- Calculus, Volume 2: Multi-Variable Calculus and Linear Algebra with Applications to Differential Equations and Probability
- Tom M. Apostol
- Second Edition
- 1991
- 978-1-119-49676-2
1.8 Bases and dimension
Definition.$\quad$ A finite set $S$ of elements in a linear space $V$ is called a finite basis for $V$ if $S$ is independent and spans $V.$ The space $V$ is called finite dimensional if it has a finite basis, or if $V$ consists of $O$ alone. Otherwise, $V$ is called infinite dimensional.
$\text{Theorem 1.6.}\quad$ Let $V$ be a finite-dimensional linear space. Then every finite basis for $V$ has the same number of elements.
Proof.$\quad$ Suppose $S$ and $T$ are two finite bases for $V,$ with $k$ and $m$ elements respectively. We know from Theorem 1.5. that for an independent set $S$ of $k$ elements in $V,$ that any set of $k + 1$ elements in its linear span (that is, in $V$) is dependent. Thus, if $T$ is an independent set of elements in $V,$ then $m \leq k.$ Likewise, if $S$ is to be an independent set of elements in $V,$ then $k \leq m.$ Combining these two results, we find that $k = m.$
Definition.$\quad$ If a linear space $V$ has a basis of $n$ elements, the integer $n$ is called the dimension of $V.$ We write $n = \dim V.$ If $V = \{O\},$ we say $V$ has dimension $0.$
$\text{Theorem 1.7.}\quad$ Let $V$ be a finite-dimensional linear space with $\dim V = n.$ Then we have the following:
(a) Any set of independent elements in $V$ is a subset of some basis of $V.$
(b) Any set of $n$ independent elements is a basis for $V.$
Proof.$\quad$
(a) Let $S = \{x_1, ..., x_k\}$ be any independent set of elements in $V$ for some finite $k.$ If the span of $S$ is $V,$ then by definition, $S$ is a finite basis for $V.$ If not, we add to $S$ a new element $y,$ not in $L(S),$ yielding a new set $S' = \{x_1, ..., x_k, y\}.$ To show that $S'$ is independent, note that if we find $k + 1$ scalars $c_1, ..., c_{k+1},$ not all zero, such that
\begin{align*}
\sum_{i = 1}^{k}c_ix_i + c_{k+1}y &= O,
\end{align*}
this would imply that $y$ is in the span of $S,$ leading to a contradiction. Now, if $L(S') = V,$ then $S'$ is a finite basis for $V$ with $\dim V = k + 1.$ And with $S$ being a subset of $S',$ this would complete the proof. If not, we repeat the process with another element $z$ in $V,$ but not in the span of $S',$ giving us a new independent set of $k + 2$ elements $S'' = \{x_1, ..., x_k, y, z\}.$ If $L(S'') = V,$ then $S''$ is a finite basis for $V$ with $S$ being a subset of $S'',$ completing the proof. Otherwise, we repeat the process until the resulting independent set of elements spans $V.$
\begin{align*}
\end{align*}
Since $V$ is finite-dimensional, there must be a finite number of independent elements (namely, $n - k,$ where $k \leq n$) that can be adjoined to the set $S$ such that the resulting union spans $V,$ otherwise this would result in a contradiction. This completes the proof of (a).
(b) From the proof of part (a), we know that for any independent set of $k$ elements in $V,$ there must be a finite number of independent elements that can be adjoined to $S$ such that the resulting union spans $V.$ But from Theorem 1.6., we know that every basis for $V$ has the same number of elements. Thus, the resulting set of independent elements that spans $V$ must have exactly $n$ elements.