Esercizi

Dimostrare che le seguenti forme sono bilineari:

1.: $f:\mathbf{R}^3 \times \mathbf{R}^3 \rightarrow \mathbf{R}, \qquad f(\mathbf{x},\mathbf{y})=x_{1}y_{3}-2x_{2}y_{2}+5x_{3}y_{1}$ ;
2.: $f:\mathbf{R}^3 \times \mathbf{R}^3 \rightarrow \mathbf{R}, \qquad f(\mathbf{x},\mathbf{y})=4x_{1}y_{1}+x_{2}y_{3}-3x_{1}y_{3}$ ;
3.: $f:\mathbf{R}^2 \times \mathbf{R}^2 \rightarrow \mathbf{R}, \qquad f(\mathbf{x},\mathbf{y})=2x_{1}y_{1}-2x_{2}y_{1}+x_{2}y_{1}+3x_{2}y_{2}$ .

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Punto 1. Prendiamo $\alpha, \beta \in \mathbf{R}, \,\, \mathbf{x},\mathbf{x}',\mathbf{y},\mathbf{y}' \in \mathbf{R}^3$ , allora

1.: $f(\alpha \mathbf{x}+\beta \mathbf{x}', \mathbf{y})= (\alpha x_{1}+\beta x_{1}')y_{3}-2(\alpha x_{2}+\beta x_{2}')y_{2}+5(\alpha x_{3}+\beta x_{3}')y_{1}=$
$=\alpha x_{1}y_{3} + \beta x_{1}'y_{3}-2\alpha x_{2}y_{2}-2\beta x_{2}'y_{2}+5\alpha x_{3}y_{1}+5\beta x_{3}'y_{1}=$
$=\alpha (x_{1}y_{3}-2x_{2}y_{2}+5x_{3}y_{1})+\beta (x_{1}'y_{3}-2x_{2}'y_{2}+5x_{3}'y_{1})=$
$=\alpha f(\mathbf{x},\mathbf{y})+\beta f(\mathbf{x}',\mathbf{y})$
$f(\mathbf{x},\alpha \mathbf{y}+\beta \mathbf{y}')= x_{1}(\alpha y_{3}+\beta y_{3}')-2x_{2}(\alpha y_{2}+\beta y_{2}')+5x_{3}(\alpha y_{1}+\beta y_{1}')=$
$=\alpha x_{1}y_{3} + \beta x_{1}y_{3}'-2\alpha x_{2}y_{2}-2\beta x_{2}y_{2}'+5\alpha x_{3}y_{1}+5\beta x_{3}y_{1}'=$
$=\alpha (x_{1}y_{3}-2x_{2}y_{2}+5x_{3}y_{1})+\beta (x_{1}y_{3}'-2x_{2}y_{2}'+5x_{3}y_{1}')=$
$=\alpha f(\mathbf{x},\mathbf{y})+\beta f(\mathbf{x},\mathbf{y}')$

Punto 2.

2.: $f(\alpha \mathbf{x}+\beta \mathbf{x}', \mathbf{y})= 4(\alpha x_{1}+\beta x_{1}')y_{1}+(\alpha x_{2}+\beta x_{2}')y_{3}-3(\alpha x_{1}+\beta x_{1}')y_{3}=$
$=4\alpha x_{1}y_{1}+4\beta x_{1}'y_{1}+\alpha x_{2}y_{3}+\beta x_{2}'y_{3}-3\alpha x_{1}y_{3}-3\beta x_{1}'y_{3}=$
$=\alpha (4x_{1}y_{1}+x_{2}y_{3}-3x_{1}y_{3})+\beta (4x_{1}'y_{1}+x_{2}'y_{3}-3x_{1}'y_{3})=$
$=\alpha f(\mathbf{x},\mathbf{y})+\beta f(\mathbf{x}',\mathbf{y})$
$f(\mathbf{x},\alpha \mathbf{y}+\beta \mathbf{y}')= 4x_{1}(\alpha y_{1}+\beta y_{1}')+x_{2}(\alpha y_{3}+\beta y_{3}')-3x_{1}(\alpha y_{3}+\beta y_{3}')=$
$=4\alpha x_{1}y_{1}+4\beta x_{1}y_{1}'+\alpha x_{2}y_{3}+\beta x_{2}y_{3}'-3\alpha x_{1}y_{3}-3\beta x_{1}y_{3}'=$
$=\alpha (4x_{1}y_{1}+x_{2}y_{3}-3x_{1}y_{3})+\beta (4x_{1}y_{1}'+x_{2}y_{3}'-3x_{1}y_{3}')=$
$=\alpha f(\mathbf{x},\mathbf{y})+\beta f(\mathbf{x},\mathbf{y}')$
Punto 3.

3.: $f(\alpha \mathbf{x}+\beta \mathbf{x}', \mathbf{y})= 2(\alpha x_{1}+\beta x_{1}'... ..._{2}')y_{1}+(\alpha x_{2}+\beta x_{2}')y_{1}+3(\alpha x_{2}+\beta x_{2}')y_{2}=$
$=2\alpha x_{1}y_{1}+2\beta x_{1}'y_{1}-2\alpha x_{2}y_{1}-2\beta x_{2}'y_{1}+\alpha x_{2}y_{1}+\beta x_{2}'y_{1}+3\alpha x_{2}y_{2}+\beta x_{2}'y_{2}=$
$=\alpha (2x_{1}y_{1}-2x_{2}y_{1}+x_{2}y_{1}+3x_{2}y_{2})+\beta (2x_{1}'y_{1}-2x_{2}'y_{1}+x_{2}'y_{1}+3x_{2}'y_{2})=$
$=\alpha f(\mathbf{x},\mathbf{y})+\beta f(\mathbf{x}',\mathbf{y})$
$f(\mathbf{x},\alpha \mathbf{y}+\beta \mathbf{y}')= 2x_{1}(\alpha y_{1}+\beta y_... ...eta y_{1}')+x_{2}(\alpha y_{1}+\beta y_{1}')+3x_{2}(\alpha y_{2}+\beta y_{2}')=$
$=2\alpha x_{1}y_{1}+2\beta x_{1}y_{1}'-2\alpha x_{2}y_{1}-2\beta x_{2}y_{1}'+\alpha x_{2}y_{1}+\beta x_{2}y_{1}'+3\alpha x_{2}y_{2}+\beta x_{2}y_{2}'=$
$=\alpha (2x_{1}y_{1}-2x_{2}y_{1}+x_{2}y_{1}+3x_{2}y_{2})+\beta (2x_{1}y_{1}'-2x_{2}y_{1}'+x_{2}y_{1}'+3x_{2}y_{2}')=$
$=\alpha f(\mathbf{x},\mathbf{y})+\beta f(\mathbf{x},\mathbf{y}')$