Dimension of Dirac $\gamma$ matrices

Let us generalize from four space-time dimensions to a $$d$$-dimensional Clifford algebra $$C$$. Define

$$p~:=~[\frac{d}{2}], \tag{1}$$

where $$[\cdot]$$ denotes the integer part. OP's question then becomes

Why must the dimension $$n$$ of a finite dimensional representation $$V$$ be a multiple of $$2^p$$?

Proof:

1. If $$C\subseteq {\rm End}(V)$$ and $$V$$ are both real, we may complexify, so we may from now on assume that they are both complex. Then the signature of $$C$$ is irrelevant, and hence we might as well assume positive signature. In other words, we assume that we are given $$n\times n$$ matrices $$\gamma_{1}, \ldots, \gamma_{d}$$, that satisfy $$\{\gamma_{\mu}, \gamma_{\nu}\}_+~=~2\delta_{\mu\nu}{\bf 1}, \qquad \mu,\nu~\in~\{1,\ldots, d\}.\tag{2}$$

2. We may define $$\gamma_{\mu\nu}~:=~ \frac{1}{2}[\gamma_{\mu}, \gamma_{\nu}]_- ~=~-\gamma_{\nu\mu}, \qquad \mu,\nu~\in~\{1,\ldots, d\}. \tag{3}$$ In particular, define $$p$$ elements $$H_1, \ldots, H_p,\tag{4}$$ as $$H_r ~:=~i\gamma_{r,p+r}, \qquad r~\in~\{1,\ldots, p\}.\tag{5}$$

3. Note that the elements $$H_1,\ldots, H_p$$, (and $$\gamma_d$$ if $$d$$ is odd), are a set of mutually commuting involutions $$[H_r,H_s]_- ~=~0, \qquad r,s~\in~\{1,\ldots, p\},\tag{6}$$ $$H_r^2 ~=~{\bf 1}, \qquad r~\in~\{1,\ldots, p\}.\tag{7}$$

4. Therefore, according to Lie's Theorem, then $$H_1,\ldots, H_p$$, (and $$\gamma_d$$ if $$d$$ is odd), must have a common eigenvector $$v$$.

5. Since $$H_1,\ldots, H_p$$ are involutions, their eigenvalues are $$\pm 1$$. In other words, $$H_1 v~=~(-1)^{j_1} v, \quad \ldots, \quad H_p v~=~(-1)^{j_p} v,\tag{8}$$ where $$j_1,\ldots, j_p~\in ~\{0,1\} \tag{9}$$ are either zero or one.

6. Apply next the $$p$$ first gamma matrices $$\gamma^{1}, \gamma^{2}, \ldots, \gamma^{p}, \tag{10}$$ to the common eigenvector $$v$$, so that $$v_{(k_1,\ldots, k_p)}~:=~ \gamma_{1}^{k_1}\gamma_{2}^{k_2}\cdots\gamma_{p}^{k_p} v, \tag{11}$$ where the indices $$k_1,\ldots, k_p~\in ~\{0,1\} \tag{12}$$ are either zero or one.

7. Next note that $$[H_r,\gamma_s]_-~=~0 \quad \text{if}\quad r~\neq~ s \mod p \tag{13}$$ and $$\{H_r,\gamma_r\}_+~=~0. \tag{14}$$ It is straightforward to check that the $$2^p$$ vectors $$v_{(k_1,\ldots, k_p)}$$ also are common eigenvectors for $$H_1,\ldots, H_p$$. In detail, $$H_r v_{(k_1,\ldots, k_p)}~=~(-1)^{k_r+j_r}v_{(k_1,\ldots, k_p)}.\tag{15}$$

8. Note that each eigenvector $$v_{(k_1,\ldots, k_p)}$$ has a unique pattern of eigenvalues for the tuple $$(H_1,\ldots, H_p)$$, so the $$2^p$$ vectors $$v_{(k_1,\ldots, k_p)}$$ must be linearly independent.

9. Since $$\gamma_{p+r}~=~ i H_r \gamma_r, \qquad r~\in~\{1,\ldots, p\}, \tag{16}$$ we see that $$W~:=~{\rm span}_{\mathbb{C}} \left\{ v_{(k_1,\ldots, k_p)} \mid k_1,\ldots, k_p~\in ~\{0,1\} \right\} \tag{17}$$ is an invariant subspace $$W\subseteq V$$ for $$C$$.

10. This shows that that any irreducible complex representation of a complex $$d$$-dimensional Clifford algebra is $$2^p$$-dimensional.

11. Finally, we believe (but did not check) that a finite dimensional representation $$V$$ of a complex Clifford algebra is always completely reducible, i.e. a finite sum of irreducible representations, and hence the dimension $$n$$ of $$V$$ must be a multiple of $$2^p$$. $$\Box$$

Intuitive explanation

Preliminary: A vector has a many components as elements of the vector space basis.

A Clifford algebra basis is generated by all (independent) products of the generators (in Dirac's equation case these are the $\gamma$'s).

