The exact relationship between Clifford algebras and certain special isomorphic k-algebras question(s):
Choose any real or complex clifford algebra $\mathcal{Cl}_{p,q}$.  It's known that there is some $A \simeq \mathcal{Cl}_{p,q}$, where $A$ is either a matrix ring $M(n,R)$ or a direct sum of matrix rings $M(n,R)\oplus M(n,R)$, for $n \geq 1$ and $R \in \{\mathbb R, \mathbb C, \mathbb H\}$ such that $\dim(\mathcal{Cl}_{p,q}) = \dim(A)$ (as k-algebras).  


*

*Given an element $X\in \mathcal{Cl}_{p,q}$, defined on a standard basis, how can I find an explicit isomorphism $f:\mathcal{Cl}(p,q)\to A$, that preserves algebraic properties such as grade interaction?  What is the "character" (qualitatively or otherwise) of such an isomorphism?  Is it part of some group like the general linear or orthogonal groups?

*Conversely, say $y^{i,j} \in R$ is an entry of some matrix $Y\in A$, and $y^{i,j}_k \in \mathbb R$ is a component of $y^{i,j}$. Let's call it a component of $A$ as well.  What is the relationship of $y^{i,j}_k$ to $\mathcal{Cl}_{p,q}$?  Does it have a single grade that can be determined by its grade in $R$? Or is there something more nuanced going on?  How is it related to other components of $A$?

*There are cases where a given $A$ is isomorphic to multiple Clifford algebras.  Then these Clifford algebras must be isomorphic to one another.  For example: $\mathcal{Cl}_{7,0} \simeq \mathcal{Cl}_{5,2} \simeq \mathcal{Cl}_{3,4} \simeq \mathcal{Cl}_{1,6} \simeq M(8,\mathbb C)$.  What's going on here? Is there a mathematical term for these kinds of special isomorphisms between Clifford algebras in general?
BONUS QUESTION:  how is this problem referred to in the academic math literature? I have scoured, with my amateur math education, the "matrix representations of Clifford algebras" stuff, and mostly found stuff about real matrix representations of real Clifford algebras with real matrix generators, etc, but that's not what I'm looking for.  How to distinguish?
context:
I have been using sage and sympy (computer algebra systems) to compute symbolic monomial representations of products for various Clifford algebras. These are then used to generate GPU code for geometric algebra usage.
It works nice for Clifford algebras with up to 7 generating dimensions, and for low grade computations like planar rotation, I can get up to 8 or 9 generating dimensions. I've been successful in pumping out reasonably fast 8 dimensional planar rotation functions in glsl.
Recently, I was reading about Bott periodicity and the classification of Clifford algebras. Using the equations given on the wikipedia page "classification of Clifford algebras", I tried generating some of these isomorphic algebras.
For whatever reason, they are much, much faster to generate, likely due to the ubiquity of matrix multiplication. But I have absolutely no idea how, in general, I would construct a generic multivector in them.  For example, generating a symbolic representation of $M(4,\mathbb H)$ is very quick in Sage. Presumably there is some isomorphism between this and the 64 dimensional $\mathcal{Cl}_{2,4}$. But how do I use it?
 A: Here is the simplest way I know of getting matrix representations of even or full Clifford algebras of vector spaces over $\def\|#1{\mathbb#1}\|R$ or $\|C$ with arbitrary nondegenerate signature. There are just two recursive rules and a trivial base case. It may also work over other fields but some details would have to be changed.
Summary: you can always factor out a copy of $\|C$, $\|H$, $\|D$ (the split-complex numbers) or $\|P$ (the split quaternions) until you get to a trivial algebra that is isomorphic to the base field. You can convert the resulting tensor product to a matrix representation using the isomorphisms $$\|D\cong\|R\oplus\|R, \quad \|P\cong M_2(\|R), \quad \|C\otimes\|C\cong\|C\otimes\|D, \quad \|C\otimes\|H\cong \|C\otimes\|P, \quad \|H\otimes\|H\cong \|P\otimes\|P$$ (and the fact that $\|R$ is the identity and $\otimes$ distributes over $\oplus$ and $M$).
Details:


*

*The full algebra in $0$ dimensions and the even algebra in ${\le}1$ dimension are isomorphic to the base field.

*If there is a unit pseudoscalar $ω$ that is not a scalar, then any element can be uniquely written as $A+Bω$ where $A$ and $B$ belong to the subalgebra of a subspace of codimension one. If additionally $ω$ commutes with all elements of the subalgebra, then the algebra is isomorphic to $\|C$ (if $ω^2=-1$) or $\|D$ (if $ω^2=1$) tensored with the subalgebra. This works in dimension at least 1, for full algebras in odd dimensions and even algebras in even dimensions. It doesn't work for full algebras in even dimensions because the pseudoscalar anticommutes with odd elements, and it doesn't work for even algebras in odd dimensions because there is no pseudoscalar.

