Graphing quadratic form, which eigenvalue should be chosen first?

Question

I just graph a quadratic function, $-4x^2_1+4x_1x_2-7x_2^2=-8$, by:

1. Find the eigenvalues of the function above, which are $\lambda_1=-8$ and $\lambda_2=-3$
2. Use the eigenvalues to make a new quadratic function in $y$ axis, which becomes $-8y^2_12-3y_2^2=-8$
3. Use the eigenvectors to shift the ellipse to the $x$ axis.
4. I checked with the exercise book and the shape is correct (tall and skinny ellipse)

What I don't understand is how to correctly create a $y$ axis quadratic formula (step 2). By choosing $-8$ before $-3$, the ellipse ($-8y^2_12-3y_2^2=-8$) is tall and skinny.

However, if I were to choose $\lambda_2$ before $\lambda_1$, and proceed with $-3y^2_12-8y_2^2=-8$, the ellipse become fat and not tall. Is there an algorithm to know which eigenvalue should come first?

Thank you for your help.

Answer 1

Hint:

Lets do a contour plot of the resulting systems.

We choose $\lambda_1=-8$ and $\lambda_2=-3$ and we get $-8y_1^2 - 3y_2^2 = -8$, so a plot is:

Next, lets choose $\lambda_1=-3$ and $\lambda_2=-8$ and we get $-3y_1^2 - 8y_2^2 = -8$, so a plot is:

What do you notice about these two plots? Look at the $y_1$ versus $y_2$ axes and compare both plots. Think rotation by? In other words, $-4x_1^2 + 4x_1x_2 - 7x_2^2 = \lambda_1 y_1^2 + \lambda_2 y_2^2$, where $\lambda_1, \lambda_2$ are the diagonal entries of $PDP^{-1}$ (recall that P is using the normalized eigenvectors). We are free to choose which eigenvalue is which since we are diagonalizing (if that is possible, using that choice of eigenvalues, that dictate our columns of $P$). The axes of the ellipse (in this case) have the directions corresponding to the eigenvectors of the matrix $A$, which define the major and minor axes of the ellipse. In the two cases, all we have done is rotated the resulting axes of the ellipse.