Basis of Dirac gamma matrices I was reading my professor's notes and this answer on Dirac matrices.
Both sources reached the conclusion that for $N$-dimensional Clifford algebra, the basis is made up of $2^N$ independent matrices.
While the notes used $4$-dimensional matrices as an example, which have $4×4 =16$ elements(and makes it kinda intuitive that it is spanned by $2^4=16$ independent matrices), this argument falls for higher dimensions.
For example, $6$-dimensional matrices have 36 elements, but are spanned by $2^6=64$ independent(?) matrices.
This feels contradictory because the matrix $6×6$ has 36 elements, so there's maximally 36 independent matrices that form that basis. Am I correct?
PS: Linear algebra is not my strongest field, so I apologize if I'm missing a trivial point.
 A: Your professor's note does address this point, and it does it the other way around: the size of the matrices is figured out by using the size of the basis!
Here is the relevant quote (brackets mine):

In $4$ dimensions we used trial and error to figure out the dimensions of Dirac matrices. 
[...]
Since the [cardinality of a] basis is $\mathcal N$, then the Dirac matrices in $D$ dimensions have $\mathcal N$ entries and as such they are $\sqrt{\mathcal N}\times \sqrt{\mathcal N}$ matrices.

A: The Clifford algebra associated with an $n$-dimensional real vector space with an inner product is $2^n$-dimensional, and if you want to represent the elements of this Clifford algebra with matrices, the size of the matrices and the type of the entries (real numbers, complex numbers, or quaternions) must be such that the dimension of that matrix space (as a real vector space) is $2^n$ too.
In the case of $\mathbf{R}^4$ with the Minkowski metric, it so happens that $4 \times 4$ real matrices work, but if you change the sign of the metric, you have to use $2 \times 2$ quaternionic matrices instead ($4$ entries with $4$ real degrees of freedom in each, so dimension $2^4=16$ again), in order for the matrix multiplication to behave like the multiplication in the Clifford algebra. And for $\mathbf{R}^3$ with the Euclidean metric, the Clifford algebra is isomorphic to the algebra of  $2 \times 2$ complex matrices (the Pauli matrices, $4$ entries with $2$ real degrees of freedom in each, so dimension $2^3=8$).
The full story is somewhat complicated; see for example Wikipedia: Classification fo Clifford algebras to get an idea. (Although there are probably more pedagogical explanations out there if you look around a bit.)
