I should start by saying that I haven't done algebra for very long time. I recently have some work related to algebra, so I need some help to speedup.

I went through a theorem in the book stating the relationship between ellipsoid's radii and the eigenvalues.

So the ellipsoid is defined as following:

$\sum_i \sum_j (x_i - u_i)(x_j-u_j)c_{ij} \leq k$

where $C=[c_ij]$ is an symmetric positive definite matrix, $x, u$ are n-dimensional vectors. The book says that the radii of this ellipsoid is $r_i = \frac{k}{\sqrt{\lambda_i}}$ where $\lambda_i$ are the eigenvalues of matrix C. However it does not give any proof for that. I've tried to google for the proof but didn't get anything useful. Could someone please help me understand it?


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    $\begingroup$ You know that a symmetric positive definite matrix $\mathbf A$ can be decomposed as $\mathbf A=\mathbf V\mathbf \Lambda\mathbf V^\top$, $\mathbf V$ being an orthogonal matrix and $\mathbf \Lambda$ being the diagonal matrix of eigenvalues? $\endgroup$ – J. M. is a poor mathematician Nov 8 '11 at 16:10
  • $\begingroup$ I think $A=V\Lambda V^{-1}$ instead of $V^T$, right? But how do you use it to prove? $\endgroup$ – chepukha Nov 8 '11 at 17:34
  • $\begingroup$ I mentioned $\mathbf V$ is orthogonal, no? $\endgroup$ – J. M. is a poor mathematician Nov 8 '11 at 17:44
  • $\begingroup$ Oh, sorry. I think I got it. So, $x^TAx=x^TV\Lambda V^Tx=y^T\Lambda y=\sum_i \lambda_i y_i^2$, right? Thanks. $\endgroup$ – chepukha Nov 8 '11 at 17:48
  • $\begingroup$ I wonder why @Heike deleted their answer. I would have upvoted it. $\endgroup$ – Rahul Jan 15 '13 at 20:59

You can start from the disc representation and gradually make the examples more complicated. Example: Let $p$ a point in the two dimensional plane $$ x^2+y^2 \leq 1 $$ can be rewritten as $$ \pmatrix{x&y}\pmatrix{1&0\\0&1}\pmatrix{x\\y} \leq 1 $$ This is the simplest case where you can see the resulting shape and read off the eigenvalues at the same time. Smashing the circle from the top and bottom would result with $$ \pmatrix{x&y}\pmatrix{1&0\\0&4}\pmatrix{x\\y} \leq 1 $$ We can see that points satisfy this can at most have a $y$ component with magnitude $0.5$ etc. We can further generalize this to circles that is centered somewhere other than the origin and also we can rotate these eccentric ellipsoids. When rotated, instead of plain $x$ and $y$ axes being principal axes you get the eigenvectors working for you and the lengths are associated with the magnitude of the eigenvalues. You can show this by eigenvalue decomposition or simply let your variables $x,y$ be the eigenvector of the largest eigenvalue and see them as rotated $(x,y)$ pairs that we have started with.

In general, we are left with inequalities in the form of $$ \pmatrix{p-p_0}C\pmatrix{p-p_0} \leq s $$ where $C$ is a positive definite matrix, $p_0$ is the center of the ellipsoid and $s$ is, roughly, the size of the ellipsoid. The positive definiteness is not crucial but often physically meaningful. If the matrix is not positive definite you allow for the reflections along some axis.


In chepukha's statement there is a small mistake. The radius must be $r_i=\sqrt{\frac{k}{\lambda_i}}$ (in his statement there is a square root missing, or, alternatively, the ellipsoid should be defined as $\sum_i\sum_j (x_i-u_i)(x_j-u_j)c_{ij}\leq k^2$).

For example, consider the scalar case (where the proof is straightforward): equation $x^2 c \leq k$ has solution $-\sqrt{\frac{k}{\lambda_i}}\leq x\leq+\sqrt{\frac{k}{\lambda_i}}$.


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