How do you find a particular representation for Grassmann numbers? This question is more general in the sense that I want to know how one finds a particular (say matrix) representation for any object. For the case of Grassmann numbers we have from Wikipedia the following representation: 

Grassmann numbers can always be represented by matrices. Consider, for example, the Grassmann algebra generated by two Grassmann numbers $\theta_1$ and $\theta_2$. These Grassmann numbers can be represented by 4×4 matrices:
  $$\theta_1 = \begin{bmatrix} 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 &
 0 & 0 & 0\\ 0 & 0 & 1 & 0\\ \end{bmatrix}\qquad \theta_2 =
 \begin{bmatrix} 0&0&0&0\\ 0&0&0&0\\ 1&0&0&0\\ 0&-1&0&0\\
 \end{bmatrix}\qquad \theta_1\theta_2 = -\theta_2\theta_1 =
 \begin{bmatrix} 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ 1&0&0&0\\ \end{bmatrix}.
 $$

How do one find these matrices? Do you guess them or is there a procedure? What about finding differenct matrix-representations for Dirac $\gamma$-matrices? How do you find them?
 A: Consider first the matrix representation of a single Grassmann number as
$\theta = \left[\begin{matrix}
  a & b \\
  c & d
 \end{matrix}\right]$
where
$\theta^2 = \left[\begin{matrix}
  a^2+bc & b(a+d) \\
  c(a+d) & d^2+bc
 \end{matrix}\right] = 0 \Leftrightarrow \theta = \left[\begin{matrix}
  i\sqrt{bc} & b \\
  c & -i\sqrt{bc}
 \end{matrix}\right]$
Notice that the easier cases ${a=b=d=0,c\neq0}$ and ${a=c=d=0,b\neq0}$ are both covered by this choice. Attempting to construct a second two dimensional Grassmann number fails since it implies that the two numbers are either proportional to each other or that one of them is zero. Now the two independent Grassmann numbers can be written on the form
$\theta_1 = \tilde{\theta}_1 \otimes M_1$ and $\theta_2 = M_2 \otimes \tilde{\theta}_2$
where the matrices $\tilde{\theta}_i$ are written on the above form (but with different values of $b_i$ and $c_i$), $\otimes$ is the outer product and $M_1,M_2$ are 2x2 matrices to be determined. The advantage of this form is that it is immediately clear that the requirement $\theta_i^2 = 0$ is fulfilled.
The anti-commutation relation $\{\theta_1,\theta_2\}=0$ amounts to
$\tilde{\theta}_1 M_2 + M_2 \tilde{\theta}_1 = 0 \;$ and/or $\; \tilde{\theta}_2 M_1 + M_1 \tilde{\theta}_2 = 0$.
Naturally this leaves quite some freedom in choosing the components but let us for simplicity focus on the choices which recover the representation you listed. Choosing $b_1=b_2=0, c_1=c_2=1$ the anti-commutator relation amounts to:
$\left[\begin{matrix}
  0 & 0 \\
  1 & 0
 \end{matrix}\right] M_{i} + M_i \left[\begin{matrix}
  0 & 0 \\
  1 & 0
 \end{matrix}\right] = \left[\begin{matrix}
  m_{12}^{(i)} & 0 \\
  m_{11}^{(i)}+m_{22}^{(i)} & m_{12}^{(i)}
 \end{matrix}\right] = 0$
So $m_{12}^{(i)} = 0$ and, in order for $\theta_{1}\theta_{2} \neq 0$, $m_{21}^{(i)}$ cannot be the only non-zero component. Hence, choosing $m_{21}^{(i)} = 0$ we are left with $m_{22}^{(i)}=-m_{11}^{(i)}$ for $i=1$ and/or $2$. Setting $m_{22}^{(2)}=-m_{11}^{(2)}=-1$ and $m_{22}^{(1)}=m_{11}^{(1)}=1$ the representations become
$\theta_1 = \tilde{\theta}_1 \otimes M_1 = \left[\begin{matrix}
  0 & 0 \\
  1 & 0
 \end{matrix}\right] \otimes \left[\begin{matrix}
  1 & 0 \\
  0 & 1
 \end{matrix}\right] = \left[\begin{matrix}
  0 & 0 & 0 & 0 \\
  1 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 1 & 0
 \end{matrix}\right] \\
\theta_2 = M_2 \otimes \tilde{\theta}_2 = \left[\begin{matrix}
  1 & 0 \\
  0 & -1
 \end{matrix}\right] \otimes \left[\begin{matrix}
  0 & 0 \\
  1 & 0
 \end{matrix}\right]  = \left[\begin{matrix}
  0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0 \\
  1 & 0 & 0 & 0 \\
  0 & -1 & 0 & 0
 \end{matrix}\right]$
A final remark would be that a similarity transformation $\theta_{i}\to S \theta_{i} S^{-1}$ leaves the algebra invariant (performed simultaneously on all the $\theta_i$).
Hope this helped! If anything was unclear then please ask.
A: Just noticed you also asked for the representation of the Dirac $\gamma$-matrices. Given a conjugate representation of the Grassmann numbers such that 
$\{\theta_i,\pi_j\} = \delta_{ij}, \quad \{\pi_{i},\pi_{j}\} = 0$
with $i=1,...,N$ then a $2N$-dimensional Clifford algebra can be built by
$\gamma_{i}=\theta_{i}+\pi_{i}\\
\gamma_{N+i}=i(\theta_{i}-\pi_{i})$
It is then 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 such that $\gamma_i\to i\gamma_i$ (and relabel a bit).
