Background:
Let $\Lambda$ be the Lorentz transformation parameterized by the asymmetric real matrix $w_{\mu \nu}$. That is, let $\Lambda = \exp(\frac{w_{\mu \nu}}{2}J^{\mu \nu})$, where $(J^{\mu \nu})_{\alpha \beta} = \delta^\mu_\alpha \delta^\nu_\beta\ - \delta^\mu_\beta \delta^\nu_\alpha$. All indices run from $0$ to $3$, and I am using the metric signature $+---$.
The spin 1/2 representation of the Lorentz group maps $\Lambda$ to $R[\Lambda]\stackrel{\mathrm{def}}{=}\exp(\frac{w_{\mu \nu}}{2}\gamma^\mu \gamma^\nu)$, where $\{ \gamma^\mu \}_{\mu = 0,1,2,3}$ are 4-by-4 complex matrices satisfying $\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu}$.
Question:
If we interpret the $\gamma^\mu$ not as matrices, but as the basis vectors of the geometric algebra $Cl(1,3)$, then we have
$R[\Lambda] x^\mu \gamma_\mu R^{-1}[\Lambda] = \Lambda^{\mu}_\nu x^\nu \gamma_\mu$.
Why is this true? I have no trouble doing the calculation - I am looking for a deeper understanding.
It seems a total coincidence to me that the matrix representation $R[\Lambda]$ happens to be also be the rotor for the Lorentz transformation $\Lambda$. Maybe this indicates that we can dispense with the matrix representation entirely, and somehow formulate the equivalent of a spin 1/2 rep. using $Cl(1,3)$ alone?
Thanks in advance for any help.