Consider $I = \left(\begin{array}{cc} 1&0\\0&1\end{array}\right)$ and $N = \left(\begin{array}{cc} 1&1\\1&1\end{array}\right) - I = \left(\begin{array}{cc} 0&1\\1&0\end{array}\right)$. We now see that $N^2 = I$, which is in correspondence with $(-1)^2 = 1$. We also see that $aIbN = aNbI = abN$. So matrix multiplication behaves as multiplication should. However we would want to be able to perform addition ( subtraction ) too.

How to do this in a nice way?

  • $\begingroup$ What do you mean by making addition/subtraction "nice"? $\endgroup$ – Omnomnomnom Jun 9 '15 at 1:00
  • $\begingroup$ probably the question is whether in the ring of $2\times 2$ matrices over $\mathbb R$ one can identify a subring (or a sub-thing) of matrices with non-negative entries which is isomorphic to the field of real numbers. OP should certainly clarify the question. $\endgroup$ – Ittay Weiss Jun 9 '15 at 1:02

Perhaps the following answers your question, which is along what you propose, but not quite so (and I think what you have in mind can't be done). consider the set $S=S_2(\mathbb R)$ of all symmetric $2\times 2$ real valued matrices. Let $f\colon S\to \mathbb R$ be given by sending $M$ to $M_{11}-M_{12}$. It is straightforward that $f$ is a surjective ring homomorphism (with the usual operations on the domain and codomain). Consequently, $S/K\cong \mathbb R$ where $K=\ker f$. Now this 'construction' of $\mathbb R$ is not quite only from the non-negative reals, and it requires a quotient construction. However, you can restrict the whole process to the subset $T$ of $S$ consisting only of those matrices with non-negative entries. This $T$ is not a subring though. It's not even a subgroup of the additive structure. It is however a subgroup of the multiplicative structure. These are two observations you made in your question. However, the quotient construction itself does restrict nicely to this $T$ and it then yields the reals, for which you can show directly to form a ring again. When doing that, your all-ones matrix corresponds to $0$, as it should.

All of this is just camouflage for the usual process of adding inverses (to $\mathbb N$ to obtain $\mathbb Z$ or to $\mathbb R_+$ to obtain $\mathbb R$) by considering pairs $(a,b)$ and a suitable equivalence relation.


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