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Vector spaces and vector subspaces.

In this tutorial, $\mathbb{K} $ denotes any field (generally $\mathbb{R} $ or $\mathbb{C} $).

Definition 5.1.1   On the set V we define two operations, one named addition and denoted by +, the other one named multiplication by scalars and denoted $\cdot$ (sometimes we can omit the dot).

The triple

\begin{displaymath}( V,+,\cdot )
\end{displaymath}

is called a vector space (actually a vector space over $\mathbb{K} $) if the following properties are fulfilled:

1.
Closure:

\begin{displaymath}\forall \overrightarrow{u} ,\overrightarrow{v}\in V, \; \overrightarrow{u} + \overrightarrow{v}\in V.
\end{displaymath}

2.
Associativity:

\begin{displaymath}\forall \overrightarrow{u} ,\overrightarrow{v} ,\overrightarr...
...verrightarrow{u} +( \overrightarrow{v} )+\overrightarrow{w} ).
\end{displaymath}

3.
Neutral element: There exists an element, called the zero vector and denoted by $\overrightarrow{0} $, such that

\begin{displaymath}\forall \overrightarrow{u}\in V, \; \overrightarrow{u} + \ove...
...\overrightarrow{0} + \overrightarrow{u} = \overrightarrow{u} .
\end{displaymath}

4.
Additive inverse: for any vector $\overrightarrow{u} $, there exists a vector denoted $-\overrightarrow{u} $ such that

\begin{displaymath}\overrightarrow{u} + (-\overrightarrow{u} ) = (-\overrightarrow{u} ) + \overrightarrow{u} = \overrightarrow{0} .
\end{displaymath}

5.
Commutativity:

\begin{displaymath}\forall \overrightarrow{u} ,\overrightarrow{v}\in V, \;
\ove...
...\overrightarrow{v} = \overrightarrow{v} + \overrightarrow{u} .
\end{displaymath}

6.
Closure:

\begin{displaymath}\forall \overrightarrow{u}\in V, \forall \alpha \in \mathbb{K} , \;
\alpha \overrightarrow{u}\in V.
\end{displaymath}

7.
$\forall \overrightarrow{u}\in V, \forall \alpha , \beta \in \mathbb{K} $,

\begin{displaymath}\alpha (\beta \overrightarrow{u} ) = (\alpha \beta ) \overrightarrow{u} .
\end{displaymath}

8.
$\forall \overrightarrow{u} ,\overrightarrow{v}\in V, \forall \alpha \in
\mathbb{K} $,

\begin{displaymath}\alpha ( \overrightarrow{u} + \overrightarrow{v} ) = \alpha \overrightarrow{u} + \alpha \overrightarrow{v} .
\end{displaymath}

9.
$\forall \overrightarrow{u}\in V, \forall \alpha , \beta \in \mathbb{K} ,$

\begin{displaymath}(\alpha + \beta) \overrightarrow{u} = \alpha \overrightarrow{u} + \beta \overrightarrow{u} .
\end{displaymath}

10.
$\forall \overrightarrow{u}\in V$,

\begin{displaymath}1 \cdot \overrightarrow{u} = \overrightarrow{u} .
\end{displaymath}

Example 5.1.2  
1.
The field $\mathbb{K} $ is a vector space over $\mathbb{K} $.
2.
The space $\mathbb{K} ^n$ is a vector space over $\mathbb{K} $.
3.
The set $V=\{ (x,y) \in a \mathbb{R} ^2 \; \vert \; x^2=y^2 \}$ is not a vector space over $\mathbb{R} $, as the first property does not hold.

Take $\overrightarrow{u} =(1,-1)$ and $\overrightarrow{v} =(1,1)$ in $\mathbb{R} ^2 $, then $\overrightarrow{u} + \overrightarrow{v} = (2,0) \notin V$.

4.
The set of all continuous functions on the interval [0,1] is a vector space over $\mathbb{R} $. For a proof, see your favorite Calculus course.

Definition 5.1.3   Let W be a subset of the vector space V. This subset is called a vector subspace of V if it is a vector space, when the addition and multiplication by scalars are the restriction to W of those of V.

Proposition 5.1.4   Let V be a vector space over $\mathbb{K} $, where $\mathbb{K} $ is any field and let W be a non empty subset of V. Then W is a (vector) subspace of V if, and only if, it fulfills the following requirements:
1.
W is closed under addition.
2.
W is closed under multiplication by scalars.

The other properties are trivially fulfilled, excepted the belonging to W of the zero vector, which is a consequence of the second requirement.


  
Figure 1: A linear combination of two vectors
\begin{figure}
\mbox{\epsfig{file=lincomb.eps,height=4.5cm} }
\end{figure}

A linear combination of the vectors

\begin{displaymath}\{ \overrightarrow{u_i} , \; i\in I \}
\end{displaymath}

is a vector of the form

\begin{displaymath}\overrightarrow{v} = \underset{i \in I}{\sum}\alpha_i \overrightarrow{u_i} ,
\end{displaymath}

where only a finite number of the coefficients is not zero.

Proposition 5.1.5   The subset W of V is a (vector) subspace of V if, and only if, it is closed under linear combinations.

The ``if'' part is obtained by setting $\alpha = \beta = 1$ for the closure under addition, and $\alpha \in \mathbb{R} , \beta =0$ for the closure under multiplication by scalars. The ``only if'' part is quite trivial.

Example 5.1.6  
1.
The set $\mathcal{F}_{odd}$ of all odd functions from $\mathbb{R} $ to itself and the set $\mathcal{F}_{even}$of all even functions from $\mathbb{R} $ to itself are vector subspaces of the space $\mathcal{F}(\mathbb{R} ,\mathbb{R} )$, which elements are all the functions from $\mathbb{R} $ to itself.
2.
The set

\begin{displaymath}E= \{ (x,y) \in \mathbb{R} ^2 \; \vert \; x^2 =y \}
\end{displaymath}

is not a vector subspace of $\mathbb{R} ^2 $.

Proposition 5.1.7   Let U and V be two vector subspaces of the same vector space W. Then $U \cap V$ is a subspace of W.

Note that the corresponding result for the union of two subspaces is false: e.g. take

\begin{displaymath}U=\{ (x,0) \in \mathbb{R} ^2 \} \qquad \text{and} \qquad
V=\{ (0,y) \in \mathbb{R} ^2 \}.
\end{displaymath}

We have $(1,0) \in V$ and $(0,1) \in W$, but (1,0)+(0,1)=(1,1) and $(1,1) \notin U \cup V$.


next up previous contents
Next: Linear dependence and independence. Up: Vector spaces. Previous: Vector spaces.
Noah Dana-Picard
2001-02-26
+00|wE