Is split-complex $j=i+2\epsilon$? In matrix representation 
imaginary unit 
$$i=\begin{pmatrix}0 & -1 \\ 1 & 0 \end{pmatrix}$$
dual numbers unit
$$\epsilon=\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$$
split-complex unit
$$j=\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}$$
Given this definition, does not it follow that
$$j=i+2\epsilon$$
and as such, one of these systems can be fully expressed through others?
 A: If, ignoring the means by which you reached this conclusion (which was well-addressed in epimorphic's answer), we supposed that
$$j=i+2\varepsilon$$
then it follows that (assuming commutativity)
$$j^2=(i+2\varepsilon)^2=i^2 + 4\varepsilon i+4\varepsilon^2$$
which, replacing each by the definition of their square:
$$1=-1 + 4\varepsilon i$$
which only works if we define $\varepsilon i = \frac{1}2$.  This is a pretty long shot from any "reasonable" definition, since our intuition about $i$ and $\varepsilon$ should certainly not lead us to this point. Moreover, making the definition $\varepsilon i = \frac{1}2$ breaks very important properties of multiplication - for instance, it makes the operation not associative since
$$(\varepsilon^2)i\neq \varepsilon(\varepsilon i)$$
$$0i \neq \varepsilon \frac{1}2$$
$$0\neq \frac{1}2\varepsilon$$
which poses a rather major difficulty for algebra. Moreover, the equation $j=i+2\varepsilon$ is not even particularly special; setting $j=i+\varepsilon$ (i.e. by making $\varepsilon$ correspond to twice the matrix representation you suggest - which is an equally valid matrix representation of the dual numbers) yields that we want $\varepsilon i = 2$ - but this doesn't solve the lack of associativity. In fact, if we want associativity, we conclude that $\varepsilon i$ must not be a linear combination of $1$, $i$, and $\varepsilon$ with real coefficients (since $\varepsilon i$ can't be invertible given that $\varepsilon^2 = 0$ and $\varepsilon i$ can't be a multiple of $\varepsilon$ as that would cause $i\cdot i \cdot \varepsilon$ to break associativity) - which implies that $(i+a\varepsilon)^2=1=j^2$ must have no solution, so we can't reasonably express $j$ in such a system.
The fundamental issue with this is that the expression $i+2\varepsilon$ doesn't even make sense without additional structure. Though we can happily add the terms together in a formal sum (i.e. where we write every number as $a+bi+c\varepsilon$ without allowing any simplification) and this sometimes yields meaningful results, this leaves us with the issue of multiplication. Eventually the term $\varepsilon i$ will come up, and, unless we wish to break important properties of multiplication, we have to consider it as an entirely new thing - and we can prove that, in any extension of $\mathbb R$ which still obeys certain algebraic properties, but contains a new element $\varepsilon$ squaring to $0$ and $i$ squaring to $-1$, there would be only two solutions to $x^2=1$, and they are $1$ and $-1$ - there is no extra root $j$, so $j$ cannot meaningfully play into that system.
A: No.
What you've done is specify injective unit-preserving homomorphisms $\phi \colon \mathbb C \hookrightarrow M_2(\mathbb R)$, $\psi \colon Cl_{0,1,0}(\mathbb R) \hookrightarrow M_2(\mathbb R)$, and $\rho \colon Cl_{1,0,0}(\mathbb R) \hookrightarrow M_2(\mathbb R)$ by
$$
\phi(i) = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix},
\quad
\psi(\epsilon) = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},
\quad
\rho(j) = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.
$$
Here $\mathbb C \cong Cl_{0,0,1}(\mathbb R)$ is the algebra of complex numbers, $Cl_{0,1,0}(\mathbb R)$ the algebra of dual numbers, $Cl_{1,0,0}(\mathbb R)$ the algebra of split-complex numbers, and $M_2(\mathbb R)$ the algebra of $2\times 2$ real matrices. $Cl_{p,q,r}(\mathbb R)$ refers to the real Clifford algebra with an orthogonal basis that consists of


*

*$1$ and

*$p$, $q$, and $r$ other elements whose squares are respectively $1$, $0$, and $-1$.


With the embeddings/homomorphisms defined this way, we do indeed have that $$\rho(j) = \phi(i) + 2 \psi(\epsilon).$$
The problem, as I mentioned in the comments to another one of your posts, is that the embeddings are not canonical: There are infinitely many ways to embed $Cl_{0,0,1}(\mathbb R)$, $Cl_{0,1,0}(\mathbb R)$, and $Cl_{1,0,0}(\mathbb R)$ into $M_2(\mathbb R)$ while preserving the multiplicative unit, with no obvious reason to prefer one over the others. Thus, we can't just claim that $\phi(i)$ is $i$ and so on. And in general, an alternate set of embeddings $\phi', \psi', \rho'$ results in $\rho'(j) \neq \phi'(i) + 2 \psi'(\epsilon)$. 
Here are some examples. For one, note that the dual numbers are "scale-invariant" in the nilpotent "unit" $\epsilon$: For any nonzero real $t$, we could just as well have called $t\epsilon$ our nilpotent "unit" instead of $\epsilon$, the reason being that $(t\epsilon)^2 = 0$. In parallel, there is a family of unit-preserving embeddings $\psi_t \colon Cl_{0,1,0}(\mathbb R) \hookrightarrow M_2(\mathbb R)$ defined by
$$
\psi_t(\epsilon) = \begin{pmatrix} 0 & t \\ 0 & 0 \end{pmatrix}.
$$
The relationship between $\phi(i)$, $\psi_t(\epsilon)$, and $\rho(j)$ is $\rho(j) = \phi(i) + \frac{2}{t}\psi_t(\epsilon)$.
For another, Wikipedia mentions a Jordan-canonical / diagonal embedding of the split-complex numbers into $M_2(\mathbb R)$, which we denote by $\rho'$, defined by
$$
\rho'(j) = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.
$$
Observe that $\phi(i)$, $\psi(\epsilon)$, and $\rho'(j)$ are linearly independent.
A: In the hindsight this question looks a bit stupid to me, but now I just want to answer it myself.
The set of the $2\times2$ matrices is isomorphic to the set of split-quaternions.
That said, in split-quaternions, $i$ is usually defined as
$$i=\begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}$$
(which has the opposite sign from the question, but the both definitions are equal up to isomorphism).
$$j=\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}$$
(as in the question).
The nilpotent element $\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix}$ indeed generates an algebra, which is isomorphic to dual numbers.
Thus, the proposed equalities are correct, but one should bear in mind that the algebra of split-quaternions is not commutative. If one would want a commutative algebra with elements of similar properties, one possibly should consider something like dual tessarines, which is $8$-dimensional.
