Dimostrazione.
Che $\mathbf{S}^{\perp f}$ sia un sottospazio vettoriale si vede facendo la verifica standard:
$ \forall \mathbf{w},\mathbf{w}' \in \mathbf{S}^{\perp f}, \,$ e $\, k \in K$, sia $\mathbf{v} \in \mathbf{S}$, allora
$f(\mathbf{v},\mathbf{w}+\mathbf{w}')=f(\mathbf{v},\mathbf{w})+f(\mathbf{v},\mathbf{w}')=
0+0=0$
$f(\mathbf{v},k\mathbf{w})=k f(\mathbf{v},\mathbf{w})=k 0 =0 $
Dimostriamo che $\mathbf{U} \subset \mathbf{W}^{\perp}$ è equivalente a verificare $\mathbf{W} \subset \mathbf{U}^{\perp}$ (per la simmetria di $f$).
$\mathbf{U} \subset \mathbf{W}^{\perp} \,\, \Leftrightarrow \,\, \forall \mathbf{u}
\in \mathbf{U},\,\, \mathbf{u} \in \mathbf{W}^{\perp}$
equivale a:
$\forall \mathbf{u} \in \mathbf{U}, \,\, f(\mathbf{u},\mathbf{w})=0 \quad \foral...

... \in \mathbf{W} \,\, \Leftrightarrow \,\, \mathbf{W} \subset \mathbf{U}^{\perp}$.
c.v.d.