The counting

There are as many $\gamma$'s as the dimension of the spacetime, and according to the definition the algebra includes a unit, $$\bigl\{\gamma^a,\gamma^b\bigr\} = 2 \eta^{ab}\mathbf{1}.$$

For any extra element the new basis consists of the previous basis elements plus the product of each of those by the extra element. This is the new basis has twice the elements. Therefore, $$\dim(\mathcal{C}\ell(n)) = 2^{n}.$$

In order to represent this algebra one needs "matrices" of $2^{n/2}\times 2^{n/2}$, which is not bad for even dimensional spacetimes.

Said that, the problem (which I don't intend to demonstrate) comes with odd dimensional spacetimes... however, intuitively again, this algebra can be represented by two copies of the co-dimension one algebra, i.e. one dimension less. This reason is why the minimal dimensionality for the representation of the $\gamma$'s is $$\dim(\gamma) = 2^{\lfloor n/2\rfloor}\times 2^{\lfloor n/2 \rfloor}.$$

If you wonder whether one can find a bigger representation of the $\gamma$'s, the answer is YES, but you will will end up with either a non-fundamental or a trivial extension.

Thats a good question. To answer this lets start with Clifford algebra generated by $$\gamma$$ matrices. $$$$\gamma_{\mu}\gamma_{\nu}+ \gamma_{\mu}\gamma_{\nu}=2\eta_{\mu\nu}$$$$ with $$\mu,\nu=0,1,2,\cdots N$$ with the metric signature $$\eta_{\mu\nu}=\text{diag}(+,-,-,-,\cdots,-)$$. Using $$I$$ and $$\gamma_{\mu}$$ we can construct a set of matrices as follow $$$$I, \gamma_{\mu},\gamma_{\mu}\gamma_{\nu}\quad(\mu<\nu), \gamma_{\mu}\gamma_{\nu}\gamma_{\lambda}\quad(\mu<\nu<\lambda),\cdots,\gamma_{1}\gamma_{2}\cdots\gamma_{N}.$$$$

There are $$$$\sum_{p=0}^{N}\binom{N}{p} = 2^{N}$$$$ such matrices. Lets call them $$\Gamma_{A}$$, where $$A$$ runs from $$0$$ to $$2^{N}-1$$. Now let $$\gamma_{\mu}$$ are $$d\times d$$ dimensional irreducible matrices. Our goal is to find a relation between $$d$$ and $$N$$. To this end lets define a matrix $$$$S = \sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1}Y\Gamma_{A}$$$$. Where $$Y$$ is some arbitary $$d\times d$$ matrix. It is follows that $$$$(\Gamma_{B})^{-1}S\Gamma_{B} = \sum_{A=0}^{2^N-1}(\Gamma_{A}\Gamma_{B})^{-1}Y\Gamma_{A}\Gamma_{B} =\sum_{C=0}^{2^N-1}(\Gamma_{C})^{-1}Y\Gamma_{C}=S$$$$ Where we have used $$\Gamma_{A}\Gamma_{B}=\epsilon_{AB}\Gamma_{C}$$, with $$\epsilon_{AB}^{2}=1$$

Hence $$$$S\Gamma_{A}=\Gamma_{A}S$$$$ Since $$S$$ commutes with all the matrices in the set, by Schur's lemma we conclude that $$S$$ must be proportional to the identity matrix so that we can write $$$$S = \sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1}Y\Gamma_{A} = \lambda I$$$$

Taking trace we get $$\begin{eqnarray} \text{Tr} S & = & \sum_{A=0}^{2^N-1} \text{Tr} Y = \lambda d\\ \Rightarrow \lambda & = & \frac{2^{N}}{d}\text{Tr} Y \end{eqnarray}$$ or $$$$\sum_{A=0}^{2^N-1}(\Gamma_{A})^{-1}Y\Gamma_{A} = \frac{2^{N}}{d}\text{Tr} Y$$$$

Taking the $$(j; m)$$ matrix element of both sides of last equation yield $$$$\sum_{A=0}^{2^N-1}((\Gamma_{A})^{-1})_{jk}(\Gamma_{A})_{km} = \frac{2^{N}}{d}\delta_{jm} \delta_{kl}$$$$ where $$j; k; l; m = 1; 2;\cdots; d$$ and we have used the fact that Y is an arbitrary $$d \times d$$ matrix. If we set $$j = k; l = m$$ and sum over these two indices, that gives $$$$\sum_{A=0}^{2^N-1} \text{Tr}[(\Gamma_{A})^{-1}] \text{Tr}[\Gamma_{A}] = 2^{N}$$$$ There are two cases to consider, namely, $$N$$ even and $$N$$ odd. For $$N = 2M$$ (even), $$\text{Tr} \Gamma_{A} = 0$$ except for $$\Gamma_{0} = 1$$ for which $$\text{Tr} \Gamma_{0} = d$$. Which gives $$$$d^2 = 2^N\qquad \text{or} \quad \boxed{d = 2^{N/2}}$$$$ This is the main result. For four dimensional Minkowski space time $$N=4$$ cosequntly the dimension of irreducible representation is $$d = 2^{4/2} =4$$.