Structure Constants for $ 2 $ by $ 2 $ Matrices The space of $2 \times 2$ matrices is $4$-dimensional. For a generic choice of $3$ matrices $X$, $Y$, and $Z$, we can take a basis to be $\{\mathrm{Id}_2, X, Y, Z\}$. The following fact seems, experimentally, to be true:
Expand the three products $YX$, $ZY$, and $XZ$ in this basis:
$$ \begin {align*}
YX &= a_{yx} \mathrm{Id} + b_{yx}X + c_{yx}Y + d_{yx}Z \\[1.2ex]
ZY &= a_{zy} \mathrm{Id} + b_{zy}X + c_{zy}Y + d_{zy}Z \\[1.2ex]
XZ &= a_{xz} \mathrm{Id} + b_{xz}X + c_{xz}Y + d_{xz}Z
\end {align*}
$$
Then it seems that we always have the following relations among these structure constants:
$$ b_{yx} = d_{zy}, ~~ c_{yx} = d_{xz}, ~~ c_{zy} = b_{xz} $$
I do not see why this is true. Does anyone know a proof or explanation?
 A: Being a linear space, those matrices cannot be that generic. Indeed, it must be
$0=aId+bX+cY+dZ\Rightarrow\,a=b=c=d=0$, i.e, linearly independent. Let's then assume that's the case.
The Jacobi identity
$$[x,[y,z]]+[y,[z,x]]+[z,[x,y]]=0\\
[x,\{yz-zy\}]+[y,\{zx-xz\}]+[z,\{xy-yx\}]=0\\
[x,yz]+[y,zx]+[z,xy]\,=\,[x,zy]+[y,xz]+[z,yx]
$$
From the LHS
$$xyz-yzx+yzx-zxy+zxy-xyz=0$$ 
and from the RHS, using those three relations of yours
$$[x,zy]=c_{zy}[x,y]+d_{zy}[x,z]\\
[y,xz]=b_{xz}[y,x]+d_{xz}[y,z]\\
[z,yx]=b_{yx}[z,x]+c_{yx}[z,y]
$$
which adding together and grouping leads to
$$0=(c_{zy}-b_{xz})[x,y]+(d_{zy}-b_{yx})[x,z]+(d_{xz}-c_{yx})[y,z]$$
As those 3 commutators have to be linearly independent, it follows the identities you were looking for.
PS: One example of such three base matrices is that of the Pauli matrices. The commutator subspace $\langle[A,B]\rangle$ is clearly $3-dimensional$ (because of the commutation relations of the pauli matrices).
Hence, the same must be true when using the basis elements $x,y,z$ for which
$$[A,B]=a[x,y]+b[z,x]+c[y,z]$$
