Let $\mathcal{U}$ be an inner product space over $F$ and let $\mathbf{u} \in \mathcal{U}$. Show that
$$
\langle\mathbf{u}, \mathbf{v}\rangle=0 \text { for every } \mathbf{v} \in \mathcal{U} \Longleftrightarrow \mathbf{u}=\mathbf{0}
$$
and (consequently)
$$
\left\langle\mathbf{u}_1, \mathbf{v}\right\rangle=\left\langle\mathbf{u}_2, \mathbf{v}\right\rangle \text { for every } \mathbf{v} \in \mathcal{U} \Longleftrightarrow \mathbf{u}_1=\mathbf{u}_2 \text {. }
$$
The symbol $\langle\mathbf{x}, \mathbf{y}\rangle_{s t}$, which is defined for $\mathbf{x}, \mathbf{y} \in \mathbb{F}^n$ by the formula
$$
\langle\mathbf{x}, \mathbf{y}\rangle_{s t}=\mathbf{y}^H \mathbf{x}=\sum_{i=1}^n \overline{y_i} x_i
$$
will be used on occasion to denote the standard inner product on $\mathbb{F}^n$. The conjugation in this formula can be dropped if $\mathbf{x}, \mathbf{y} \in \mathbb{R}^n$. It is important to bear in mind that there are many other inner products that can be imposed on $\mathbb{F}^n$ :