Bases

A spanning set might contain redundant vectors; an independent set might not reach everything. A basis is a set that does both jobs with nothing wasted. It combines spanning sets for subspaces with independence.

Key idea

A basis of a subspace $H$ is a set of vectors in $H$ that is linearly independent and spans $H$. Then every vector in $H$ can be written as a combination of the basis vectors in exactly one way.

Why it works

Why exactly one way? Spanning gives at least one. Suppose $\mathbf{x}$ $= c_1\mathbf{b}_1 + \cdots + c_p\mathbf{b}_p$ $= d_1\mathbf{b}_1 + \cdots + d_p\mathbf{b}_p$. Subtract: $(c_1 - d_1)\mathbf{b}_1 + \cdots + (c_p - d_p)\mathbf{b}_p$ $= \mathbf{0}$. Independence forces every $c_i - d_i = 0$, so the two ways were the same. Those unique weights are the coordinates of $\mathbf{x}$ relative to the basis.

The standard basis of $\mathbb{R}^n$: $\mathbf{e}_1, \dots, \mathbf{e}_n$.
$\{1, t, t^{2}\}$ is a basis of $\mathbb{P}_2$: every polynomial $a + bt + ct^{2}$ is a unique combination.
Any $n$ independent vectors in $\mathbb{R}^n$ form a basis (by the Invertible Matrix Theorem, they also span).

Two independent vectors are a basis of the plane. The dots are combinations with whole-number weights; every point is reached by exactly one pair of weights.

Worked example: Not a basis

Is $\left\{\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix}\right\}$ a basis of $\mathbb{R}^3$?

Three vectors in $\mathbb{R}^3$: they form a basis exactly when the matrix with these columns has 3 pivots (equivalently, nonzero determinant).
$\det\begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 1 & 1 & 1 \end{bmatrix}$ $= 1(1 - 3) - 0 + 2(2 - 1)$ $= -2 + 2$ $= 0$.
So they are dependent: indeed $\begin{bmatrix} 2 \\ 3 \\ 1 \end{bmatrix} = 2\begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} - \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix}$.

Answer: No.

Worked example: A basis for a plane

Find a basis for the plane $x + y - z = 0$.

Solve for $x$: $x = -y + z$, with $y = s$ and $z = t$ free.
$(x, y, z)$ $= s(-1, 1, 0) + t(1, 0, 1)$.
The two vectors span the plane and are not multiples of each other, so they are a basis.

Answer: $\{(-1, 1, 0),\ (1, 0, 1)\}$

Trimming a spanning set. If $H = \operatorname{Span}\{\mathbf{v}_1, \dots, \mathbf{v}_p\}$ and some $\mathbf{v}_k$ is a combination of the others, removing it leaves the span unchanged (anything built with $\mathbf{v}_k$ can be rebuilt without it). Keep removing until the set is independent: what is left is a basis. With a matrix, this is just "keep the pivot columns".

Bases also exist in other vector spaces. In $\mathbb{P}_2$, the set $\{1, t, t^{2}\}$ works, and so does $\{1,\ 1 + t,\ (1 + t)^{2}\}$; the second is handy for describing polynomials near $t = -1$. Choosing a basis is choosing a coordinate system, which is the subject of change of basis.

A basis is a smallest spanning set and a largest independent set. Add any vector of $H$ to a basis and it becomes dependent; remove any vector and it stops spanning.

Common mistake

Calling a set a basis after checking only one condition. A spanning set can be dependent (too many vectors), and an independent set can fail to span (too few). For $n$ vectors in $\mathbb{R}^n$ one check is enough, but in general you need both.

Bases

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