I) Yes, if OP insists of the presence of Grassmann variables$^1$, it is possible to represent the fermionic operators as matrices with the caveat that the fermionic Fock space of states is a super vector space, and the matrices are super matrices.
If we have 2 creation operators $\hat{c}^{\dagger}_{\sigma}$, $\sigma\in \{\uparrow,\downarrow\}$, then there are:
2 bosonic states (1 vacuum state $\left|0\right>$ and 1 two-particle state $\left|\uparrow\downarrow\right>$), and
2 fermionic single-particle states, $\left|\uparrow\right>$ and $\left|\downarrow\right>$.
See also e.g. my Phys.SE answer here. Let us represent the 4 states
$$ \left|0\right>
=\begin{pmatrix} 1\\ 0 \\ 0 \\ 0 \end{pmatrix}, \qquad
\left|\uparrow\downarrow\right>
=\begin{pmatrix} 0\\ 1 \\ 0 \\ 0 \end{pmatrix}, \qquad
\left|\uparrow\right>
=\begin{pmatrix} 0\\ 0 \\ 1 \\ 0 \end{pmatrix}, \qquad
\left|\downarrow\right>
=\begin{pmatrix} 0\\ 0 \\ 0 \\ 1 \end{pmatrix}.\tag{1}$$
as 4 basis vectors in the super vector space $\mathbb{C}^{2|2}$.
In other words, the Fock space is isomorphic to $\mathbb{C}^{2|2}$.
The fermionic operators are represented by $(2+2)\times (2+2)$ supermatrices in ${\rm End}(\mathbb{C}^{2|2})=L(\mathbb{C}^{2|2},\mathbb{C}^{2|2})$. They will have four $2\times 2$ blocks. For instance:
$$ \hat{c}_{\uparrow}
~=~\begin{pmatrix} 0&0&1&0\\ 0&0&0&0 \\ 0&0&0&0 \\ 0&1&0&0 \end{pmatrix}, \qquad
\hat{c}^{\dagger}_{\uparrow}
~=~\begin{pmatrix} 0&0&0&0\\ 0&0&0&1 \\ 1&0&0&0 \\ 0&0&0&0 \end{pmatrix}. \tag{2}
$$
Notice that both the above supermatrices are Grassmann-odd despite the fact that all non-zero matrix elements are Grassmann-even. This is because the non-zero matrix elements sit in the off-diagonal $2\times 2$ Bose-Fermi blocks.
Moreover, one may check that the anticommutator of the two above supermatrices (2) is the $4\times 4$ identity matrix, as it should be to mimic the CAR algebra.
The operators $\hat{c}_{\downarrow}$ and $\hat{c}^{\dagger}_{\downarrow}$ have similar representations in terms of supermatrices. We leave it as an exercise to the reader to work them out.
Be aware that the equal symbols '$=$' in eqs. (1) & (2) mean are represented by rather than are equal to. In particular, be aware that Grassmann numbers still commute/anticommute with the operators/states based on their Grassmann-parity.
II) If there are no Grassmann variables but only fermionic operators and states, then we can represent the fermionic Fock space as an exterior algebra $\bigwedge\!{}^{\bullet}V$ generated by the space of 1-particle states
$$V~:=~{\rm span}_{\mathbb{C}}\{\left|\uparrow\right>,\left|\downarrow\right>\}~\cong~\mathbb{C}^2.\tag{3} $$
The vacuum state $\left|0\right>$ is implemented as
$$ \bigwedge\!{}^{0}V~\cong~\mathbb{C}~\cong~\mathbb{C}\left|0\right>.\tag{4}$$
The 2 creation & 2 annihilation operators generate a Clifford algebra $Cl(W)\cong\mathbb{C}^{16}$, where
$$ W~:=~{\rm span}_{\mathbb{C}}\{\hat{c}_{\uparrow},\hat{c}^{\dagger}_{\uparrow},\hat{c}_{\downarrow},\hat{c}^{\dagger}_{\downarrow}\}
~=~~{\rm span}_{\mathbb{C}}\{\hat{\gamma}_{\mu}| \mu=1,2,3,4\}
~\cong~\mathbb{C}^4,\tag{5} $$
where
$$ \begin{align}
\hat{\gamma}_{1}~=~& \hat{c}_{\uparrow}+\hat{c}^{\dagger}_{\uparrow},\qquad
\hat{\gamma}_{2}~=~ \hat{c}_{\downarrow}+\hat{c}^{\dagger}_{\downarrow} \cr
\hat{\gamma}_{3}~=~& \frac{\hat{c}_{\uparrow}-\hat{c}^{\dagger}_{\uparrow}}{i},\qquad
\hat{\gamma}_{4}~=~ \frac{\hat{c}_{\downarrow}-\hat{c}^{\dagger}_{\downarrow}}{i},
\end{align}\tag{6} $$
so that
$$\{\hat{\gamma}_{\mu},\hat{\gamma}_{\nu}\}_{+}
~=~2\delta_{\mu\nu}\hat{\bf 1},\tag{7} $$
cf. e.g. Ref. 1. It is well-known that the Clifford algebra $Cl(W)$ can be represented by $4 \times 4$ Dirac matrices.
References:
- M.B. Green, J.H. Schwarz & E. Witten, Superstring theory, Vol. 1, 1986; Appendix 5.A.
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$^1$ NB: In this answer we do not try to also make matrix representations of the Grassmann numbers.
