One option is to start out with the matrix representation for two sets of conjugate Grassmann numbers (see previous thread), $\theta_i, \pi_i$ with $i=1,...,N$, such that
$\{\theta_i,\theta_j\}=0,\quad\{\pi_{i},\pi_{j}\} = 0, \quad \{\theta_i,\pi_j\} = \delta_{ij}$
Then a $2N$-dimensional Clifford algebra can be built by
$\gamma_{i}=\theta_{i}+\pi_{i}\\
\gamma_{N+i}=i(\theta_{i}-\pi_{i})$
Given the above anti-commutation relations it is straightforward to verify that $\{\gamma_{i},\gamma_{j}\}=2\delta_{ij}\mathbf{1}$. For a odd number of dimensions the last $\gamma$-matrix can be found by considering the product
$\gamma_{2N+1} = i^N\prod_{i=1}^{2N}\gamma_{i} = i^N\gamma_{1}\gamma_2...\gamma_{2N}$
To get a representation of the Dirac algebra $\{\gamma_{\mu},\gamma_\nu\}=2g_{\mu\nu}\mathbf{1}$ with signature (+,-,-,...,-) simply rotate all but one of the matrices in the representation such that $\gamma_i\to i\gamma_i$ (and relabel a bit).
This approach enables one to derive general representations of the gamma matrices from Grassmann numbers.
However, another option exists, namely to start out with a lower dimensional representation of the Clifford algebra (which can be computed by the method described above). A well-known case of a lower-dimensional representation, which was also known to Weyl & Dirac, would be the Pauli matrices:
$\sigma_{1} = \left[\begin{matrix}
0 & 1 \\
1 & 0
\end{matrix}\right], \quad
\sigma_{2} = \left[\begin{matrix}
0 & -i \\
i & 0
\end{matrix}\right], \quad
\sigma_{3} = \left[\begin{matrix}
1 & 0 \\
0 & -1
\end{matrix}\right] $
From these matrices outer products, $\rho_i = \mathbf{1}\otimes \sigma_i$ and $\eta_i = \sigma_i \otimes \mathbf{1}$, can be formed. It is then clear that $[\rho_i,\eta_j]=0$ which makes it possible to choose five matrices from the set $\{\rho_i,\eta_j,\rho_i\eta_j\}$ which fulfill the Clifford algebra.
To make this approach a bit more explicit, consider starting with a diagonal matrix from the initial set for simplicity - let us choose $\rho_3$ ($\eta_3$ would have been another option). This leaves us with two potential sets of matrices, namely $\{\rho_1,\rho_2\eta_1,\rho_2\eta_2,\rho_2\eta_3\}$ and $\{\rho_{2},\rho_{1}\eta_{1},\rho_{1}\eta_{2},\rho_{1}\eta_{3}\}$. Since $\rho_1$ is real I choose the first set, making the matrices:
\begin{align}
\gamma_0 &= \left[\begin{matrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{matrix}\right] = \rho_3,
&&\gamma_1 = \left[\begin{matrix}
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & -1 & 0 & 0 \\
-1 & 0 & 0 & 0
\end{matrix}\right] = i\rho_2\eta_1 \\
\gamma_2 &= \left[\begin{matrix}
0 & 0 & 0 & -i \\
0 & 0 & i & 0 \\
0 & i & 0 & 0 \\
-i & 0 & 0 & 0
\end{matrix}\right] = i\rho_2\eta_2,
&&\gamma_3 = \left[\begin{matrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -1 \\
-1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0
\end{matrix}\right] = i\rho_2\eta_3\\
\gamma_5 &= \left[\begin{matrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0
\end{matrix}\right] = \rho_1
\end{align}
where I have taken the liberty to rotate three of them as described above. In this way the Dirac representation is found. Notice that a few choices were made along the way but that several of them can be motivated by the search for a simple representation (choosing diagonal and/or real when possible).
This approach can naturally be generalized to generate higher dimensional representations as well.
As previous answers have correctly noted gamma matrices do not forma a basis of $M(4,\mathbb{C})$. Nevertheless you can construct one from them in the following way
- 1 the identity matrix $\mathbb{1}$
- 4 matrices $\gamma^\mu$
- 6 matrices $\sigma^{\mu\nu}=\gamma^{[\mu}\gamma^{\nu]}$
- 4 matrices $\sigma^{\mu\nu\rho}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho]}$
- 1 matrix $\sigma^{\mu\nu\rho\delta}=\gamma^{[\mu}\gamma^{\nu}\gamma^{\rho}\gamma^{\delta]}=i\epsilon^{\mu\nu\rho\delta}\gamma^5$
these 16 matrices form the basis that we were looking for.
Furthermore they are used to construct the spinor bilinears multiplying by $\bar{\psi}$ on the left and $\psi$ on the right, which transform in the Lorentz indices as follows
- $\bar{\psi}\psi$ scalar
- $\bar{\psi}\gamma^\mu\psi$ vector
- $\bar{\psi}\sigma^{\mu\nu}\psi$ 2nd rank (antisymmetric) tensor
- $\bar{\psi}\sigma^{\mu\nu\rho}\psi$ pseudovector
- $\bar{\psi}\gamma^5\psi$ pseudoscalar
the fact that they form a basis of of $M(4,\mathbb{C})$ is very important because these are then the only independent spinor bilinears (i.e $\bar{\psi}M\psi$) that can be constructed, any other can be expressed a linear combination of these. A different issue is if it would make any sense to sum any of these since they are different types of tensors under Lorentz group transformations.
Best Answer
The answer is negative. If you have a representation of the generators of the algebra made of matrices $\gamma^\mu$ and $S$ is an invertible matrix, Dirac's commutation relations $$\{\gamma^{\mu},\gamma^{\nu}\}=2g^{\mu\nu}I$$ are valid also replacing $\gamma^\mu$ for $\gamma'^\mu := S\gamma^\mu S^{-1}$.
However, if $S$ is not unitary your second commutation relation generally fails as it is generally false that $$(S\gamma^\mu S^{-1})^\dagger = S(\gamma^\mu)^\dagger S^{-1}$$ for $S$ generic. This proves that the commutation relations $$(\gamma^{\mu})^{\dagger}\gamma^{0}=\gamma^{0}\gamma^{\mu}$$ are not consequence of the Dirac's ones, thus they are not valid in general, but they depend on the choice of the representation.