Linearly Independent Sets and Bases

A spanning set describes a subspace but may carry redundant vectors. A linearly independent set carries no redundancy but may fail to reach every vector. A basis is both at once: independent and spanning. It is the most efficient description of a subspace, and every vector in the space is a unique combination of it. The definitions of independence and spanning transfer from $R^{n}$ to any vector space without change.¹

Linear independence in a vector space

The small cases match $R^{n}$ : a single vector ${v}$ is independent iff $v \neq = 0$ ; a pair is dependent iff one vector is a multiple of the other; and any set containing the zero vector is dependent. The general test is a chain condition.

When the vectors are not columns of numbers, the dependence equation usually cannot be turned into a matrix equation $A x = 0$ , so the definition and the dependence-and-predecessors theorem are the working tools.

Bases

Taking $H = V$ recovers the notion of a basis for the whole space. The two most common bases are the standard ones:

The standard basis for $R^{n}$ is ${e_{1}, \dots, e_{n}}$ , the columns of the $n \times n$ identity matrix. Any invertible $n \times n$ matrix's columns also form a basis for $R^{n}$ , since they are independent and span by the Invertible Matrix Theorem.
The standard basis for $P_{n}$ is ${1, t, t^{2}, \dots, t^{n}}$ . It spans by definition, and it is independent because a polynomial of degree at most $n$ with more than $n$ roots must be the zero polynomial, forcing every coefficient to zero.

The standard basis of

R^{3}

: three mutually independent axis vectors, each a unit step along one coordinate direction.

The standard basis of

P_{2}

drawn as functions: the constant

1

, the line

t

, and the parabola

t^{2}

. They are independent because a nonzero polynomial of degree at most

2

has at most two roots.

The spanning-set theorem

A basis can be built from any spanning set by throwing away vectors that contribute nothing new. Discarding a vector that is a linear combination of the others leaves the span unchanged.

Part (a) is the trimming step; part (b) is its repeated application. As long as the spanning set is dependent, some vector depends on the others and can be deleted without shrinking the span. The process stops precisely when the set becomes independent, which is a basis.

A three-vector spanning set of a plane with

v_{3} = 5 v_{1} + 3 v_{2}

; the redundant

v_{3}

is discarded, and the independent pair

{v_{1}, v_{2}}

remains a basis for the same plane.

Bases for the null space and column space

The two subspaces of a matrix each come with a standard basis-finding procedure.

Null space. The spanning set produced by solving $A x = 0$ — one vector per free variable — is automatically linearly independent, as shown when the null space was described explicitly. So that spanning set is already a basis for $Nul A$ .

Column space. The pivot columns of $A$ form a basis for $Col A$ . Each nonpivot column is a linear combination of pivot columns to its left, so the spanning-set theorem discards it; the pivot columns that remain are independent.

The proof uses that row reduction preserves the linear dependence relations among columns: $A x = 0$ and $B x = 0$ have the same solutions when $B$ is row equivalent to $A$ . So a dependence among the columns of $B$ is a dependence among the columns of $A$ in the same positions, and conversely.

Algorithm:

\textsc{ColumnBasis}(A)

— a basis for

\operatorname{Col} A

1
row-reduce $A$ to an echelon form $B$
2
identify the pivot positions of $B$
3
for each pivot column index $j$ do
4
select column $a_j$ of the original matrix $A$
5
return the selected columns of $A$

For example, if the pivots of an echelon form of $A$ sit in columns $1, 2, 4$ , then columns $1, 2, 4$ of $A$ are a basis for $Col A$ , no matter how different the echelon form looks.

An echelon form marks pivots in columns 1, 2, 4; the basis for

Col A

is columns 1, 2, 4 of the original

A

, not of the echelon form.

Two views of a basis

A basis is optimal from opposite directions.

A spanning set as small as possible. When the spanning-set theorem trims a set down, deletion must stop once the set is independent. Removing one more vector loses the span, because that vector was not a combination of the rest.
An independent set as large as possible. If $S$ is a basis for $V$ and any vector $w$ from $V$ is added, the enlarged set is dependent: $w$ is already a combination of $S$ , giving a nontrivial dependence relation.

These two views bracket the size of every basis: an independent set can be grown to a basis and a spanning set shrunk to one, and both processes terminate at the same count. That common count is the dimension of the space, and the unique coefficients a basis assigns to each vector are its coordinates.

Lay, Linear Algebra and Its Applications, §4.3 — Linearly Independent Sets; Bases: Theorem 4 (dependence via predecessors), the basis definition, Theorem 5 (spanning-set theorem), and Theorem 6 (pivot columns as a basis for $Col A$ ). ↩