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Let $A$ be a finite-dimensional Jordan algebra over $\mathbb{R}$, i.e. a finite-dimensional real vector space with a commutative bilinear product $\circ: A \times A \rightarrow A$ satisfying $(a^2 \circ b) \circ a = a^2 \circ (b \circ a)$ for all $a, b \in A$. Let $Q = \{ a^2 \mid a \in A\}$ be the cone of squares.

I want to show: $Q$ is cone (i.e. closed under addition and multiplication with non-negative reals and $Q \cap -Q = \{0\}$).

The only thing I don't know how to show: If $a, b \in Q$, then $a + b = c^2$ for some $c \in Q$. How can I show this?

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    I'm not seeing anything about this in McCrimmon's book. Where did you get this question?2012-12-06
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    I've seen this claim several times, e.g. in chapter III.2 in "Analysis on Symmetric Cones" by Faraut & Koranyi, or in "Perspectives on the Formalism of Quantum Theory" (http://uwspace.uwaterloo.ca/handle/10012/7017), the latter being my primary source of interest.2012-12-06

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