Identifying the algebra In order to solve an obscure (physics) problem I have been considering whose details are not important, I am looking for elements (I am thinking in terms of matrices and their products but this may probably be relaxed to something more abstract) $A$, $B$ and $C$ which satisfy
$C^2 = 0$,
$A \cdot C + C \cdot A = I$
and
$B \cdot C + C \cdot B = -A^2$
where $I$ is the identity element (matrix) and $0$ is the zero element (matrix).
They can also be written more systematically as requirements on the anticommutators with $C$:
$A \cdot C + C \cdot A = I$,
$B \cdot C + C \cdot B = -A^2$,
$C\cdot C + C \cdot C = 0$,
such that 
$\{ A,C \} = I$, $\{ B,C\} = -A^2$ and $\{C,C\} = 0$ where $\{X,Y\} = X \cdot Y + Y \cdot X$.
I have little knowledge about abstract algebras and even less on their representation by matrices.
Do you recognise any interesting structure in this? My first thought was that it could somehow be related to Clifford algebras.
I am a physics student and apologise for any description of mathematics that is ambiguous or wrong.
 A: The free algebra $A$ generated by three letters $a$, $b$ and $c$ subject to the relations 
$$cc=0, \qquad ca=1-ac, \qquad cb=-aa-bc$$
has as a basis the set of non-commutative monomials in the three letters in which $c$ does not appear to the left of any letter. It contains the free algebra $B$ on $a$ and $b$, and in fact $A$ is a free left $B$-module with basis $\{1,c\}$. In particular, the algebra $A$ is infinite dimensional.
Suppose we have a representation of the algebra $A$ on a vector space $V$. Let $K=\ker c$. We have $aK\cap K=0$: if $v$ is in this intersection, it is equal to $aw$ for some $w\in K$ and $cv=0$, and it follows that $v=cav+acv=caaw=aacw=0$, because the relations imply that $c$ and $aa$ commute. On the other hand, if $v\in V$, we have $v=acv+cav$ and $cav\in K$ and $cv\in K$, so that $c\in aK+K$. We conclude that $V=aK\oplus K$. As $K$ intersects the kernel of $a$ trivially, the map $k\in K\mapsto ak\in aK$ is an isomorphism of vector spaces, whose inverse is $x\in aK\mapsto cx\in K$.
We have $aaK\subseteq K$. The module $V$ is in fact completely determined by the linear map $\phi:x\in aK\mapsto ax\in K$. This showws that one can classify modules over the algebra $B$ using Jordan canonical forms of size half the dimension of $V$ (for the map $aa:K\to K$). Maybe someone with more energy can see what happens if you include $b$ into this, but it does not look primising.
A: If you want an explicit representation in terms of matrices, you can just brute force the given relations as $2*2$ matrices and take into account that since you wanted $C$ to be a nilpotent matrix, there exists a basis such that $C=((0,1),(0,0))$ (sorry about the typesetting). So without loss of generality you can take this to be your $C$ and just calculating the rest gives solutions $A=((a,b),(1,-a))  B=((x,y),(-a^2-b,-x))$ if I calculated correctly. However no idea if this helps you at all. 
