\big]
_{1\leq i,j\leq n}\in A$, then $$\operatorname{Trd}(\sigma(a)a)=\sum_{i,j=1}^n a_{ij}^2$$ is obviously a sum of (hermitian) squares in $F$.

The reader will have no problems extending this example to the case $K=F(\sqrt{-1})$ and $A=M_n(K)$ endowed with the conjugate transpose involution ${{\overline}{\phantom{x}}}t$.

The Counterexamples {#sec:counter}
===================

When the transpose involution in the previous example is replaced by an arbitrary orthogonal involution ${\sigma}$ on $M_n(F)$ (i.e., an involution which is adjoint to a quadratic form over $F$), the equivalence between totally ${\sigma}$-positive elements and sums of hermitian squares is in general no longer true, as we proceed to show in this section. We assume throughout that $F_0$ is a formally real field.

\[lem3.1\] Let $F=F_0(\!(X)\!)(\!(Y)\!)$, the iterated Laurent series field in two commuting variables $X$ and $Y$. The quadratic form $$q={\langle}X,Y,XY{\rangle}$$ does not weakly represent $1$ over $F$. In fact this is already true over the rational function field $F_0(X,Y)$.

Assume for the sake of contradiction that $m{\times}q$ represents $1$ for some positive integer $m$. Then the form $${\varphi}:={\langle}1{\rangle}\perp m{\times}{\langle}-X,-Y,-XY{\rangle}$$ is isotropic over $F$. This leads to a contradiction by repeated application of Springer’s theorem on fields which are complete with respect to a discrete valuation, cf. [@Lam Chapter VI, §1]. Since $F_0(X,Y)$ embeds 