TITLE: When is a multivector in Cl(n) invertible?
QUESTION [1 upvotes]: My question
Let $\mathrm{Cl}(n)$ be the Clifford algebra over $\mathbb{R}^n$ with the usual inner product. That is, it's the quotient of the tensor algebra over $\mathbb{R}^n$ by the ideal generated by
$$\{ v^2 - v \cdot v : v \in \mathbb{R}^n \}.$$
Given an element of $\mathrm{Cl}(n)$, is there a "simple" procedure to decide whether it is invertible? For example, it's easy to check whether a matrix is invertible by checking whether its determinant is nonzero.
What I know
I know that every Clifford algebra is isomorphic to a direct sum of one or two matrix algebras, so I could use that isomorphism and then check the determinant. But I'm hoping that you know a way that doesn't involve quaternion matrix determinants and five cases depending on the dimension. :)
Maybe the determinant formulae for each case will end up identical; I haven't tried it yet.

REPLY [1 votes]: If anything else fails, you always have this tedious but straightforward method at hand.
Let $A$ be any finite-dimensional unital algebra over a field $F$, say $\dim A=n$. Then, $A$ acts on itself by multiplication from the left, providing an algebra homomorphism $\varphi:A \rightarrow \operatorname{End}(A)$ to the algebra of $F$-linear endomorphisms of $A$. Explicitly, $\varphi(a)x = ax$.
The kernel of $\varphi$ consists of elements $a\in A$ such that $\forall x\in A : ax=0$. Substitute $x=1$ to get $a=0$. Thus, $\varphi$ is injective, and $A$ is isomorphic to its image $\varphi(A) \subset \operatorname{End}(A)$. So, $a\in A$ is invertible iff $\varphi(a)$ is.
Finally, $\varphi(a)$ is a linear operator on $A$, thus an $n\times n$-matrix in some basis of $A$, and invertibility of $\varphi(a)$ can be decided by computing the determinant $\det\varphi(a)$.

Consider the case of $A = \operatorname{Cl}^{1,0}(\mathbb R)$ the Clifford algebra of a one-dimensional Euclidean space. It has a basis $\{1, e\}$ with $e^2 = 1$. An element $a + be \in A$ corresponds (in this basis) to a matrix
$$\begin{pmatrix}a & b \\ b & a\end{pmatrix}$$
with determinant $a^2-b^2$. Indeed, whenever $a^2-b^2\neq 0$, we have
$$(a+be)^{-1} = \frac{a-be}{a^2-b^2}$$
If $a^2-b^2 = 0$, then $(a+be)(a-be) = a^2 - b^2 = 0$, so $a+be$ is a zero divisor and cannot be invertible.

As another example, take $A = \operatorname{Cl}^{0,1}(\mathbb R)$ the Clifford algebra of a one-dimensional space with negative-definite form. It is known that $A \cong \mathbb C$, the complex numbers, so every nonzero element should be invertible.
Take the basis $\{1, e\}$ with $e^2=-1$. Now $a+be$ corresponds to
$$\begin{pmatrix}a & -b \\ b & a\end{pmatrix}$$
with determinant $a^2+b^2$, which is zero iff $a=0$ and $b=0$.
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