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.

# Definition

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

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

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.

# Examples

## Determinant of a linear function

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

## Rotors

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.}