Dimostrazione.
Ricordiamo che dalla (2) deriva che

\begin{displaymath}f(\mathbf{v},\mathbf{w})= f(\sum_{i=1}^{n} x_{i}\mathbf{v}_{i...
...n} x_{i}f(\mathbf{v}_{i},\sum_{j=1}^{n} y_{j}\mathbf{v}_{j})=
\end{displaymath}


\begin{displaymath}f(\mathbf{v},\mathbf{w})= f(\sum_{i=1}^{n} x_{i}\mathbf{v}_{i...
...n} x_{i}f(\mathbf{v}_{i},\sum_{j=1}^{n} y_{j}\mathbf{v}_{j})=
\end{displaymath}


\begin{displaymath}=\sum_{i=1}^{n} x_{i} \sum_{j=1}^{n} y_{j} f(\mathbf{v}_{i},\mathbf{v}_{j})=
\sum_{i,j=1}^{n}x_{i}y_{j}a_{ij}.
\end{displaymath}

Inoltre avremo:

\begin{displaymath}X^{t} Mat(f,\mathcal{B}) Y= (x_{1},\ldots,x_{n})
\begin{arra...
...egin{array}({c})
y_{1}\\
\vdots \\
y_{n}
\end{array}
=
\end{displaymath}


\begin{displaymath}= (x_{1},\ldots,x_{n})
\begin{array}({c})
a_{11}y_{1}+ \cdo...
...y}
= \sum_{i=1}^{n} x_{i}(a_{i1}y_{1}+ \cdots +a_{in}y_{n})=
\end{displaymath}

$= \sum_{i,j}^{n}x_{i}y_{j}a_{ij}$.
c.v.d.