Ellipsoid coefficient normalization to get semi-axes

eigenvalues-eigenvectorsmatricesquadrics

I have an ellipsoid. Its quadric equation is :

$Ax^2 + By^2 + Cz^2 + Dxy + Eyz + Fxz + Gx + Hy + Iz + J = 0$

The quadratic part of the quadric equation can be expressed with a matrix :

\begin{bmatrix}A&D/2&F/2\\D/2&B&E/2\\F/2&E/2&C\end{bmatrix}

I read here https://en.wikipedia.org/wiki/Ellipsoid that the eigenvalues of the above matrix are the reciprocals of the squares of the semi-axes of my ellipsoid.

But because the quadric equation is homogeneous, the coefs (A, B, C…) are not important, only their ratio is. But different values of the coefs lead to different values of the eigenvalues (but the eigenvectors remain always the same).

Have I to normalize the coefs to something to get the right eigenvalues ?

Thanks

Best Answer

You already noticed that there is a problem with the normalization. Let's look at the simple case, where only $A,B,C,J$ are non-zero. In that case, the equation for your ellipse is $$Ax^2+By^2+Cz^2+J=0$$ You want to get to $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$$ That means you need to move $J$ top the right hand side, then divide everything by $-J$. In that case $$\frac1{a^2}=-\frac AJ$$ and similar for the other coefficients.

The problem is just a little more complicated when your ellipse is off center. If you write $$(\mathbf x-\mathbf v)^TM(\mathbf x-\mathbf v)=1$$ you notice that the center $\mathbf v$will shift the value of $J$

Related Solutions

[Math] Does an Implicit equation for an infinite Cylinder exist ? i.e f(x,y,z)=0

If the axis of the cylinder is the line through $\vec{x}_0 = (x_0,y_0,z_0)$ and in the direction of $\vec{v} = (a,b,c)$ with radius $R$, assuming $|\vec{v}| = 1$, we can write the equation of the cylinder as the set of all points $\vec{x} = (x,y,z)$ satisfying:

$$|\vec{x} - \vec{x}_0|^{2} = R^{2} + [(\vec{x} - \vec{x}_0)\bullet \vec{v}]^{2}$$

[Math] Plotting an ellipsoid using eigenvectors and eigenvalues

Turns out I was indeed doing two things wrong!

1. Axes order
Like I suspected — and mentioned in the OP as an important note — the frame defined by the axes of unit eigenvectors $x$ was sometimes switching direction, and thus not respecting the right-hand rule and world frame convention.

To overcome this, I just reassigned the 3rd eigenvector as the cross product between the 1st and 2nd eigenvectors:

x[:, 2] = np.cross(x[:, 0], x[:, 1])

Since the eigenvectors $x$ returned by numpy.linalg.eig are already normalized (i.e., unit eigenvectors), there is no need to normalise the result of the cross product.

2. Rotation matrix
The second thing I was not doing properly was during the construction of the $3$x$3$ rotation matrix. I was missing a transpose operation.

To solve this I just changed the indexing of the eigenvectors $x$ to:

rot_matrix = np.array([x[0, :],
                       x[1, :],
                       x[2, :]])

Using numpy.transpose would have worked too.

Below is the correct expected final result:

final-result

Special thanks to Vladimir Ivan (a.k.a. "Slovak")!

Best Answer

Related Solutions

[Math] Does an Implicit equation for an infinite Cylinder exist ? i.e f(x,y,z)=0

[Math] Plotting an ellipsoid using eigenvectors and eigenvalues

Related Question