Although we will not use the following definitions as a starting point for discussing vectors, later on there are some aspects which will need a more general point of view. For sake of completeness and correctness we therefor will already introduce here some formal definitions.
Definition 11 A commutative group \((G,+)\) is defined by
\[\begin{array}{ll} \mbox{Closure:} & \mbox{For all $a$ and $b$ in $G$, $a + b$ belongs to $G$.}\\ \mbox{Associativity:} & \mbox{For all $a$, $b$, and $c$ in $G$, $(a + b)+c = a+(b+c)$.}\\ \mbox{Neutral element:} & \mbox{There is an element $e$ in $G$ such that for all $a$ in $G$, $e + a = a + e = a$.}\\ \mbox{Inverse element:} & \mbox{For each $a$ in $G$, there is an element $b$ in $G$ such that $a + b = b + a = e$,}\\ & \mbox{where $e$ is the neutral element from the previous axiom.}\\ \mbox{Commutative}& \mbox{For all $a$ and $b$ in $G$, $a + b = b + a$}. \end{array}\] |
Definition 12 Let \((G,+)\) be a commutative group and \( a, b \in G\). Let \((K,+,*)\) be a field and \(\alpha, \beta \in K\). \(V\) is a vector space, if
\[\begin{array}{ll} \mbox{Distributive 1:} & \alpha (a + b) = \alpha a + \alpha b.\\ \mbox{Distributive 2:} & (\alpha + \beta) a = \alpha a + \beta a.\\ \mbox{Multiplications are compatible:} & (\alpha \; \beta)\; a = \alpha \;(\beta \; a).\\ \mbox{Scalar multiplication has an identity element:} & 1\;a = a. \end{array}\] |
Definition 13 Let \(V\) be a vector space. A transformation \(A: \; V\;\rightarrow\;V\) is a linear function, if for all \(f_1,\;f_2\in V\) and all \(\lambda \in \mathbb{R}\;(\mathbb{C})\)
\[\begin{array}{ll} \mbox{Linearity 1:} & A\;(f_1\; +\; f_2) = A\; f_1\; +\; A \;f_2.\\ \mbox{Linearity 2:} & A\;(\lambda \;f_1) = \lambda \;A (f_1). \end{array}\] |
Definition 14 Let \(V\) be a vector space over a field \(K\). A transformation \(\langle \; ,\; \gt: \; V \times V\;\rightarrow\;K\) is a scalar product, if
\[\begin{array}{ll} \mbox{Nearly Commutative:} & \langle x,y\rangle = \langle y,x\rangle ^*.\\ \mbox{Linearity 1:} & \langle x_1\; +\; x_2,y\rangle = \langle x_1,y\rangle \; +\; \langle x_2,y\rangle .\\ \mbox{Linearity 2:} & \langle x , \alpha \;y\rangle = \alpha \;\langle x , y\rangle .\\ \mbox{Positive:} & \langle x,x\rangle \in \mathbb{R}^+ \mbox{ for $x \neq 0$}.\\ \end{array}\] |
This last definition holds for vectors with complex components. For real numbers the \(*\) in the commutative law can be omitted.
© J. Carstensen (Math for MS)