Vectors, Spaces, and Basis

A vector is an arrow, but the real content of linear algebra is what you can build from a handful of them using only two operations: scaling and adding. The set of everything you can build is a span, an independent set that wastes nothing is a basis, and its size is the dimension. This part makes those four words precise.

Linear combinations and span

Given vectors $\mathbf{v}_1,\dots,\mathbf{v}_k \in \mathbb{R}^n$, a linear combination is any weighted sum

$$\mathbf{u} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_k\mathbf{v}_k, \qquad c_i \in \mathbb{R}.$$

Their span is the set of all such combinations,

$$\operatorname{span}\{\mathbf{v}_1,\dots,\mathbf{v}_k\} = \Big\{\, c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k \;:\; c_i \in \mathbb{R} \,\Big\}.$$

A span is never a stray collection of points: it always passes through the origin (take all $c_i = 0$) and is closed under addition and scaling. Any set $S \subseteq \mathbb{R}^n$ with that closure property, meaning $\mathbf{x},\mathbf{y}\in S$ and $a\in\mathbb{R}$ imply $\mathbf{x}+\mathbf{y}\in S$ and $a\mathbf{x}\in S$, is a subspace. Spans are exactly the subspaces of $\mathbb{R}^n$.¹

Independence

A set is linearly independent when the only way to write the zero vector is the trivial way:

$$c_1\mathbf{v}_1 + \cdots + c_k\mathbf{v}_k = \mathbf{0} \;\Longrightarrow\; c_1 = \cdots = c_k = 0.$$

Equivalently, no vector in the set is a linear combination of the others. If one of them were, say $\mathbf{v}_j = \sum_{i\neq j} a_i\mathbf{v}_i$, it would contribute nothing new to the span, and moving it to one side gives a nontrivial relation equal to $\mathbf{0}$. Dependence means redundancy.²

Basis and dimension

A basis of a subspace $V$ is an independent set that spans $V$. Independence and spanning pull in opposite directions: spanning wants enough vectors to reach everything, independence forbids any wasted ones, and a basis is the exact balance. Its payoff is uniqueness. If $\{\mathbf{b}_1,\dots,\mathbf{b}_m\}$ is a basis, every $\mathbf{x}\in V$ has one and only one coordinate tuple $(c_1,\dots,c_m)$ with $\mathbf{x} = \sum_i c_i\mathbf{b}_i$.

Every basis of $V$ has the same number of elements; that number is the dimension $\dim V$. So $\mathbb{R}^2$ has dimension $2$, a line through the origin has dimension $1$, and the trivial subspace $\{\mathbf{0}\}$ has dimension $0$.

The 2D test: a determinant

In $\mathbb{R}^2$ the whole story reduces to a single number. Stack $\mathbf{v}=(v_1,v_2)$ and $\mathbf{w}=(w_1,w_2)$ as columns and take the determinant,

$$A = \begin{bmatrix} v_1 & w_1 \\ v_2 & w_2 \end{bmatrix}, \qquad \det A = v_1 w_2 - v_2 w_1.$$

Then $\det A \neq 0$ if and only if $\mathbf{v},\mathbf{w}$ are independent, if and only if $\operatorname{span}\{\mathbf{v},\mathbf{w}\} = \mathbb{R}^2$, so the pair is a basis. When $\det A = 0$ the vectors are parallel: one is a multiple of the other, and the span collapses from the whole plane down to a single line through the origin (or to $\{\mathbf{0}\}$ if both are zero). Drag the two arrows and watch the shaded span switch between the plane and a line as the determinant crosses zero:

Span of two vectors in the planeinteractive

v

(1.00, 0.00)

w

(1.00, 1.00)

det [v w]

1.00

relation

independent

span

ℝ²

c₁ = 1.00c₂ = 1.00

Drag the arrowheads. The green point is c₁v + c₂w. When det ≠ 0 the combination can reach the whole plane; when det = 0 it is trapped on the line.

Worked example

Take $\mathbf{v} = (1,2)$ and $\mathbf{w} = (2,4)$. The determinant is

$$\det\begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix} = (1)(4) - (2)(2) = 0,$$

so the pair is dependent, which we can see directly: $\mathbf{w} = 2\mathbf{v}$. Their span adds nothing beyond $\mathbf{v}$ alone,

$$\operatorname{span}\{\mathbf{v},\mathbf{w}\} = \{\, t\,(1,2) : t \in \mathbb{R} \,\},$$

a one-dimensional line, not the plane. Now take $\mathbf{v} = (1,0)$ and $\mathbf{w} = (1,1)$:

$$\det\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} = (1)(1) - (0)(1) = 1 \neq 0.$$

This pair is independent, so $\operatorname{span}\{\mathbf{v},\mathbf{w}\} = \mathbb{R}^2$ and $\{\mathbf{v},\mathbf{w}\}$ is a basis. Uniqueness is concrete here: solving $c_1(1,0) + c_2(1,1) = (x,y)$ gives $(c_1 + c_2,\, c_2) = (x,y)$, hence

$$c_2 = y, \qquad c_1 = x - y,$$

the one coordinate pair that reproduces any target $(x,y)$. For instance $(3,1)$ is $2\mathbf{v} + 1\mathbf{w}$, and nothing else works.

Footnotes

1. Strang, Introduction to Linear Algebra (5th ed., 2016). ↩

2. Axler, Linear Algebra Done Right (3rd ed., 2015). ↩