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Math Help - matrix forms of quadratic equations

  1. #1
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    matrix forms of quadratic equations

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

    Identify and sketch the graph of the quadratic equation
    4x + 10xy + 4y = 9

    This gives the matrix form:
    \begin{pmatrix} 4 & 5  \\<br />
5 & 4 \\<br />
\end{pmatrix}
    Now we find the eigenvalues:
    Det(A – xI) = \begin{pmatrix} (4-x) & 5  \\<br />
5 & (4-x) \\<br />
\end{pmatrix}
    = x – 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.
    Last edited by Dr Zoidburg; May 17th 2008 at 11:17 PM.
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  2. #2
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    Quote Originally Posted by Dr Zoidburg View Post
    I have a problem with determining eigenvalues. This is what I've got thus far:

    Identify and sketch the graph of the quadratic equation
    4x + 10xy + 4y = 9

    [This gives the matrix form:
    \begin{pmatrix} 4 & 5  \\<br />
5 & 4 \\<br />
\end{pmatrix}
    Now we find the eigenvalues:
    Det(A xI[font=Arial]) = \begin{pmatrix} (4-x) & 5  \\<br />
5 & (4-x) \\<br />
\end{pmatrix}
    = x 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
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  3. #3
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    Quote Originally Posted by Dr Zoidburg View Post
    I have a problem with determining eigenvalues. This is what I've got thus far:

    Identify and sketch the graph of the quadratic equation
    4x + 10xy + 4y = 9

    This gives the matrix form:
    \begin{pmatrix} 4 & 5  \\<br />
5 & 4 \\<br />
\end{pmatrix}
    Now we find the eigenvalues:
    Det(A – xI) = \begin{pmatrix} (4-x) & 5  \\<br />
5 & (4-x) \\<br />
\end{pmatrix}
    = x – 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.
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  4. #4
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    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.
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  5. #5
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    Quote Originally Posted by Isomorphism View Post
    Hmm.. I think that depends on how you wrote the matrix form in the first step
    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.
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