Eigenvalues of a symmetric $2 \times 2$ matrix

Question

Consider a real symmetric matrix

$$ A=\begin{bmatrix}m_1^2&\mu^2\\\mu^2&m_2^2\end{bmatrix}. $$

Then we find that, the eigenvalues are:

$$ \frac{1}{2}(m_1^2+m_2^2)\pm\frac{1}{2}\sqrt{(m_1^2+m_2^2)^2-4\mu^4} $$

which is evidently complex if $4\mu^4>(m_1^2+m_2^2)^2$

So, what am I missing here? It must be something trivial.

Thanks

Answer 1

No, the discriminant should be $\operatorname{tr}(A)^2-4\det(A)^2=(m_1^2+m_2^2)^2-4(m_1^2m_2^2-\mu^4)=(m_1^2-m_2^2)^2+4\mu^4$.