Finite-Dimensional Linear Spaces
Finite-Dimensional Linear Spaces
Recall from the Linearly Independent Sets and Spanning Sets in Linear Spaces page that if $X$ is a linear space and if $\{ x_1, x_2, ..., x_n \} \subseteq X$ then:
- The set $\{ x_1, x_2, ..., x_n \}$ is said to be linearly independent if the equation $a_1x_1 + a_2x_2 + ... + a_nx_n = 0$ implies that $a_1 = a_2 = ... = a_n = 0$, and linearly dependent otherwise.
- The span of $\{ x_1, x_2, ..., x_n \}$ is defined as:
\begin{align} \quad \mathrm{span} (x_1, x_2, ..., x_n) = \{ a_1x_1 + a_2x_2 + ... + a_nx_n : a_1, a_2, ..., a_n \in \mathbb{R} (\mathrm{or} \: \mathbb{C}) \} \end{align}
- And $\{ x_1, x_2, ..., x_n \}$ is said to span $X$ or be a spanning set of $X$ if $\mathrm{span} (x_1, x_2, ..., x_n) = X$.
We are now ready to define what it means for a linear space to be finite-dimensional.
Definition: Let $X$ be a linear space. Then $X$ is said to be Finite-Dimensional if there exists a finite subset $\{ x_1, x_2, ..., x_n \} \subseteq X$ such that $\mathrm{span} (x_1, x_2, ..., x_n) = X$. |
Definition: Let $X$ be a finite-dimensional linear space. Then $X$ is said to have Finite Dimension $n$ denoted $\mathrm{dim} (X) = n$ if there exists a finite subset of size $n$, call it $\{ x_1, x_2, ..., x_n \} \subseteq X$, such that: 1) $\mathrm{span} (x_1, x_2, ..., x_n) = X$. 2) $\{ x_1, x_2, ..., x_n \}$ is linearly independent in $X$. Such a set is called a Basis of $X$. If no such finite subset of $X$ exists, then $X$ is said to be Infinite-Dimensional. |
It is easy to prove that the dimension of a linear space is unique.
It can be shown that $\mathbb{R}^n$ is finite-dimensional for every $n \in \mathbb{N}$ and that $\mathrm{dim} (\mathbb{R}^n) = n$. The standard basis of $\mathbb{R}^n$ is $\{ e_1, e_2, ..., e_n \}$ where for each $k \in \{ 1, 2, ..., n \}$, $e_k$ is the $n$-tuple containing all $0$s except in the $k^{\mathrm{th}}$ component which is $1$.