# matrix forms of quadratic equations

• May 17th 2008, 09:34 PM
Dr Zoidburg
I have a problem with determining eigenvalues. This is what I've got thus far:

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

This gives the matrix form:
$\begin{pmatrix} 4 & 5 \\
5 & 4 \\
\end{pmatrix}$

Now we find the eigenvalues:
Det(A – xI) = $\begin{pmatrix} (4-x) & 5 \\
5 & (4-x) \\
\end{pmatrix}$

= x² – 8x – 9
= (x – 9)(x + 1)
eigenvalues are $\lambda1 = 9 and + \lambda1 = -1$

From there, it's pretty simple solving:
$\lambda1x'^2 + \lambda2y'^2 = 9$

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.
• May 17th 2008, 09:48 PM
Isomorphism
Quote:

Originally Posted by Dr Zoidburg
I have a problem with determining eigenvalues. This is what I've got thus far:

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

[This gives the matrix form:
$\begin{pmatrix} 4 & 5 \\
5 & 4 \\
\end{pmatrix}$

Now we find the eigenvalues:
Det(A – xI[font=Arial]) = $\begin{pmatrix} (4-x) & 5 \\
5 & (4-x) \\
\end{pmatrix}$

= x² – 8x – 9
= (x – 9)(x + 1)
eigenvalues are $\lambda1 = 9 and + \lambda1 = -1$

From there, it's pretty simple solving:
$\lambda1x'^2 + \lambda2y'^2 = 9$

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.

Hmm.. I think that depends on how you wrote the matrix form in the first step(Thinking)
• May 18th 2008, 12:11 AM
KGene
Quote:

Originally Posted by Dr Zoidburg
I have a problem with determining eigenvalues. This is what I've got thus far:

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

This gives the matrix form:
$\begin{pmatrix} 4 & 5 \\
5 & 4 \\
\end{pmatrix}$

Now we find the eigenvalues:
Det(A – xI) = $\begin{pmatrix} (4-x) & 5 \\
5 & (4-x) \\
\end{pmatrix}$

= x² – 8x – 9
= (x – 9)(x + 1)
eigenvalues are $\lambda1 = 9 and + \lambda1 = -1$

From there, it's pretty simple solving:
$\lambda1x'^2 + \lambda2y'^2 = 9$

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.

The key point is tracing back the transformation that you have made by diagonalizing the symmetric matrix $A$.
Since $\lambda_1=9,\lambda=-1$ are eigenvalues, we find their corresponding eigenvectors. If we do so, we can choose $v_1=\left [\begin{array}{c}1\\ 1\end{array}\right ], v_2=\left [\begin{array}{c}1\\ -1\end{array}\right ]$ where the indices correspond. Now form the matrix $P$ using $v_1,v_2$, i.e., $P=\left [\begin{array}{cc}1&1\\ 1&-1\end{array}\right ]$. Let $D=\left [\begin{array}{cc}9&0\\ 0&-1\end{array}\right ]$, so $PAP^{-1}=D$. Any solution $w=\left [\begin{array}{cc}x&y \end{array}\right ]$ satisfies the equation $wAw^t=9$. Note that $P^{-1}=1/2P^t$, so $wAw^t=wP^{-1}DPw^t=1/2wP^tDPw^t=1/2(wP^t)D(wP^t)^t=9$. So solutions $w$ correspond to solutions of $zDz^t=18$, i.e., the equation $9{z_1}^2-{z_2}^2=18$, via the transformation $wP^t=z$.

Hope this helps.
• May 18th 2008, 12:13 AM
KGene
To answer your question, the choice of $\lambda_1,\lambda_2$ is not "crucial", because it will be taken care by the transformation matrix $P$.
• May 18th 2008, 05:06 AM
Dr Zoidburg
Quote:

Originally Posted by Isomorphism
Hmm.. I think that depends on how you wrote the matrix form in the first step(Thinking)

I got it from the quadratic equation: $\begin{pmatrix} a & b \\b & c \\\end{pmatrix}$
where a,b,c,d are
$ax^2 + 2bxy + cy^2 = d$

I should explain more:
once the equation
$\lambda1x'^2 + \lambda2y'^2 = d$
has been reached, I need to state what graphic function it is and plot the graph. So it would appear that knowing what $\lambda1$ and $\lambda2$ are important as I get either
$9x'^2 - y'^2 = 9$ giving $x'^2 - y'^2/9 = 1$
or
$-x'^2 + 9y'^2 = 9$ giving $-x'^2/9 + y'^2 = 1$
which lead to vastly different graphs. As I said, there's nothing mentioned in my lecture notes about this, and because I'm doing this paper by correspondence I can't go speak to the lecturer.