Dimostrazione.
Dimostriamo che $\mathbf{U}^{\perp} \subset \mathbf{u}_{1}^{\perp} \cap \ldots \cap \mathbf{u}_{t}^{\perp}$.
Se $\mathbf{v} \in \mathbf{U}^{\perp}$, per definizione $f(\mathbf{u},\mathbf{v})=0, \quad \forall \mathbf{u} \in \mathbf{U}$.
In particolare $f(\mathbf{v},\mathbf{u}_{i})=0, \quad \forall i=1,\ldots,t.\,$ Quindi $\, \mathbf{v} \in \mathbf{u}^{\perp}$.
Dimostriamo che $\mathbf{U}^{\perp} \supset \mathbf{u}_{1}^{\perp} \cap \ldots \cap \mathbf{u}_{t}^{\perp}$.
Sia $\mathbf{v} \in \mathbf{u}_{1}^{\perp} \cap \ldots \cap \mathbf{u}_{t}^{\perp}$, dovremo dimostrare che $f(\mathbf{v},\mathbf{w})=0, \quad \forall \mathbf{w} \in \mathbf{U}.$
Se $\mathbf{w} \in \mathbf{U},$ allora $\mathbf{w}=\sum_{i=1}^{t}a_{i}\mathbf{u}_{i}$, quindi
$f(\mathbf{v},\mathbf{w})= f(\mathbf{v},\sum_{i=1}^{t}a_{i}\mathbf{u}_{i})=\sum_{i=1}^{t}a_{i}\underbrace{f(\mathbf{v},\mathbf{u}_{i})}_{=0}=0$.
c.v.d.