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Math Help - Sketching an Ellipsoid using a matrix

  1. #1
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    Sketching an Ellipsoid using a matrix

    How would I sketch the ellipsoids x^t \Sigma^{-1} x <= 1 for

    \Sigma =\begin{bmatrix}5 &4\\ 4 &5\end{bmatrix} and  \Sigma=\begin{bmatrix}3 & 0\\ 0 &3\end{bmatrix}

    clearly indicating the principal axes of the ellipsoid and the intersection of the ellipoids with the axes.

    I know that the equation for an ellipsoid \frac{x^2}{a} + \frac{y^2}{b} + \frac{z^2}{c} = 1 but not sure how or if it even applies in this problem.

    Any help would be greatly appreciated.
    Last edited by statmajor; September 27th 2010 at 06:22 AM.
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  2. #2
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    Are you after the confidence ellipsoid for \mu

    \displaystyle n(\bar{x}-\mu)^TS^{-1}(\bar{x}-\mu) \leq \frac{p(n-1)}{n-p}\times F_{p,n-p,\alpha}

    ?
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  3. #3
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    The matrix \Sigma= \begin{bmatrix}5 & 4 \\ 4 & 5 \end{bmatrix} has eigenvalues 9 and 1 with corresponding eigenvectors \begin{bmatrix}1 \\ 1\end{bmatrix} and \begin{bmatrix}1 \\ -1\end{bmatrix} respectively. If we let P= \begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}, the matrix having those eigenvalues as columns, then P^{-1}=\begin{bmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{bmatrix} and P^{-1}\Sigma P= \begin{bmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{bmatrix}\begin{bmatrix}5 & 4 \\ 4 & 5 \end{bmatrix}\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}= \begin{bmatrix}9 & 0 \\ 0 & 1\end{bmatrix}, a diagonal matrix having the eigenvalues of \Sigma on its diagonal.

    Blast! I did this whole thing without realizing it was \Sigma^{-1} you wanted rather than Sigma! Fortunately, we can still use all of that. If P^{-1}\Sigma P= \begin{bmatrix}9 & 0 \\ 0 & 1\end{bmatrix}, then taking the inverse of both sides, P^{-1}\Sigma^{-1}P= D^{-1}= \begin{bmatrix}\frac{1}{9} & 0 \\ 0 & 1\end{bmatrix}. And so P\begin{bmatrix}\frac{1}{9} & 0 \\ 0 & 1\end{bmatrix}P^{-1}= \Sigma^{-1}

    The original equation was x^T\Sigma^{-1}x= 1. Which is the same as x^T(P\begin{bmatrix}\frac{1}{9} & 0 \\ 0 &  1\end{matrix}P^{-1}x= 1. Let x'= P^{-1}x so that x'^T= (P^{-1}x)^T= x^TP and the equation becomes x'^T\begin{bmatrix}\frac{1}{9} & 0 \\ 0 & 1\end{bmatrix}x'= 1.

    If x'= \begin{bmatrix}x' \\ y'\end{bmatrix} then that equation is \frac{x'^2}{9}+ y'^2= 1. That is an ellipse with axis of length 1/3 along the x'-axis and matrix of length 1 along the y' axis. That choice for x' means that the x'-axis is the same as the y= x line and the y' axis is the same as the y= -x line.

    Summarizing, since the original ellipse, \Sigma^{-1}, has eigenvalues 1/3 and 1, with eigenvectors <1,1> (along the y= 1 line) and <1, -1> (along the y= -x line), respectively, the ellipse has semi-minor axis of length 1/3 along the line y= x and semi-major axis of length 1 along the y= -x line. The center is still at (0, 0), of course.

    By the way, since this is in two dimensions, this is an ellipse, not an ellips[b]oid[b]. An "ellipsoid" is the three dimensional analog of the ellipse.

    \Sigma= \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix} has determinant 0 and so does NOT have an inverse.
    Last edited by HallsofIvy; September 27th 2010 at 07:14 AM.
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    Thanks for your help HallsofIvy. Just got a quick question. What's the different between semi-minor axis and semi-major axis? Is the major bigger than the minor?

