$XY= \operatorname{Trd}({\sigma}(b) b)$ for $$b=\left[\begin{smallmatrix}
0&X&0\\
0&0&0\\
0&0&0
\end{smallmatrix}\right].$$

Next we show that $XY$ is not a sum of hermitian squares in $(A,{\sigma})=(M_3(F), \operatorname{ad}_q)$. We identify $XY$ with $XYI_3$ in $M_3(F)$, where $I_3$ denotes the $3{\times}3$ identity matrix. Assume for the sake of contradiction that $XYI_3$ is a sum of elements of the form ${\sigma}(a)a$ with $a=[a_{ij}]_{1\leq i,j\leq 3} \in M_3(F)$. Recall that $${\sigma}(a)a =\operatorname{ad}_q(a)a=
\left[\begin{smallmatrix}
X & & \\
& Y & \\
& & XY
\end{smallmatrix}\right] \cdot a^t\cdot 
\left[\begin{smallmatrix}
X & & \\
& Y & \\
& & XY
\end{smallmatrix}\right]^{-1} \cdot a.$$ The $(3,3)$-entry of ${\sigma}(a)a$ is equal to $$Ya_{13}^2+X a_{23}^2 + a_{33}^2.$$ By our assumption there are $s_1, s_2, s_3 \in \sum F^{{\times}2}$ such that $$XY=Y s_1+ Xs_2 + s_3,$$ which is equivalent with $$1= X^{-1}s_1 + Y^{-1} s_2 + X^{-1} Y^{-1} s_3.$$ Thus, $1$ is weakly represented by the quadratic form $${\langle}X^{-1}, Y^{-1}, X^{-1} Y^{-1}{\rangle}\simeq {\langle}X, Y, XY{\rangle}=q,$$ which is impossible by Lemma \[lem3.1\]. This finishes the proof.

The previous theorem gives us a counterexample to the PS Conjecture. It shows that the conjecture is in general not true for full matrix algebras equipped with an orthogonal involution. In contrast, when we equip a full 