Eigenvectors in non-orthogonal coordinate system

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I have a general ellipsoid in the matrix form of

$$X=\left[\begin{matrix}x_{11}&x_{12}&x_{13}\\x_{12}&x_{22}&x_{23}\\x_{13}&x_{23}&x_{33}\end{matrix}\right],$$

(symmetrical around the main diagonal) and this ellipsoid is situated in a non-orthogonal coordinate system (angle between $x$ and $y = 120^{\circ},$ others $90^{\circ}$), see screenshot link. X is expressed relative to this coordinate system. What is the proper way to calculate the eigenvectors and values from it?

ellipsoid in non-orthogonal coordinates

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You go about finding eigenvalues and eigenvectors the same way regardless of the coordinate system you’re working in, but the results will be specific to that coordinate system: in general, eigenvalues in different coordinate systems will be different and eigenvectors will not be related to each other via the coordinate transformation. Therefore, if you want to know the principal axes in the standard coordinate system, you’ll have to transform the ellipsoid’s matrix first.

A simple two-dimensional example to illustrate: Consider the ellipse given by $x^2/25+y^2/9=1$ in the standard basis. The corresponding matrix is $$C=\begin{bmatrix}\frac1{25}&0\\0&\frac19\end{bmatrix}.$$ Its eigenvalues are of course $1/25$ and $1/9$, with the standard basis vectors for the corresponding eigenvectors. If we rotate the $y$-axis thirty degrees so that the angle between the positive axis directions is $120°$, the corresponding change-of-basis matrix is $$M=\begin{bmatrix}1&\frac1{\sqrt3}\\0&\frac2{\sqrt3}\end{bmatrix}$$ and the matrix of the ellipse relative to this new basis is $$C'=\begin{bmatrix}\frac1{25}&-\frac1{50}\\-\frac1{50}&\frac7{75}\end{bmatrix}.$$ Its eigenvalues are $1/10$ and $1/30$, with respective eigenvectors $(1,-3)^T$ and $(3,1)^T$, which, even if normalized, are clearly not the images under $M$ of the eigenvectors of $C$.