I refer to the article Matrix representation of conic sections in Wikipedia. It is shown that a conic section in 2D will satisfy a second-degree polynomial equation: $$ Q(x, y) = Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 $$ This can be then expressed in matrix form: $$ f = \begin{pmatrix} x& y\end{pmatrix} \begin{pmatrix} A & B/2\\ B/2 & C \end{pmatrix} \begin{pmatrix} x\\ y\end{pmatrix} + \begin{pmatrix} D & E \end{pmatrix} \begin{pmatrix} x\\ y\end{pmatrix} + F = 0 $$ or in homogeneous coordinates: $$ f = \begin{pmatrix} x & y & 1\end{pmatrix} \begin{pmatrix} A & B/2 & D/2\\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{pmatrix} \begin{pmatrix} x\\ y \\ 1\end{pmatrix} = 0 $$
Let $A_{33} = \begin{pmatrix} A & B/2\\ B/2 & C \end{pmatrix}$, and $A_Q = \begin{pmatrix} A & B/2 & D/2\\ B/2 & C & E/2 \\ D/2 & E/2 & F \end{pmatrix}$.
If $det(A_{33}) > 0$, then the equation is an ellipse. This equation can then be converted to the canonical form of a centered ellipse as such: $$ \lambda_1 x^2 + \lambda_2 y^2 = -\frac{det(A_Q)}{det(A_{33})} $$
$$ a^2 = -\frac{det(A_Q)}{\lambda_1 det(A_{33})},\; b^2 = -\frac{det(A_Q)}{\lambda_2 det(A_{33})} $$ where $\lambda$ are the eigenvalues of $A_{33}$, and $a$ and $b$ are the lengths of the axes of the ellipse.
My question is, does this generalize to $n$-Dimensions hyper-ellipsoids? In the $n$-D case, $A_{33}$ will be a $n \times n$ matrix and $A_Q$ a $(n+1) \times (n+1)$ matrix.
Firstly, does a positive $det(A_{33})$ indicate an hyper-ellipsoid rather than other types of hyper-conics?
Secondly, will the lengths, $L_i$ of the $n$ axes of the hyper-ellipsoid also be found as $L_i^2 = -\frac{det(A_Q)}{\lambda_i det(A_{33})}$?
In $n$-dimensions, let $\mathbf{x} \in \mathbb{R}^n $, then you be given the equation of the hyper-conic as
$ \mathbf{x}^T {Q} \mathbf{x} + \mathbf{x}^T \mathbf{b} + c = 0 $
Assuming $Q$ is invertible, then you can find the center of this hyper-conic as follows
$ \mathbf{x}_0 = - \dfrac{1}{2} Q^{-1} \mathbf{b} $
And the equation can be re-written using this center as follows
$ ( \mathbf{x} - \mathbf{x}_0 )^T Q (\mathbf{x} - \mathbf{x}_0) = {\mathbf{x}_0}^T Q \mathbf{x}_0 - c $
The next thing to do, is diagonalize $Q$ and write it as
$ Q = R D R^T $
where $R$ is a rotation matrix. Substituting this gives
$ ( \mathbf{x} - \mathbf{x}_0 )^T R D R^T (\mathbf{x} - \mathbf{x}_0) = {\mathbf{x}_0}^T Q \mathbf{x}_0 - c $
If Q is positive definite, i.e. all the diagonal entries of $D$ are positive, and the right hand side of the above equation is positive, then define the matrix $E$ as follows
$ E = \dfrac{D}{{\mathbf{x}_0}^T Q \mathbf{x}_0 - c} $
Then
$ ( \mathbf{x} - \mathbf{x}_0 )^T R E R^T (\mathbf{x} - \mathbf{x}_0) = 1 $
Further define the vector $ \mathbf{v} = R^T (\mathbf{x} - \mathbf{x}_0) $
Then
$ \mathbf{v}^T E \mathbf{v} = 1 $
And this is the standard equation of the hyper-conic, in this case, hyper-ellipse. The diagonal entries of the diagonal matrix $E$ are the reciprocal of the squares of the semi-axes lengths of the hyper-ellipse. That is,
$ E = \begin{bmatrix} \dfrac{1}{s_1^2} && 0 && \dots && 0 \\0 && \dfrac{1}{s_2^2} && \dots && 0 \\ \vdots && \vdots && \ddots && \vdots \\ 0 && 0 && \dots && \dfrac{1}{s_n}^2 \end{bmatrix} $