*Let $\{e_1, \ldots, e_n\}$ be an orthonormal basis for the vector space, and define (if possible) $i=e_1e_2,\;j=e_2\cdots e_n,\;k=ij$. Note that regardless of signature, $i^2j^2k^2=-i^4j^4=-1$. If $i$, $j$, $k$ commute with all elements of the subalgebra of the subspace spanned by $\{e_3, \ldots, e_n\}$, then the algebra is isomorphic to $\|H$ (if $i^2,j^2,k^2$ are all negative) or $\|P$ (if two of them are positive) tensored with the subalgebra. (You may have to permute $i,j,k$ to get a standard basis where $j^2=k^2=ijk$.) This works in dimension at least 2, for full algebras in even dimensions and for even algebras in odd dimensions. It doesn't work for full algebras in odd dimensions because $j$ and $k$ anticommute with odd elements, and it doesn't work for even algebras in even dimensions because $j$ and $k$ don't exist.
Note that precisely one of these rules applies in any given situation. However, you have a choice of subspaces when applying the recursive rules, so there are many possible factorizations.
As for the isomorphisms:


*

*$\|D\cong\|R\oplus\|R{:}\;1\leftrightarrow(1,1),\;i\leftrightarrow(1,-1)$

*$\|P\cong M_2(\|R){:}\; 1,i,j,k \leftrightarrow \def\M#1{\bigl(\begin{smallmatrix}#1\end{smallmatrix}\bigr)} \M{1&0\\0&1}, \M{0&1\\-1&0}, \M{0&1\\1&0}, \M{1&0\\0&-1}$

*$\|C\otimes\|C\cong\|C\otimes\|D{:}\; i' \leftrightarrow ii'$

*$\|C\otimes\|H\cong \|C\otimes\|P{:}\; j', k' \leftrightarrow ij', ik'$

*$\|H\otimes\|H\cong \|P\otimes\|P{:}\; j, k, j', k' \leftrightarrow ji', ki', ij', ik'$

Here's a worked example: the even algebra of $\|R^{1,3}$.
The pseudoscalar squares to $-1$ so we factor out $\|C$, leaving us with either $\|R^{0,3}$ or $\|R^{1,2}$. I'll pick the latter.
Let $\{\hat t,\hat x,\hat y\}$ be an orthonormal basis for this subspace with $\hat t^2=1$. Let $i=\hat x\hat y,\;j=\hat y\hat t,\;k=\hat t\hat x$. We factor out $\|P$, leaving us with $\|R^{0,1}$ or $\|R^{1,0}$.
The even Clifford algebra of $\|R^{0,1}$ or $\|R^{1,0}$ is just $\|R$, so our original algebra was isomorphic to $\|C\otimes\|P \cong M_2(\|C)$.
Explicitly, we can take $\hat t\hat x\hat y\hat z = \M{i&0\\0&i},\ \hat x\hat y = \M{0&1\\-1&0},\ \hat y\hat t = \M{0&1\\1&0}$, and this generates the rest of the algebra.

A weakness of this method is that it doesn't give you a chiral basis for the full algebra in even dimensions. You can work around that by using a third recursive rule:


*

*Let $\{e_1, \ldots, e_n\}$ be an orthonormal basis for the vector space, and define (if possible) $i=e_1,\;j=e_2\cdots e_n,\;k=ij$. If $i^2j^2k^2=-1$, then the algebra is isomorphic to $\|H$ or $\|P$ tensored with the even subalgebra of the subspace spanned by $\{e_2, \ldots, e_n\}$. This works in nonzero even dimension for full algebras only. It doesn't work in odd dimensions because $i^2j^2k^2=1$.




$\mathcal{Cl}_{7,0} \simeq \mathcal{Cl}_{5,2} \simeq \mathcal{Cl}_{3,4} \simeq \mathcal{Cl}_{1,6} \simeq M(8,\mathbb C)$.  What's going on here?

$M(8,\mathbb C)$ only makes sense as a Clifford algebra if you also supply an embedding of the underlying vector space, and these embeddings will necessarily be different for nonisomorphic vector spaces, so I wouldn't say that these algebras are isomorphic in any particularly interesting sense.
For what it's worth, the factorization in this answer can get you explicit isomorphisms fairly easily. You will get for each of these algebras a factor of $\|C$ and three factors of $\|H$ or $\|P$, and after converting the $\|H$s to $\|P$s (or vice versa), the map taking generators to their counterparts in another algebra extends to an isomorphism of the algebras.
