Opposite Clifford-Algebra for a symmetric bilinearform $\beta$ on a $\mathbb{K}$-vectorspace $V$ the associated Clifford Algebra $Cl(\beta)$ is the associative algebra with unit subject to the relations $$v\cdot v=\beta(v,v)\cdot 1\qquad\forall v\in V.$$It is then often said that $Cl(-\beta)$ is isomorphic to the opposite Algebra $Cl(\beta)^\text{op}$. Why is that?
Cheers, Robert

There are different conventions for Clifford algebras, which is confusing.
If I understand your question correctly.
If the  convention for the bilinear form is
$
v . v=\beta(v,v)\cdot 1\qquad\forall v\in V
$
then the algebra is denoted as $\mathcal{Cl}_{p,q}$.
In the opposite convention
$v . v= - \beta(v,v)\cdot 1\qquad\forall v\in V
$
it is equivalent to the algebra $\mathcal{Cl}_{q,p}$.
