is a positive semidefinite matrix in $A\otimes_F R 
=M_n(R)$ for any real closed field $R$ containing $F$ (equivalently: for any real closure of $F$).

Indeed, if $a$ is totally ${\sigma}$-positive, then for all $x\in A$, $\operatorname{tr}(x^{t}ax)$ is positive with respect to every ($\sigma$-)ordering of $F$, i.e., $\operatorname{tr}(x^{t}ax)\in{\ensuremath{\sum{F}^{2}}}$. A diagonalisation of the quadratic form $\operatorname{tr}(x^{t}ax)$ will contain only sums of squares in $F$ (as it would otherwise violate the total ${\sigma}$-positivity). Hence this quadratic form remains positive semidefinite under every ordered field extension of $F$.

The converse implication is also easy: if $a$ is positive semidefinite over $M_n(R)$ for every real closed field $R\supseteq F$, then the trace of $x^{t}ax$ for $x\in A$ is nonnegative under the ordering of $R$ and hence under all orderings of $F$. By definition, this means that $a$ is totally ${\sigma}$-positive.

Moreover, every totally ${\sigma}$-positive element of $(A,\sigma)$ is a sum of hermitian squares. Essentially, this goes back to Gondard and Ribenboim [@GR] and has been reproved several times [@Djo; @FRS; @HN; @KS]. It also follows easily from Theorem \[thm:NCh17\] for it suffices to show that the trace of a hermitian square is a sum of hermitian squares. But this is clear: if $a=
\big[
a_{ij}
