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Why are the eigenvectors the axes of an ellipse?

  1. Nov 19, 2015 #1
    I'm almost there in terms of understanding it, but I need to go beyond the text.

    Here is the example problem:

    UMj55tF.jpg

    imgur link: http://i.imgur.com/UMj55tF.jpg

    I can see that where we have [itex]1 = \vec{x}^T A \vec{x} = \lambda \vec{x}^T \vec{x}[/itex] that [itex]1=\lambda \vec{x}^T \vec{x} = \lambda ||\vec{x}||^2[/itex]

    so [tex]||\vec{x}|| = \frac{1}{\sqrt{\lambda}}[/tex]

    and I understand that the length of the vector [itex]||\vec{x}||[/itex] will have min and max values when pointing along the minor and major axis'.

    What I cannot prove to myself, or see the reason behind, is why the eigenvectors that you would solve for would be pointing in the direction of the minor and major axis'. Why is that? What is the reason for that?
     
  2. jcsd
  3. Nov 19, 2015 #2

    HallsofIvy

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    Staff Emeritus
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    Using the Eigenvectors as a basis, the linear transformation corresponds to a diagonal matrix. Another way of putting that is that, creating the matrix S, having the eigenvectors as columns, such that [itex]A= S^TAS[/itex] and where D is a diagonal matrix, having the eigenvalues on the diagonal. Then [itex]x^TAx= x^T(S^TD)x= (Sx)^TD(Sx)= 1[/itex]. So the equation becomes [tex]\begin{bmatrix}x & y \end{bmatrix}\begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_2\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \lambda_1x^2+ \lambda_2^2= 1[/tex].
     
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