Properties of diagonalization of a conic matrix

Question

I have an ellipse described by the conic matrix

$$ C= \begin{bmatrix} A &\frac B2 & \frac D2 \\ \frac B2 & C & \frac E2 \\ \frac D2 & \frac E2 & F \end{bmatrix} $$

with $B^2 - 4AC < 0$. I compute the eigenvalues $\lambda_1, \lambda_2, \lambda_3$, the corresponding eigenvectors $\vec{e}_1,\vec{e}_2,\vec{e}_3$, and the diagonalized matrix $C_\lambda$ such that $$C = P^TC_\lambda P$$

where

$$ C = \begin{bmatrix} \lambda_1 & & \\ & \lambda_2 & \\ & & \lambda_3 \end{bmatrix} \qquad\text{and}\qquad P = \begin{bmatrix} & & \\ \vec{e}_1 & \vec{e}_2 & \vec{e}_3 \\ & & \end{bmatrix}$$

My question is: which properties should I expect from the diagonalized matrix $C_\lambda$, given what I know about $C$? Can I say it is still an ellipse (i.e., should be $\lambda_1 \lambda_2 > 0$)?

Answer 1

(I’m going to use $Q$ for the matrix of the ellipse instead of $C$ to avoid confusion with the coefficient $C$ in the ellipse’s equation.)

We can view the conic given by $(x,y,1)\,Q\,(x,y,1)^T=0$ as the intersection of the plane $z=1$ with the degenerate quadric surface $\mathbf x^TQ\mathbf x=0$, where here $\mathbf x\in\mathbf R^3$. Diagonalizing $Q$ as you’ve done amounts to rotating this quadric so that it is axis-aligned and perhaps scaling along its principal axes, neither of which changes the nature of the surface.

What is this quadric surface, though? Every ellipse can be obtained from the unit circle via a nonsingular affine transformation. If $M$ is the matrix of this transformation, then the equation of the resulting ellipse is given by the equation $$(M^{-1}\mathbf p)^T\operatorname{diag}(1,1,-1)\,(M^{-1}\mathbf p)=\mathbf p^T(M^{-1})^T\operatorname{diag}(1,1,-1)\,(M^{-1})\,\mathbf p=\mathbf p^TQ\mathbf p=0$$ with $\mathbf p=(x,y,1)^T$. The transformation matrix $M^{-1}$ has the canonical form $$M^{-1}=\pmatrix{a&b&c\\d&e&f\\0&0&1}\tag1$$ therefore we have $$\det Q=-(\det M^{-1})^2\lt0.$$ This means that the signature of this quadratic form is either $++-$ or $---$. If it’s the latter, then the only solution to $\mathbf x^TQ\mathbf x=0$ is the origin, which is a bit too degenerate for this problem, so we must have the former signature, that of an elliptic cone. More generally, $M^{-1}$ might have some other non-zero value in its lower-right corner, but that is equivalent to multiplying the matrix (1) by some non-zero scalar value $k$, which will then multiply $Q$ by $k^2$, but that doesn’t change the possible signatures.

Finally, after diagonalizing $Q$ into the matrix $\Lambda=\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3)$, we have the conic $(x,y,1)\,\Lambda\,(x,y,1)^T=\lambda_1x^2+\lambda_2y^2+\lambda_3=0$, which is the intersection of the rotated and scaled cone $\lambda_1x^2+\lambda_2y^2+\lambda_3z^2=0$ with the plane $z=1$. This will be either an ellipse or hyperbola depending on the order of the eigenvalues in $\Lambda$.

Answer 2

Edit Appears as if I failed to grasp the OP's intention.
As I currently have no time available, revision is postponed to a later time.

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Your given "The conic $C$ describes an ellipse." is equivalent to "$C$ is a positive-definite matrix.", supposed an ellipsoid is meant whose three principal axes have positive length.
This is supported by the statement $B^2-4AC < 0$ saying that the second leading principal minor of $C$ is positive (which is part of Sylvester's criterion for positive-definiteness).

The diagonal matrix $$P^TCP\; =\;\begin{pmatrix} \lambda_1\\ & \lambda_2\\ & & \lambda_3 \end{pmatrix}$$ displays the eigenvalues $\,\lambda_k>0\,$ indicating the lengths of the principal axes. The directions of the latter are given by the unit eigenvectors which build up the orthogonal transformation matrix $P$.