Characteristic polynomial of self-adjoint map A is self-adjoint map on $R^3$. I need to determine if 
$$-\lambda^3 + 3\lambda^2 - 2$$
$$-\lambda^3 - 3\lambda - 2$$
may be its characteristic polynomials?
The only relevant fact I know about such a map is that it should have real $\lambda$. Thus, I tried to find roots of the first polynomial:
$$-\lambda \cdot \lambda^2 + \lambda^2 + 2\lambda^2 - 2 = -\lambda^2 (\lambda - 1) + 2(\lambda - 1)(\lambda + 1) = $$
$$=(\lambda - 1)(-\lambda^2 + 2\lambda + 2) \Rightarrow \lambda = \{1; 1 \pm \sqrt3\}$$
The roots are real, as required. But is it enough?
Btw, I don't know how to find roots of the second polynomial.
 A: For the first one, yes it is enough that they are real.  To see this it is enough to consider the special case of diagonal matrices.   The diagonal entries are the roots of the characteristic polynomial, and such maps are self-adjoint when the diagonal entries are real. 
For the second one, you do not need to find explicit formulas for the roots to discover that there are nonreal roots.  Think of $f(\lambda)=-\lambda^3-3\lambda -2$ as a function from $\mathbb R$ to $\mathbb R$.  Then $f'(\lambda)=-3\lambda^2-3\leq -3$.  Because $f'$ is always negative on $\mathbb R$, $f$ is strictly decreasing, hence has at most one real root.  Because $f'$ is never $0$ on $\mathbb R$, there is no real root of multiplicity greater than $1$.  Thus $f$ has at most $1$ real root counted with multiplicity (in fact exactly $1$), and because it has degree three it must have $2$ nonreal roots.
(A similar approach can be used in general for determining how many real roots a degree $3$ polynomial $p$ with real coefficients has, because the derivative $p'$ is quadratic and therefore easier to work with.  If you find where $p$ is increasing and decreasing on $\mathbb R$, and evaluate $p$ at the real critical points, you will have enough to determine the amount and  multiplicity of the real zeros.)
