Basis of Dirac gamma matrices

Question

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.

Answer 1

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.