I have been trying to learn Clifford algebra from Lawson & Michelsohn's Spin Geometry. There is a geometric explanation for the adjoint representation formula ($V$ is a vector space, $q$ is a quadratic form on $V$, $\text{Cl}(V,q)$ is the associated Clifford algebra)
Let $v\in V\subset\text{Cl}(V,q)$ be an element with $q(v)\neq 0$, then for all $w\in V$, the following equation holds $$-\text{Ad}_v(w)=w-2\frac{q(v,w)}{q(v)}v,$$
which says
We observe now that the right-hand side of the equation is just the map $\rho_v:V\to V$ given by reflection across the hyperplane $v^\perp=\{w\in V: q(w,v) = 0\}$. That is, the map $\rho_v$ fixes this hyperplane and maps $v$ to $-v$. Unfortunately, there is a minus sign on the left in equation. This means, for example, that if dim$V$ is odd, then $\text{Ad}_v$ is always orientation preserving.
I don't understand the last remark. Because of the sign on the LHS, $\text{Ad}_v$ would simply be the identity ($v\mapsto v$) in the hyperplane, why does it have to do with dimension of $V$? It's probably just a simple thing I'm not understanding.
We have $-\operatorname{Ad}_v = \rho_v$, so $\operatorname{Ad}_v = -\rho_v$ is the negative identity on $v^{\perp}$, and the identity on the span of $v$. If $\{v_1, \dots, v_{n-1}\}$ is a basis for $v^{\perp}$, then $\{v_1,\dots, v_{n-1}, v\}$ is a basis for $V$. With respect to this basis, the map $\operatorname{Ad}_v$ is represented by the matrix $\operatorname{diag}(-1, -1, \dots, -1, 1)$ which has determinant $(-1)^{n-1}$. Therefore, if $n$ is odd, the map $\operatorname{Ad}_v$ is orientation-preserving.