JanezK said:
Proving elipsoide radius of axis i equals 1/sqrt(λi)
How is radius related to eigenvalues? I can't find the connection ... :S
Hi JanezK! Welcome to MHB! :)
For an ellipsoid around the origin, with axes that are aligned with the coordinate axes, we have the equation:
$$\frac {x^2}{a^2} + \frac {y^2}{b^2} + \frac {z^2}{c^2} = 1$$
where $a,b,c$ are the radii of the ellipsoid.
Or writing it in matrix form:
$$\begin{bmatrix}x&y&z\end{bmatrix}
\begin{bmatrix}\frac 1{a^2}&0&0 \\
0&\frac 1{b^2}&0\\
0&0&\frac 1{c^2}\end{bmatrix}
\begin{bmatrix}x\\y\\z\end{bmatrix} = 1
$$
The more general form of an ellipsoid, or any quadric, is given by:
$$\mathbf x^TA\mathbf x + B\mathbf x + C = 0$$
The "trick" is that we can diagonalize A, leading to the form:
$$A=BDB^T$$
where $B$ is an orthogonal matrix identifying a basis transformation (a rotation in this case), and $D$ is a diagonal matrix.
It so happens that the diagonal matrix $D$ contains exactly the eigenvalues of $A$.
So each eigenvalue corresponds to the inverse square of a radius.
