I have an ellipsoid defined by the semiaxes $a,b,c$ and the orthonormal vectors $v_x, v_y, v_z$ describing the directions in which the axes point. The matrix $\Lambda=\left(\begin{matrix}a&0&0\\0&b&0\\0&0&c\end{matrix}\right)$ then simply represents the ellipsoid in its "native" coordinate system while $A=T\cdot\Lambda$ with $T=\left(v_x,v_y,v_z\right)$ represents the ellipsoid in the "laboratory" coordinate system in which it is arbitrary oriented.
Now: Shouldn't the Eigenvalues of $\Lambda$ and $A$ be the same and be related to the shape of the ellipsoid irrespective of the ellipsoid orientation given by $T$, i.e. $EigenValues(A)=(a,b,c)$?
When i create the rotation matrix $T=\left(\begin{matrix}Cos(\Theta)&Sin(\Theta)&\\Sin(\Theta)&Cos(\Theta)&0\\0&0&1\end{matrix}\right)$ and calculate the Eigenvalues of $T \cdot \Lambda$ for different values of $\Theta$ with mathematica, i get varying results.
Angle, Eigenvalue1, Eigenvalue2, Eigenvalue3
0 3, 2, 1
45°, 3, 1.06+0.93j, 1.06-0.93j
180°, 3, 0+1.41j, 0-1.41j
275°, 3, -1.06+0.93j, -1.06-0.93j
Where's my mistake?
You've forgotten to multiply $T\cdot\Lambda $ by $T^{-1}$ on the right - when we change bases/coordinates we have to make sure that our linear map is expressing a transformation in this new basis. The $T^{-1}$ above ensures this description is complete. The resulting matrix $T\cdot\Lambda \cdot T^{-1}$is then $similar$ to our original matrix $\Lambda$. As similar matrices represent the same linear map just with respect to different bases, we should expect that they respect qualities of linear maps that are invariant under a change of basis, i.e. eigenvalues, dimensions of eigenspaces, the characteristic polynomial, determinant and trace to name but a few.