Characteristic polynomials of matrices 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!
 A: 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? 
A: 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.)
