eigenvector of $\lambda_i$ then

$A^2v_i=A(\lambda v_i)=\lambda Av_i=\lambda^2v_i.$

For instance if the eigenvalues are purely imaginary the number of real eigenvalues increases. To take an example from the comments

$A=\begin{pmatrix}0 & -1 \\ 1 & 0\end{pmatrix}$

then $\mathsf{char}(A)=x^2+1$ which has imaginary roots $\pm i$ but $\mathsf{char}(A^2)=(x+1)^2$ has real repeated root $-1$.

Are there some more conditions on your matrix perhaps?

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    The question was about different (unique) roots, so if I changed it to complex roots, then: $char(A) = x^2 +1$ has 2 unique complex roots, while $char(A^2) = (x+1)^2$ has one unique (complex) root, which is less.2012-05-10
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Since the roots of these polynomials are eigenvalues, let's just think in terms of eigenvalues.

The eigenvalues of $A^2$ are just the eigenvalues of $A$, squared (right? It's tempting to believe, and scribbling on a napkin made it seem so but I could be overlooking something silly).

In that case, the real eigenvalues of $A$ would stay real after squaring, and some complex eigenvalues might become real after squaring.

This would show that $A^2$ has at least as many real eigenvalues as $A$. (This is not the original question, but it seems like this might have been the intended question.)