Are you after the confidence ellipsoid for
How would I sketch the ellipsoids for
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 but not sure how or if it even applies in this problem.
Any help would be greatly appreciated.
The matrix has eigenvalues 9 and 1 with corresponding eigenvectors and respectively. If we let , the matrix having those eigenvalues as columns, then and , a diagonal matrix having the eigenvalues of on its diagonal.
Blast! I did this whole thing without realizing it was you wanted rather than ! Fortunately, we can still use all of that. If , then taking the inverse of both sides, . And so
The original equation was . Which is the same as . Let so that and the equation becomes .
If then that equation is . 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, , 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.
has determinant 0 and so does NOT have an inverse.
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?
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 .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)?
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. and if we write , then or . That "ellipse" is actually a circle with center at (0, 0) and radius .
is an ellipse with sem-axis of length a along the x- axis and semi-axis of length b along the y axis.
With , so that, taking , becomes
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 and in the equation of the ellipse, , then determine to make the coefficient of x'y' equal to 0. You should get .