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Eigenblades of linear functions

An eigenblade of a linear function is a non-zero blade of arbitrary grade that is only rescaled by a real number when the linear function is applied to it.


Let f: \mathbb{R}^n \rightarrow \mathbb{R}^m be a linear function mapping vectors from \mathbb{R}^n to \mathbb{R}^m. The function f can be extended to act on multivectors using the outermorphism relation f(e_1) \wedge f(e_2) \wedge f(e_3) \wedge ... = f(e_1 \wedge e_2 \wedge e_3 \wedge ...).

A blade B is an eigenblade of f with eigenvalue \lambda \in \mathbb{R} if

f(B) = \lambda B

Relation to eigenvectors

Eigenvectors of f are related to eigenblades by outermorphism, for example for a linear function with two eigenvectors x_1, x_2 with corresponding eigenvalues \lambda_1, \lambda_2

f(x_1) = \lambda_1 x_1, f(x_2) = \lambda_2 x_2

we can see that

f(x_1 \wedge x_2) = f(x_1) \wedge f(x_2) = \lambda_1 \lambda_2 (x_1 \wedge x_2) = \lambda (x_1 \wedge x_2)

meaning f has the eigenblade x_1 \wedge x_2 with eigenvalue \lambda = \lambda_1 \lambda_2.

Finding eigenblades and eigenvalues

Characteristic polynomial

As in conventional linear algebra we can use the determinant to obtain a polynomial equation in \lambda with which we can find the eigenvalues.

Starting from

\begin{aligned} f(B)- \lambda B = 0 \\ \left[ f - \lambda \right](B) = 0 \end{aligned}

a new function f - \lambda acting on the blade B is formed. The determinant of a function f is the eigenvalue of the pseudoscalar of the space I, and therefore we compute

\begin{aligned} 0 &= \det ( f - \lambda ) \\ 0 &= \left[ f - \lambda \right](I)I^{-1} \\ 0 &= \left(\left[ f(e_1) - \lambda e_1 \right] \wedge \cdots \wedge \left[ f(e_n) - \lambda e_n \right] \right) I^{-1} \\ 0 &= \left(f(I) - \lambda e_1 \wedge f(e_2) \wedge \cdots \wedge f(e_n) + \ldots + (-\lambda)^n I \right) I^{-1} \end{aligned}

The equation above is called the characteristic polynomial of f, and its roots are the eigenvalues \lambda. Once the eigenvalues are known, their corresponding eigenblades can be found by inserting them into the original eigenblade equation.


Determinant of a linear function

The determinant det f is an eigenvalue for the eigenblade I (the pseudoscalar) of a linear function f.


Rotors can be constructed as R = e^{B \frac{\phi}{2}} where B is a 2-blade. Applying them to a vector with the sandwich product R x \widetilde{R} will rotate the vector around the origin in the plane represented by B by \phi. B is an eigenblade of this sandwich product with eigenvalue 1.

In 2D, for a rotor R = e^{e_{12} \frac{\phi}{2}} we conclude from above that e_{12} is an eigenblade of the sandwich product with eigenvalue 1. However in this case no eigenvectors with non-zero eigenvalue exist as no vector is left unchanged by the rotation.

In 3D, the same rotor sandwich has eigenvectors with eigenvalue 1 in the direction orthogonal to the rotation plane as these are unchanged by the rotation, for example: e^{e_{12} \frac{\phi}{2}} ze_3 e^{-e_{12} \frac{\phi}{2}} = z e_3.

The red points on the line orthogonal to the rotation plane are left unchanged.



M is not an eigenblade of the sandwiching, because it is not a blade. If M is rhe exp of a 2blade B, that blade B is an eigenblade.

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Thanks for the correction, clearly I need to read up more on this.

We had a discussion in the BiVector discord to create a wiki and this was my first (attempt at) an article. The purpose of it (in my opinion) would be to collect the information that is scattered in different articles and videos / lectures right now and present it in an easy to understand way. So, anyone in the community, feel free to correct anything in the articles and add to them or create new ones (or comment I suppose, since not everyone has the necessary rights to edit the wiki posts).

// Edit: rewrote the sandwich / rotor part, hopefully correct now

In general, you will get complex eigenvalues, what to do with them? The important thing is that in a real geometric algebra we need real eigenvalues.
However, we immediately have two eigenblades, real numbers (the eigenvalue 1) and pseudoscalars (the eigenvalue detF = I F(I)). Now we just need to check by grades. For example, in 3D, we can take vectors and write F (v) = λ v, and then bivectors F (B) = λ B. We will get systems of equations that are easy to solve by methods of geometric algebra.
When we have a linear transformation in the sandwich form, we can usually find eigenblades without calculations: we just look for objects that commute. Consider a rotor in 3D with the bivector e1e2. The blades that commute with e1e2 are e3, e1e2, and I. This gives the eigenblades e3, e1e2, and I, all with the (real) eigenvalue 1.

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