Let $\mathbf{v}_1,\mathbf{v}_2$ and $\mathbf{u}$ be a multivector of $Cl_4(\mathbb{C})$:
Example #1 :
$\mathbf{v}_1=t\gamma_0+x\gamma_1+y\gamma_2+z\gamma_3$
The matrix representation of $\mathbf{v}_1$, using the gamma matrices is:
$$ \pmatrix{t&0&z&x-iy\\0&t&x+iy&-z\\-z&-x+iy&-t&0\\-x-iy&z&0&-t} $$
The two unique eigenvalues of this matrix are $\lambda=\pm\sqrt{t^2-x^2-y^2-z^2}$, which we recognize as the interval of special relativity.
Example #2 :
$\mathbf{v}_2=E_1\gamma_0\gamma_1+E_2\gamma_0\gamma_1+E_3\gamma_0\gamma_3+B_1\gamma_2\gamma_3+B_2\gamma_1\gamma_3+B_3\gamma_1\gamma_2$
The four unique eigenvalues of its matrix representation are $\lambda=\pm\sqrt{-B_1^2-B_2^2-B_3^2+E_1^2+E_2^2+E_3^2\pm2\sqrt{-(B_1 E_1+B_2E_2+B_3E_3)^2}}$. It is also quite interesting, because if we write $\mathbf{E}=(E_1,E_2,E_3)$ and $\mathbf{B}=(B_1,B_2,B_3)$, we can write the eigenvalues as:
$$ \lambda =\pm \sqrt{-||\mathbf{B}||^2+||\mathbf{E}||^2\pm 2i \mathbf{B}\cdot \mathbf{E}} $$
These are simply the two Lorentz invariants of electromagnetism.
So in each case, there is a physical interpretation to the eigenvalues of the multivector.
Now, I am trying to look at more complicated cases and see if there is even more physics to be found. Specifically, I am looking at the interference pattern found within the eigenvalues of $\mathbf{u}=\mathbf{v}_1+\mathbf{v}_2$, or:
$$ \mathbf{u}=t\gamma_0+x\gamma_1+y\gamma_2+z\gamma_3+E_1\gamma_0\gamma_1+E_2\gamma_0\gamma_1+E_3\gamma_0\gamma_3+B_1\gamma_2\gamma_3+B_2\gamma_1\gamma_3+B_3\gamma_1\gamma_2 $$
The matrix representation of $\mathbf{u}$, using the gamma matrices, is:
$$ \left( \begin{array}{cccc} t-i \text{B3} & -\text{B2}+(-i) \text{B1} & \text{E3}+z & \text{E1}-i \text{E2}+x-i y \\ \text{B2}-i \text{B1} & t+i \text{B3} & \text{E1}+i \text{E2}+x+i y & -\text{E3}-z \\ \text{E3}-z & \text{E1}-i \text{E2}-x+i y & -t-i \text{B3} & -\text{B2}+(-i) \text{B1} \\ \text{E1}+i \text{E2}-x-i y & z-\text{E3} & \text{B2}-i \text{B1} & -t+i \text{B3} \\ \end{array} \right) $$
and the eigenvalues are too verbose and Mathematica nearly gives up, but manually, we can find a pattern. Let $||\mathbf{s}||^2$ be the interval of special relativity. Then eigenvalues of $\mathbf{u}$ are:
$$ \lambda=\pm\sqrt{-||\mathbf{B}||^2+||\mathbf{E}||^2+||\mathbf{s}||^2\pm 2i\sqrt{(\mathbf{B}\cdot\mathbf{E})^2 + \text{interference-pattern}}} $$
As we can see, it is a unification of the two previous results, but with the addition of a complicated "interference-pattern" between the two. My question is what is the physical meaning of this interference pattern?
Let me give the interference pattern explicitly:
$$ -B_1^2 t^2 - B_2^2 t^2 - B_3^2 t^2 + 2 B_3 E_2 t x - 2 B_2 E_3 t x + B_1^2 x^2 - E_2^2 x^2 - E_3^2 x^2 - 2 B_3 E_1 t y + 2 B_1 E_3 t y + 2 B_1 B_2 x y + 2 E_1 E_2 x y + B_2^2 y^2 - E_1^2 y^2 - E_3^2 y^2 + 2 B_2 E_1 t z - 2 B_1 E_2 t z + 2 B_1 B_3 x z + 2 E_1 E_3 x z + 2 B_2 B_3 y z + 2 E_2 E_3 y z + B_3^2 z^2 - E_1^2 z^2 - E_2^2 z^2 $$
Trying to "organize" it a little bit I get (I pose $\mathbf{x}=(x,y,z)$):
$$ -t^2 ||\mathbf{B}||^2 + (\mathbf{x}\cdot \mathbf{B})^2 - (E_2 x-E_1 y)^2-(E_3 x- E_1z )^2-(E_3 y -E_2z)^2 - 2t \det \pmatrix{x&y&z\\B_1&B_2&B_3\\E_1&E_2&E_3} $$
Your first example v$_1$ is straightforward, and perhaps you'd wish to see a pattern in it to generalize to the rest.
Gamma matrices are special, in that any odd power of them is traceless, and the traces of its square and 4th power are trivial, so for your $ {\mathbf v}_1=X\cdot \gamma$, the corresponding traces are trivial functions of the interval,
$$ \operatorname{Tr} {\mathbf v}^2_1 = 4 X^2 , \qquad \operatorname{Tr} {\mathbf v}^4_1 = 4 (X^2)^2. $$
These 4×4 matrices then have a simple characteristic polynomial found by the virtually collapsing Faddeev–LeVerrier algorithm, $$ p(\lambda)= \det (\lambda I - {\mathbf v}_1 )= \det {\mathbf v}_1 -\frac{\lambda^2}{2} \operatorname{Tr} {\mathbf v}^2_1 +\lambda^4 , $$ while the determinant of the leading term is famously evident from the algorithm -- but also the trace of the C-H theorem expression for a general 4×4 matrix , $$ 4\det {\mathbf v}_1= \tfrac{1}{2} (\operatorname{Tr} {\mathbf v}^2_1)^2 - \operatorname{Tr} {\mathbf v}^4_1, $$ so that $$ p(\lambda) = (\lambda^2 - X^2)^2, $$ with the double roots you found.
Perhaps this might inspire you for the further cases. The general theme is that traces of powers are simpler to evaluate than other invariants: we are back to Cayley's theory of invariants and Hilbert's molding of it!