Comparing eigenvalues of a matrix

Question

I want to prove that the matrix $$\begin{pmatrix} 8q_1^2-4q_2&-4q_1\\-4q_1&2(1+\epsilon) \end{pmatrix}$$ for all $q=(q_1,q_2)\in\mathbb{R}^2$ has two eigenvalues $\lambda_1(q)$ and $\lambda_2(q)$ one positive and the other negative for all $q\in\mathbb{R}^2$

Please help me to do so. Thanks

Answer 1

The characteristic polynomial is a second degree polynomial that you can solve using the general formula. Check whether the discriminant is always greater than 0

Answer 2

The characteristic polynomial is: $$((8q_1^2-4q_2)-\lambda)(2(1+\epsilon) -\lambda) -16q_1^2=\lambda ^2-((8q_1^2-4q_2^2)+2(1+\epsilon))\lambda +(8q_1^2-4q_2)(2(1+\epsilon) -16q_1^2$$.

This is a quadratic polynomial with discriminant $((8q_1^2-4q_2)+2(1+\epsilon))^2-4(16q_1^2\epsilon-4q_2(2(1+\epsilon))=(8q_1^2-4q_2)^2+2(8q_1^2-4q_2)(2(1+\epsilon) +(2(1+\epsilon)) ^2 -4(16q_1\epsilon-8q_2(1+\epsilon) )=(8q_1^2-4q_2)^2+(2(1+\epsilon ))^2+32q_1^2-16q_2+4\epsilon (8q_1^2-4q_2)-64q_1\epsilon +32q_2(1+\epsilon) =(8q_1^2-4q_2)^2+4(1+\epsilon) ^2+32(1+\epsilon) q_1^2+16(1+\epsilon) q_2-64q_1\epsilon$.

This will be positive if $q=(q_1,q_2)$ is in the second quadrant... in which case we get two distinct real eigenvalues...

Here are expressions for the eigenvalues using Wolfram's eigenvalue calculator...