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Inner product.

Definition 12.1.1   Let $\mathcal{V}$ be an $\mathbb{R} -$ vector space and let $\phi$ be a mapping

\begin{displaymath}\mathcal{V} \times \mathcal{V} \longrightarrow \mathbb{R}\end{displaymath}

The mapping $\phi$ is an inner product on $\mathcal{V}$ if it verifies the following properties:
1.
Linearity in the first variable:
(a)
$\forall \overrightarrow{u} ,\overrightarrow{v} ,\overrightarrow{w}\in \mathcal{...
...rrow{u} ,\overrightarrow{w} )+ \phi ( \overrightarrow{v} ,\overrightarrow{w} ).$
(b)
$\forall \overrightarrow{u} ,\overrightarrow{v}\in \mathcal{V}, \;
\forall \alph...
... ,\overrightarrow{v} ) =
\alpha \phi ( \overrightarrow{u} ,\overrightarrow{v} )$.
2.
Linearity in the second variable:
(a)
$\forall \overrightarrow{u} ,\overrightarrow{v} ,\overrightarrow{w}\in \mathcal{...
...rrow{u} ,\overrightarrow{w} )+ \phi ( \overrightarrow{u} ,\overrightarrow{w} ).$
(b)
$\forall \overrightarrow{u} ,\overrightarrow{v}\in \mathcal{V}, \;
\forall \alph...
...a \overrightarrow{v} ) =
\alpha \phi ( \overrightarrow{u} ,\overrightarrow{v} )$.
3.
Symmetry: $\forall \overrightarrow{u} ,\overrightarrow{v}\in \mathcal{V}, \;
\phi ( \over...
...row{u} ,\overrightarrow{v} ) = \phi ( \overrightarrow{v} ,\overrightarrow{u} ).$
4.
$\phi$ is positive: $\forall \overrightarrow{u}\in \mathcal{V}, \;
\phi ( \overrightarrow{u} ,\overrightarrow{u} ) \geq 0$.
5.
$\phi$ is definite: $\forall \overrightarrow{u}\in \mathcal{V}, \;
\phi ( \overrightarrow{u} ,\overrightarrow{u} ) =0 \Longleftrightarrow \overrightarrow{u} = \overrightarrow{0} $.

Generally, a given inner product is denoted $\langle \overrightarrow{u} ,\overrightarrow{v}\rangle$ instead of $\phi
( \overrightarrow{u} ,\overrightarrow{v} )$.

A mapping which is linear in the first variable and in the second variable is called bilinear. We summarize the above definition by saying that an inner product is a bilinear symmetric definite positive from $\mathcal{V} \times
\mathcal{V} $ to $\mathbb{R} $.

Remark 12.1.2   If a mapping from $\mathcal{V} \times
\mathcal{V} $ to $\mathbb{R} $ is linear in one variable and symmetric, then it is bilinear.

Example 12.1.3   Let $\mathcal{V}=\mathbb{R} ^2$. For $\overrightarrow{x} = \begin{pmatrix}x_1 \\ x_2 \end{pmatrix}$ and $\overrightarrow{y} = \begin{pmatrix}y_1 \\ y_2 \end{pmatrix}$, define

\begin{displaymath}\langle \overrightarrow{x} , \overrightarrow{y}\rangle= x_1y_1 + x_2 y_2 .
\end{displaymath}

Then this mapping is an inner product on $\mathbb{R} ^2 $.

Example 12.1.4       

   
$\langle \quad \cdot \quad , \quad \cdot \quad \rangle$ IS an inner product $\langle \quad \cdot \quad , \quad \cdot \quad \rangle$ IS NOT an inner product
   
   
$\langle \overrightarrow{x} ,\overrightarrow{y}\rangle=
x_1y_1+x_2y_2 \quad , \quad \overrightarrow{x} ,\overrightarrow{y}\in \mathbb{R} ^2$ $\langle \overrightarrow{x} ,\overrightarrow{y}\rangle=
x_1^2y_1+x_2^2y_2 \quad , \quad \overrightarrow{x} ,\overrightarrow{y}\in
\mathbb{R} ^2$
   
   
$\langle \overrightarrow{x} ,\overrightarrow{y}\rangle=
x_1y_1+x_2y_2+x_3y_3 \quad , \quad \overrightarrow{x} ,\overrightarrow{y}\in
\mathbb{R} ^3$ $\langle \overrightarrow{x} ,\overrightarrow{y}\rangle=
x_1^2y_1^2+x_2^2y_2^2+x_3^2y_3^2 \quad , \quad \overrightarrow{x} ,\overrightarrow{y}\in \mathbb{R} ^3$
   
   
$\langle \overrightarrow{x} ,\overrightarrow{y}\rangle=
\overset{i=n}{\underset{...
...} x_iy_i \quad , \quad \overrightarrow{x} ,
\overrightarrow{y}\in \mathbb{R} ^n$ $\langle \overrightarrow{x} ,\overrightarrow{y}\rangle=
x_1y_1+x_2y_2 \quad , \quad \overrightarrow{x} ,\overrightarrow{y}\in \mathbb{R} ^3$
   
   
$\langle \overrightarrow{x} ,\overrightarrow{y}\rangle=
\overset{i=n}{\underset{...
...d
\overrightarrow{x} ,\overrightarrow{y}\in \mathbb{R} ^n, \quad \lambda_i > 0$ $\langle \overrightarrow{x} ,\overrightarrow{y}\rangle=
x_1^2+x_2^2-y_1^2-y_2^2 \quad , \quad \overrightarrow{x} ,\overrightarrow{y}\in \mathbb{R} ^2$
   
   
  $\langle \overrightarrow{x} ,\overrightarrow{y}\rangle=
\overset{i=n}{\underset{...
...uad
\overrightarrow{x} ,\overrightarrow{y}\in \mathbb{R} ^n, \quad \lambda_1=0$
   
   
$\langle f,g \rangle=
\int_a^b \quad f(x)g(x)dx \quad , \quad f,g \in C(a,b)$ $\langle f,g \rangle=
\int_a^b \quad \begin{vmatrix}f(x)g(x) \end{vmatrix} dx \quad , \quad f,g \in C(a,b)$
   
   
  $\langle \overrightarrow{x} ,\overrightarrow{y}\rangle=
\vert x_1\vert y_1+x_2y_2 \quad , \quad \overrightarrow{x} ,\overrightarrow{y}\in \mathbb{R} ^2$
   


next up previous contents
Next: Verification of the axioms Up: Euclidean spaces. Previous: Euclidean spaces.
Noah Dana-Picard
2001-02-26