Review of General Vector Spaces
Review of General Vector Spaces
- Recall from the Vector Spaces over the Field of Real or Complex Numbers page that a Vector Space is a set $E$ equipped with operations of addition and scalar multiplication which satisfy the following axioms:
Axioms of a Vector Space | |
(1) | $x + y = y + x$ for all $x, y \in E$. |
(2) | $x + (y + z) = (x + y) + z$ for all $x, y, z \in E$. |
(3) | There exists a vector $o \in E$ such that $x + o = x$ for all $x \in E$. |
(4) | For each $x \in E$ there exists a vector $-x \in E$ with $x + (-x) = o$. |
(5) | $(\lambda \mu)x = \lambda (\mu x)$ for all $\lambda, \mu \in \mathbf{F}$ and for all $x \in E$. |
(6) | $(\lambda + \mu)x = \lambda x + \mu x$ for all $\lambda, \mu \in \mathbf{F}$ and for all $x \in E$. |
(7) | $\lambda(x + y) = \lambda x + \lambda y$ for all $\lambda \in \mathbf{F}$ and for all $x, y \in E$. |
(8) | $1x = x$ for all $x \in E$. |
- A Vector Subspace of $E$ is a subset $M \subseteq E$ such that $(x + y) \in M$ for all $x, y \in M$, and $\lambda x \in M$ for all $x \in M$ and for all $\lambda \in \mathbf{F}$. We noted that the set containing just the origin is always a vector subspace, and any intersection of subspaces is a subspace.
- On the Spanning Sets of Vectors page, if $A \subseteq E$, we defined a Linear Combination of points in $A$ to be a vector of the form:
\begin{align} \quad \lambda_1 x_1 + \lambda_2 x_2 + ... + \lambda_n x_n \end{align}
- with $x_1, x_2, ..., x_n \in A$ and $\lambda_1, \lambda_2, ..., \lambda_n \in \mathbf{F}$. The Span of $A$ denoted by $\mathrm{span}(A)$ is defined to be the set of all linear combinations of vectors in $A$.
- On the Linearly Independent Sets of Vectors page, we noted that if $A \subseteq E$ then $A$ is said to be Linearly Independent if for all $n \in \mathbb{N}$ and for all $x_1, x_2, ..., x_n \in A$ we have that the equation $\lambda_1 x_1 + \lambda_2x_2 + ... + \lambda_n x_n = o$ implies that $\lambda_1 = \lambda_2 = ... = \lambda_n = 0$.
- On the Bases for a Vector Space page we said that a Base for $E$ is a subset $A$ of $E$ that is linearly independent and spans $E$, and said that a vector space is Finite-Dimensional if it has a base that is a finite-set, and Infinite-Dimensional otherwise.
- We noted that every maximal linearly independent set is a base and every minimal spanning set is a base.
- On the Every Vector Space has a Base page we proved using the maximal axiom that every vector space has a base.