Math Core

A plane has many different bases, but every one of them has exactly two vectors. That number does not depend on which basis you choose, so it measures something intrinsic about the space itself: its dimension. Dimension turns vague phrases like "a line's worth" or "a plane's worth" of solutions into precise numbers.

Every basis has the same size

The whole idea rests on one fact about independence.

Too many vectors are dependent

If a vector space $V$ has a basis with $n$ vectors, then any set in $V$ containing more than $n$ vectors is linearly dependent. Consequently, every basis of $V$ has exactly $n$ vectors.

Here is why. Let $\mathcal{B} = \lbrace \mathbf{b}_1, \dots, \mathbf{b}_n \rbrace$ be a basis and take vectors $\mathbf{u}_1, \dots, \mathbf{u}_p$ in $V$ with $p > n$ . Each $\mathbf{u}_j$ is a combination of the basis vectors, so write $\mathbf{u}_j = a_{1j}\mathbf{b}_1 + \cdots + a_{nj}\mathbf{b}_n$ and collect the coefficients into an $n \times p$ matrix $M = [a_{ij}]$ . Because $M$ has more columns than rows, $M\mathbf{c} = \mathbf{0}$ has a nontrivial solution $\mathbf{c}$ . Then

c_1\mathbf{u}_1 + \cdots + c_p\mathbf{u}_p = \sum_{i=1}^{n} \Big( \sum_{j=1}^{p} a_{ij}c_j \Big)\mathbf{b}_i = \mathbf{0},

a nontrivial dependence. For the second claim, if $\mathcal{C}$ is another basis, it can't have more than $n$ vectors (it is independent), and by the same argument with the roles swapped $\mathcal{B}$ can't have more vectors than $\mathcal{C}$ . So both have $n$ .

Definition

Dimension

If $V$ is spanned by a finite set, $V$ is finite-dimensional, and its dimension, written $\dim V$ , is the number of vectors in any basis of $V$ . The dimension of the zero space $\lbrace \mathbf{0} \rbrace$ is defined to be $0$ . If $V$ is not spanned by any finite set, it is infinite-dimensional.

Counting standard bases gives the dimensions of the familiar spaces:

$\dim \mathbb{R}^n = n$ (the basis $\mathbf{e}_1, \dots, \mathbf{e}_n$ ).
$\dim \mathbb{P}_n = n + 1$ (the basis $1, t, \dots, t^n$ has $n + 1$ vectors).
$\dim M_{m \times n} = mn$ (one matrix with a single $1$ in each position).
The space of all polynomials is infinite-dimensional: any finite set of polynomials has a maximum degree, so it can't span.

Common mistake

$\dim \mathbb{P}_n$ is $n + 1$ , not $n$ . The constant term counts. For example, $\mathbb{P}_3$ has the basis $\lbrace 1, t, t^2, t^3 \rbrace$ , so $\dim \mathbb{P}_3 = 4$ .

Subspaces of R^3

Dimension sorts the subspaces of $\mathbb{R}^3$ cleanly:

0-dimensional: only $\lbrace \mathbf{0} \rbrace$ .
1-dimensional: spanned by one nonzero vector, so lines through the origin.
2-dimensional: spanned by two independent vectors, so planes through the origin.
3-dimensional: three independent vectors in $\mathbb{R}^3$ span all of $\mathbb{R}^3$ .

A subspace $H$ of a finite-dimensional space $V$ always satisfies $\dim H \le \dim V$ , and any independent set in $H$ can be extended to a basis of $H$ .

The basis theorem

In general you must check both independence and spanning to confirm a basis. Once you know the dimension, one of them is enough.

The basis theorem

Let $V$ have dimension $p \ge 1$ .

Any linearly independent set of exactly $p$ vectors in $V$ is a basis.
Any set of exactly $p$ vectors that spans $V$ is a basis.

For instance, you showed in the previous lesson that three polynomials in $\mathbb{P}_2$ were independent. Since $\dim \mathbb{P}_2 = 3$ , they automatically span $\mathbb{P}_2$ .

Dimensions of Nul A and Col A

The bases you built in the last lesson make these dimensions easy to count.

Counting with pivots

For an $m \times n$ matrix $A$ :

$\dim \operatorname{Nul} A$ is the number of free variables in $A\mathbf{x} = \mathbf{0}$ .
$\dim \operatorname{Col} A$ is the number of pivot columns of $A$ .

Since every column is either a pivot column or corresponds to a free variable, these two numbers add up to $n$ . The next lesson makes this into a theorem.

Worked example: Dimension of a solution set

Find the dimension of $H = \lbrace (x_1, x_2, x_3, x_4) : x_1 + x_2 - x_3 + 2x_4 = 0 \rbrace$ .

$H$ is $\operatorname{Nul} A$ for the $1 \times 4$ matrix $A = \begin{bmatrix} 1 & 1 & -1 & 2 \end{bmatrix}$ . There is one pivot and three free variables ( $x_2, x_3, x_4$ ), so $\dim H = 3$ . Geometrically, $H$ is a 3-dimensional "hyperplane" in $\mathbb{R}^4$ .

Worked example: Dimension of a parametrized subspace

Find the dimension of the subspace $H$ of $\mathbb{R}^4$ consisting of all vectors

\begin{bmatrix} a + 2b + 3c \\ 2a + 4b + 6c \\ b + c \\ a + b + 2c \end{bmatrix}, \qquad a, b, c \text{ real}.

Split by parameter: $H = \operatorname{Span}\lbrace \mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3 \rbrace$ with

\mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \\ 0 \\ 1 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 2 \\ 4 \\ 1 \\ 1 \end{bmatrix}, \quad \mathbf{v}_3 = \begin{bmatrix} 3 \\ 6 \\ 1 \\ 2 \end{bmatrix}.

Three parameters do not guarantee dimension $3$ . Here $\mathbf{v}_3 = \mathbf{v}_1 + \mathbf{v}_2$ , so $\mathbf{v}_3$ can be dropped. The remaining two vectors are independent (neither is a multiple of the other: every multiple of $\mathbf{v}_1$ has third entry $0$ , and the only multiple of $\mathbf{v}_2$ with third entry $0$ is the zero vector). So $\lbrace \mathbf{v}_1, \mathbf{v}_2 \rbrace$ is a basis and $\dim H = 2$ .

Tip

When a subspace is described by parameters, the dimension is at most the number of parameters. It equals that number only if the spanning vectors are independent. Row reducing the matrix $[\,\mathbf{v}_1 \;\cdots\; \mathbf{v}_p\,]$ and counting pivots settles it.