Q:

The eigenvalues of a multivector

Since multivectors have a matrix representation, then they should have eigenvalues. In the case of matrices, the characteristic polynomial is defined as:
$$
\det[\lambda I-M]=0
$$
How can I define the characteristic polynomial of a multivector, without referencing the notion of matrices? What is the general procedure to construct the characteristic polynomial without leaving the geometric algebra world?

Edit: 
For instance, in $Cl_3(\mathbb{R})$, represented by the Pauli matrices, the multivector:
$$
\mathbf{v}=a+x\sigma_x+y\sigma_y+z\sigma_z
$$
has the following matrix representation:
$$
V=\pmatrix{a+z&x-iy\\x+iy&a-z}
$$
and its eigenvalues are $\lambda=a\pm\sqrt{x^2+y^2+z^2}$.
The question then becomes, how do I construct, using geometric operations, a polynomial such that its roots are $\lambda$ starting from $\mathbf{v}$ and without converting to matrices, for any multivector of any geometric algebra. Also, how do I adapt the definition $\det[\lambda I-M]=0$ to refer only to geometry, without matrices?
For example, taking the geometric product I get:
$$
\mathbf{v}\mathbf{v}=(a+x\sigma_x+y\sigma_y+z\sigma_z)(a+x\sigma_x+y\sigma_y+z\sigma_z)\\
=a^2+x^2+y^2+z^2+2a(x\sigma_x+y\sigma_y+z\sigma_z)\\
=a^2+x^2+y^2+z^2+2a(x\sigma_x+y\sigma_y+z\sigma_z+a-a)\\
=a^2+x^2+y^2+z^2+2a(\mathbf{v}-a)\\
=-a^2+x^2+y^2+z^2+2a\mathbf{v}\\
\implies \mathbf{v}^2-2a\mathbf{v}+a^2-x^2-y^2-z^2=0\\
\implies \mathbf{v}=a\pm\sqrt{x^2+y^2+z^2}
$$
The procedure creates the characteristic polynomial at the line $\mathbf{v}^2=a^2+x^2+y^2+z^2+2a(\mathbf{v}-a)$, when all basis elements are erased or hidden behind $\mathbf{v}$ of different degrees.
I am able to work it out in a handful of specific cases.
In $Cl_4$ specifically, the polynomial would have to be of degree 4.

A:

The determinants of low dimensional matrices, especially your 2×2 ones, are easiest gotten from traces through the combinatorics of the Jacobi formula. 
In particular, by the  Faddeev–LeVerrier algorithm, you get, for such 2×2 matrices,  a characteristic polynomial
$$
p(\lambda)= \lambda^2 -\lambda \operatorname{Tr} M +\frac{1}{2} ( (\operatorname{Tr} M)^2 - \operatorname{Tr} M^2).
$$
So, for $M= \mathbf v$, you have just 
$$
p(\lambda) = (\lambda-a)^2-r^4, 
$$
etc, in an evident recursive structure. I assume you can work it out?

