4 Linear algebra 4.8 Bases 4.10 Basis and dimension examples

4.9 Dimension

4.9.1 Basis size

We are going to define the dimension of a finite-dimensional vector space $V$ as the size of a basis of $V$ . But as we’ve seen, a vector space can have many different bases. So we have some proving to do before this definition makes sense. We need to know that any two bases have the same size.

4.9.2 Spanning sequences are at least as big as linearly independent sequences (Steinitz exchange)

Theorem 4.9.1.

Let $V$ be a vector space and suppose $\mathbf{s}_{1},\ldots,\mathbf{s}_{m}$ spans $V$ and $\mathbf{l}_{1},\ldots,\mathbf{l}_{n}$ is linearly independent. Then $m\geqslant n$ .

Proof.

Assume for a contradiction that $m<n$ . Since the $\mathbf{s}_{i}$ span $V$ we can write each $\mathbf{l}_{j}$ as a linear combination of the $\mathbf{s}_{i}$ so there are scalars $a_{ij}$ such that $\mathbf{l}_{j}=\sum_{i=1}^{m}a_{ij}\mathbf{s}_{i}$ . Let $A$ be the $m\times n$ matrix $(a_{ij})$ , which has more columns that rows.

By Corollary 3.11.1, the matrix equation $A\mathbf{x}=\mathbf{0}$ has at least one nonzero solution $\mathbf{v}=\begin{pmatrix}v_{1}\\ \vdots\\ v_{n}\end{pmatrix}$ . Because $A\mathbf{v}=\mathbf{0}$ , for any $i$ we have $\sum_{j=1}^{n}a_{ij}v_{j}=0$ . Now

	$\displaystyle\sum_{j=1}^{n}v_{j}\mathbf{l}_{j}$	$\displaystyle=\sum_{j=1}^{n}v_{j}\sum_{i=1}^{m}a_{ij}\mathbf{s}_{i}$
		$\displaystyle=\sum_{i=1}^{m}\left(\sum_{j=1}^{n}a_{ij}v_{j}\right)\mathbf{s}_{j}$
		$\displaystyle=\sum_{i=1}^{m}0\mathbf{s}_{i}$
		$\displaystyle=\mathbf{0}_{V}$

and since the $v_{j}$ are not all zero, this contradicts the linear independence of $\mathbf{l}_{1},\ldots,\mathbf{l}_{n}$ . ∎

4.9.3 All bases of a finite-dimensional vector space have the same size

To make life slightly easier, we are going to work only with finite-dimensional vector spaces. A vector space is called finite-dimensional if it contains a finite spanning sequence.

Theorem 4.9.2.

Any two bases of a finite-dimensional vector space $V$ have the same size.

Proof.

$V$ has a finite spanning sequence $\mathbf{s}_{1},\ldots,\mathbf{s}_{m}$ because it is finite-dimensional. Therefore every linearly independent sequence has size at most $m$ , so is finite, so every basis is finite. (We haven’t actually shown that a basis exists, but this will follow from something we prove later).

Let $\mathbf{b}_{1},\ldots,\mathbf{b}_{k}$ and $\mathbf{c}_{1},\ldots,\mathbf{c}_{l}$ be bases of $V$ . Then $k\leqslant l$ (as the $\mathbf{b}_{i}$ s are linearly independent and the $\mathbf{c}_{i}$ s span). By the same argument with the two bases swapped, $l\leqslant k$ . Therefore $k=l$ . ∎

Now that we know any two bases have the same size, we can make our definition of dimension:

Definition 4.9.1.

The dimension of a vector space $V$ , written $\dim V$ , is the size of any basis of $V$ .

There’s a special case: the dimension of the zero vector space $\{0\}$ is defined to be 0. If you want you can talk yourself into believing that the empty set is a basis of the zero vector space, so that this is covered by the definition above, but it’s easier just to think of this as a special case.