    \Sigma= \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix} should have determinant 9 (not 0) and the inverse should exist. So would I apply the same process as the first matrix (even though this matrix only has 1 eigenvalue)?
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  5. #5
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    Quote Originally Posted by statmajor View Post
    Thanks for your help HallsofIvy. Just got a quick question. What's the different between semi-minor axis and semi-major axis? Is the major bigger than the minor?
    Yes, that's the only difference.

    \Sigma= \begin{bmatrix}3 & 0 \\ 0 & 3\end{bmatrix} should have determinant 9 (not 0) and the inverse should exist. So would I apply the same process as the first matrix (even though this matrix only has 1 eigenvalue)?
    For some reason, I was seeing \Sigma= \begin{bmatrix}3 & 0 \\ 3 & 0\end{bmatrix}.

    Actually, you don't need to go through that process- basically, I was "diagonalizing" the matrix and this matrix is already diagonal. Not every 2 by 2 matrix having only one eigenvalue can be diagonalized- it depends on the number of independent eigenvectors. (Every symmetric matrix can be diagonalized.)

    Here, 3 is the eigenvalue and independent eigenvectors are given by <1, 0> and <0, 1>, along the x and y axes. \Sigma^{-1}= \begin{bmatrix}\frac{1}{3} & 0 \\ 0 & \frac{1}{3}\end{bmatrix} and if we write = x= \begin{bmatrix}x \\ y \end{bmatrix}, then x^T\Sigma^{-1}x= \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix} \frac{1}{3} & 0 \\ 0 & \frac{1}{3}\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix} = \frac{x^2}{3}+ \frac{y^2}{3}= 1 or x^2+ y^2= 3. That "ellipse" is actually a circle with center at (0, 0) and radius \sqrt{3}.
    Last edited by HallsofIvy; September 28th 2010 at 12:25 AM.
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    Just wondering what the point of diagonilzing the matrix is. Couldnt I just find the inverse of the matrix, let x = (y x) and just plug it back into the equation x^t \Sigma^{-1} x <= 1?
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  7. #7
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    Quote Originally Posted by statmajor View Post
    Just wondering what the point of diagonilzing the matrix is. Couldnt I just find the inverse of the matrix, let x = (y x) and just plug it back into the equation x^t \Sigma^{-1} x <= 1?
    The point is to get the equation in "standard form" for an ellipse:
    \frac{x^2}{a^2}+ \frac{y^2}{b^2}= is an ellipse with sem-axis of length a along the x- axis and semi-axis of length b along the y axis.

    With \Sigma= \begin{bmatrix}5 & 4 \\ 4 & 5\end{bmatrix}, Sigma^{-1}= \begin{bmatrix}\frac{5}{9} & -\frac{4}{9} \\ -\frac{4}{9} & \frac{5}{9}\end{bmatrix} so that, taking x= \begin{bmatrix}x\\ y\end{bmatrix}, x^t\Sigma^{-1}x<= 1 becomes
    \begin{bmatrix}x & y\end{bmatrix}\begin{bmatrix}\frac{5}{9} & -\frac{4}{9} \\ -\frac{4}{9} & \frac{5}{9}\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \frac{5}{9}x^2- \frac{8}{9}xy+ \frac{5}{9}y^2\le 1.

    That "xy" term signals an ellipse whose axes are NOT along the x and y axes.

    By determining the eigenvectors for the matrix, as I did before, you see that the axes are the lines y= x and y= -x.

    A more tedious, less sophisticated, way of doing the same thing would be to let x= x'cos(\theta)- y sin(\theta) and y= x' sin(\theta)+ y' cos(\theta) in the equation of the ellipse, \frac{5}{9}x^2- \frac{8}{9}xy+ \frac{5}{9}y^2= 1, then determine \theta to make the coefficient of x'y' equal to 0. You should get \theta= \pi/4.
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