To give an explicit example, consider the representation of the two Grassmann numbers above. The conjugate representation is then
$\pi_1=\left[\begin{matrix}
  0 & 1 & 0 & 0 \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 1 \\
  0 & 0 & 0 & 0
 \end{matrix}\right]\\
\pi_2=\left[\begin{matrix}
  0 & 0 & 1 & 0 \\
  0 & 0 & 0 & -1 \\
  0 & 0 & 0 & 0 \\
  0 & 0 & 0 & 0
 \end{matrix}\right]$
which leads to the 4-dimensional representation of the Dirac algebra (rotating $i=1,2,3$ and relabeling $\gamma_4 \to \gamma_0$)
$\gamma_0=\left[\begin{matrix}
  0 & 0 & -i & 0 \\
  0 & 0 & 0 & i \\
  i & 0 & 0 & 0 \\
  0 & -i & 0 & 0
 \end{matrix}\right],\quad
\gamma_1=\left[\begin{matrix}
  0 & i & 0 & 0 \\
  i & 0 & 0 & 0 \\
  0 & 0 & 0 & i \\
  0 & 0 & i & 0
 \end{matrix}\right], \quad
\gamma_2=\left[\begin{matrix}
  0 & 0 & i & 0 \\
  0 & 0 & 0 & -i \\
  i & 0 & 0 & 0 \\
  0 & -i & 0 & 0
 \end{matrix}\right]\\
\gamma_3=\left[\begin{matrix}
  0 & 1 & 0 & 0 \\
  -1 & 0 & 0 & 0 \\
  0 & 0 & 0 & 1 \\
  0 & 0 & -1 & 0
 \end{matrix}\right], \quad \gamma_{5} =\left[\begin{matrix}
  -1 & 0 & 0 & 0 \\
  0 & 1 & 0 & 0 \\
  0 & 0 & 1 & 0 \\
  0 & 0 & 0 & -1
 \end{matrix}\right]$
Notice that a five dimensional representation is recovered if $\gamma_5 \to i\gamma_5$.
Unfortunately the explicit representation found coincides with neither the Weyl nor the Dirac basis (does not really matter though - it is still a valid representation).
A: There is a general expression for the matrix representation of $N$ Grassmann numbers (called the Clifford-Jordan-Wigner representation), and it is intimately related to the matrix representation of the Euclidean $\gamma$-matrices in $D=2N$ dimensions ($\gamma$-matrices in $D=2N+1$ and Lorentzian $\gamma$-matrices are simple to find from these). I will simply quote the results, but arriving at them follows the spirit of AltLHC's answer, namely to exploit Kronecker products to automatically retain the nilpotency of all matrices. The expression is $$\theta_m=\left(\otimes_{n=1}^{m-1}\sigma_3\right)\otimes\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}\otimes\left(\otimes_{n=m+1}^NI_2\right)$$ and for the $\gamma$-matrices we have $$\gamma_m=\left(\otimes_{n=1}^{m-1}\sigma_3\right)\otimes\sigma_1\otimes\left(\otimes_{n=m+1}^NI_2\right)$$ $$\gamma_{N+m}=\left(\otimes_{n=1}^{m-1}\sigma_3\right)\otimes\sigma_2\otimes\left(\otimes_{n=m+1}^NI_2\right)$$ $$\gamma_{2N+1}=\otimes_{i=1}^N\sigma_3$$ The $\gamma_{2N+1}$ matrix is included if $D=2N+1$, and the Lorentzian $\gamma$-matrices are obtained from the Euclidean ones by the replacing $\gamma_{2N}$ with $\gamma_0=i\gamma_{2N}$ in $D=2N$ or $\gamma_{2N+1}$ with $\gamma_0=i\gamma_{2N+1}$ in $D=2N+1$ The relationship between the representation of the Grassmann numbers and the $\gamma$-matrices is $$\theta_m=\frac{1}{2}\left(\gamma_m-i\gamma_{N+m}\right)$$ The conjugate Grassmann numbers to $\theta_m$ are $$\pi_m=\frac{1}{2}\left(\gamma_m+i\gamma_{N+m}\right)=\left(\otimes_{n=1}^{m-1}\sigma_3\right)\otimes\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\otimes\left(\otimes_{n=m+1}^NI_2\right)$$ such that $\{\theta_m,\pi_n\}=\delta_{mn}$.
A: The representation of Grassmann numbers you mentioned can be obtained as follows (for higher number of generators the construction is the same). Let $G_2$ be the Grassmann algebra in two generators $\theta_1,\theta_2$. Then $1,\theta_1,\theta_2,\theta_1\wedge\theta_2$ is a basis of $G_2$ . Let $\tilde\theta_i$, $i=1,2$, be the operator $G_2\to G_2$ of left multiplication by $\theta_i$, namely $\tilde\theta_i(x)=\theta_i\wedge x$ for any $x\in G_2$. It is clear that the operators $\tilde\theta_1,\tilde\theta_2$ satisfy the same relations as $\theta_1,\theta_2$. Then in the above basis of $G_2$ the matrices of $\tilde\theta_1,\tilde\theta_2$ are exactly the same as you wrote.
