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Are there real matrix rings which are not hypercomplex number systems? Is there a canonical form of a real matrix ring?

By a hypercomplex number system I mean a finite-dimensional, unital, associative algebra over $\mathbb{R}$ with a distinguished basis (containing $1 \in \mathbb{R}$) such that the square of each basis vector is either $-1$, $0$, or $1$. (Wikipedia drops the associativity condition.)

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    $\begingroup$ I didn't downvote, but I don't think this question can be answered because there is no precise notion of "hypercomplex numbers." Or if there is one you know of, you should put it in the question... $\endgroup$
    – Sam Hopkins
    Commented Aug 22 at 21:20
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    $\begingroup$ The definition on Wikipedia (en.wikipedia.org/wiki/Hypercomplex_number) is this: "A definition of a hypercomplex number is given by Kantor & Solodovnikov (1989) as an element of a unital, but not necessarily associative or commutative, finite-dimensional algebra over the real numbers. Elements are generated with real number coefficients $(a_0,\ldots,a_n)$ for a basis $\{1,i_1,\ldots,i_n\}$. Where possible, it is conventional to choose the basis so that $i_k^2 \in \{-1,0,1\}$. A technical approach to hypercomplex numbers directs attention first to those of dimension two." $\endgroup$
    – Sam Hopkins
    Commented Aug 22 at 21:37
  • $\begingroup$ OK so do you agree that there is a question ? $\endgroup$
    – Lefevres
    Commented Aug 22 at 21:44
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    $\begingroup$ So your question is “does any finite-dimensional (unitary, associative) $\mathbb{R}$-algebra admit a basis of elements whose squares are in $\{-1,0,1\}$?”, is it? Because if this is it, then it would be better to phrase it that way instead of using the term “hypercomplex number”. $\endgroup$
    – Gro-Tsen
    Commented Aug 22 at 21:58
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    $\begingroup$ Wikipedia does not require that last condition; it says "where possible..." $\endgroup$
    – Qiaochu Yuan
    Commented Aug 22 at 23:15

2 Answers 2

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It's not clear what you mean by "real matrix ring," which could either mean a ring of the form $M_n(\mathbb{R})$ or a real subalgebra of $M_n(\mathbb{R})$; if the latter, this is the same as saying "finite-dimensional associative real algebra" (by Cayley's theorem for rings).

If that's what you're asking about, then the answer is no, and for example $\mathbb{R}[x]/x^3$ is not hypercomplex in this sense. It has no elements that square to $-1$, the only elements that square to $1$ are $\pm 1$, and the only elements that square to $0$ are the multiples of $x^2$. So $x$ can't be written as a linear combination of such elements.

As for canonical forms, we can at least say that if $A$ is a f.d. associative real algebra then the quotient $A/J(A)$ of $A$ by its Jacobson radical is semisimple, hence by the Artin-Wedderburn theorem is a finite product of matrix algebras over $\mathbb{R}$, $\mathbb{C}$, or $\mathbb{H}$. We also have that $J(A)$ is nilpotent. So, very roughly, $A$ is an extension of a semisimple ring by a bunch of nilpotents.

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    $\begingroup$ This example is nice because it even refutes the weaker form of the question, “does any finite-dimensional (unitary, associative) $\mathbb{R}$-algebra admit a generating set (as an algebra) whose squares are in $\{-1,0,1\}$?” which might have been another possible interpretation of what OP asked. $\endgroup$
    – Gro-Tsen
    Commented Aug 22 at 23:57
  • $\begingroup$ On the other hand, if one is concerned with the whole matrix algebra $M_n(\mathbb{R}$, the answer to the question is rather trivially "yes". Concretely, one can take the basis that consists of all off-diagonal matrix units $E_{ij}$ [which all square to $0$], the identity matrix $I_n$, and the matrices $A_i=\mathrm{diag}(1,1,\ldots,-1,\ldots,-1)$, that is, $i$ copies of $1$ followed by $n-i$ copies of $-1$, for $i=1,\ldots,n-1$ [which all square to $I_n$], since these $n$ diagonal matrices clearly form a basis in the space of all diagonal matrices. $\endgroup$
    – Vladimir Dotsenko
    Commented Oct 22 at 18:18
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Any matrix ring can be considered a hypercomplex number system in your definition if you amend it to allow all nilpotent elements, such that $x^n=0$, not only those which square to zero.

For instance, the matrices

$1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$; $\epsilon=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \\ \end{array} \right)$; $\varepsilon=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)$

form a basis to generate a commutative associative unitial algebra that I would absolutely call "a hypercomplex system", but $\epsilon^2=\varepsilon$, so it does not satisfy your conditions. Yet, $\epsilon^3=0$. It is actually the same example as in the other answer, but I wanted to stress that even a system that does not satisfy your conditions, otherwise perfectly looks like a hypercomplex system (it even includes a subset isomorphic to the dual numbers).

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    $\begingroup$ I think $\epsilon$ and $\varepsilon$ look so much alike that they should not both be in the same formula. $\endgroup$
    – Gerald Edgar
    Commented Oct 22 at 14:47

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