# Conic formula Eigenvalues

1. Mar 27, 2013

### squenshl

1. The problem statement, all variables and given/known data
We have the following conic formula $ax^2 + 2bxy + cy^2 + dx + ey =$ constant which corresponds to a ellipse, hyperbola or parabola. The second order terms of the corresponding PDE $$a\frac{\partial^2 u}{\partial x_1^2} + 2b\frac{\partial^2 u}{\partial x_1\partial x_2} + c\frac{\partial^2 u}{\partial x_2^2} + d\frac{\partial u}{\partial x_1} + e\frac{\partial u}{\partial x_2} + gu = f(x_1,x_2)$$ can be written as $$\sum_{i,j=1}^2 a_{ij} \frac{\partial^2 u}{\partial x_i\partial x_j}$$ where $a_{ij}$ are the entries of the symmetric matrix $$A = \begin{pmatrix} a & b \\ b & c \end{pmatrix}.$$ Show that the eigenvalues have the same sign if $b^2-ac > 0$, opposite signs if $b^2-ac < 0$ and one is zero if $b^2-ac = 0$.

2. Relevant equations

3. The attempt at a solution
I know that the eigenvalues of a symmetric matrix are always real. Also I know that the PDE is said to be elliptic, hyperbolic or parabolic depending on whether $b^2-ac$ is positive, negative or zero. Not too sure what to do next. Please help. The determinant of $A$ is $ac-b^2$.

Last edited: Mar 27, 2013
2. Mar 27, 2013

### voko

Given matrix A, how are its eigenvalues computed?

3. Mar 27, 2013

### HallsofIvy

A 2 by 2 matrix can always be put in "Jordan Normal form" or "diagonalized" with its eigenvalues on the diagonal. That is, the derterminant of a matrix is always the product of its eigenvalues.

You are aware that $b^2- ac= -(ac- b^2)$ aren't you?

4. Mar 28, 2013

### squenshl

So we have $$(\lambda - a)(\lambda - c) -b^2 = \lambda^2 -(a+c)\lambda + ac - b^2$$ which is the characteristic polynomial with the coefficient of $\lambda$ being the trace of $A$ and the last term being the determinant of $A$.

Last edited: Mar 28, 2013
5. Mar 28, 2013

### vela

Staff Emeritus
Do you understand what HallsofIvy is getting at?

6. Mar 28, 2013

### squenshl

I do. But not in showing the signs of eigenvalues in each case for $b^2-ac$.

7. Mar 28, 2013

### vela

Staff Emeritus
If $\lambda_1 \lambda_2 > 0$, what can you say about the relative signs of $\lambda_1$ and $\lambda_2$?

8. Mar 28, 2013

### squenshl

Then $\lambda_1$ and $\lambda_2$ must either be both positive or negative, i.e. the same sign. Same argument for < 0 and = 0. Cheers.