Matrix forms of quadratic equations

[Dr Zoidburg]

I have a problem with determining eigenvalues. This is what I've got thus far:

Homework Statement

Identify and sketch the graph of the quadratic equation
4x² + 10xy + 4y² = 9

The Attempt at a Solution

We put it in the matrix form:
\begin{pmatrix} 4 &amp; 5 \\<br /> 5 &amp; 4 \\<br /> \end{pmatrix}

Now we find the eigenvalues:

Det(A – xI) = \begin{pmatrix} (4-x) &amp; 5 \\<br /> 5 &amp; (4-x) \\<br /> \end{pmatrix}

= x² – 8x – 9
= (x – 9)(x + 1)
eigenvalues are \lambda1 = 9 & \lambda2 = -1

From there, it's pretty simple solving:
\lambda1x&#039;^2 + \lambda2y&#039;^2 = 9

My problem here is: How do I know which eigenvalue is which? It obviously makes quite a bit of difference to the final result. Nothing in my textbook says.

 

[Defennder]

What do you mean "Which eigenvalue is which" ? You have already found the eigenvales you need. The next step would be to find the eigenvectors associated with the eigenvalues you have found. When you do that, you would have obtained a a square matrix P which orthogonally diagonalises the matrix \left ( \begin{array}{cc}4&amp;5\\5&amp;4\end{array}\right ).

From there, once you have P you can find \textbf{y} = \textbf{Px}, where y is the new coordinate system with respect to the original coordinate system x by means of transition matrix P.

 

[Dr Zoidburg]

sorry I should have stated the problem a bit more in detail.
Once I've got the eigenvalues, I'm to put them into the equation
\lambda1x^2 + \lambda2y^2 = 9
so knowing which is which is important as swapping produces vastly different graphs with either:
9x^2 - y^2 = 9 giving x^2 - y^2/9 = 1
or
-x^2 + 9y^2 = 9 giving -x^2/9 + y^2 = 1

There's nothing in the lecture notes about working out which is which, and I'm doing this by correspondence so can't go see the lecturer.

 

[Defennder]

I don't know how you've been taught for this, but that's not how I was taught to identify conic sections. What do your notes say?

I'll just show you how I would do this problem. First we have to find P, which is the transition matrix from the standard coordinate axes to the 'new' coordinate axes. Once you have the eigenvalues, you must next find the associated eigenvectors. Let \textbf{y} be the new coordinate axes with \textbf{x} being the old coordinate axes. P is an orthogonal matrix and consists of the eigenvectors (as column vectors in P). Hence P^T is the inverse matrix of P.

The quadratic equation may be expressed in the form x^TAx = f where f=9. A is the matrix which is diagonalisable by P. Let \textbf{x} = \textbf{Py} and the equation becomes y^T P^T APy = 9. You'll get the equation of a hyperbola in the new coordinate axes. By \textbf{y} = \textbf{Px}, you can sketch the new coordinate axes relative to the old one. Then you can plot the equation of the hyperbola with respect to the new axes.

 

[HallsofIvy]

You have not yet found the eigenvectors. There will be an eigenvector coresponding to each eigenvalue. Choosing one of the eigenvectors as x' axis and the other as y' axis determines which eigenvalue multiplies x' and which y'.