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MATH
[1251]

Theorems from Linear independence

All theorems from Linear independence.

Back to Linear independence

The homogeneous-system test

Let S={u1,…,un}⊆RmS=\{u_1,\ldots,u_n\}\subseteq\mathbb R^m and let

A=[ u1 ⋯ un ].A=[\,u_1\ \cdots\ u_n\,].

Then SS is linearly independent if and only if the homogeneous system

Ax=0Ax=0

has only the zero solution. Equivalently, after row reduction, every column of AA contains a pivot. If a column has no pivot, the free variable in that column produces a non-trivial dependence relation among the original columns.

If m<nm<n, there cannot be a pivot in every column, so any nn vectors in Rm\mathbb R^m with m<nm<n are dependent.

When SS is dependent, the original vectors in the pivot columns form an independent subset with the same span as SS. This works because row operations left-multiply by an invertible matrix, so they preserve every linear relation among the columns. Each non-pivot column is therefore a combination of the pivot columns, not necessarily only the earlier ones.

  1. [1]:Linear independence
  2. [2]:Linear dependence
  3. [3]:A linear map, bases and a discrete-time system
  4. [4]:Seen Questions
  5. [5]:A basis of R3\mathbb R^3
  6. [6]:A basis for a generated subspace
  7. 11 more

Independence and uniqueness

Suppose

v=λ1u1+⋯+λnun.v=\lambda_1u_1+\cdots+\lambda_nu_n.

The coefficients λ1,…,λn\lambda_1,\ldots,\lambda_n are unique if and only if {u1,…,un}\{u_1,\ldots,u_n\} is linearly independent.

The span gives existence of a representation; independence gives its uniqueness.

  1. [1]:Linear independence
  2. [2]:Span
  3. [3]:Representation in a basis
  4. [4]:Coordinate vector
  5. [5]:Proof
  6. [6]:Dependence gives non-unique coefficients

Independence and the span of the other vectors

A finite set SS is linearly independent if and only if no vector in SS belongs to the span of the other vectors.

If uju_j is a combination of the other vectors, moving it to one side gives a non-trivial relation equal to zero. Conversely, in a non-trivial relation, choose a vector with a non-zero coefficient and solve for it in terms of the others.

  1. [1]:Linear independence
  2. [2]:Span
  3. [3]:Dependence in P1P_1

Adding a vector to a span

For a finite set SS and a vector vv,

span⁡(S∪{v})=span⁡(S)\operatorname{span}(S\cup\{v\})=\operatorname{span}(S)

if and only if

v∈span⁡(S).v\in\operatorname{span}(S).

Since S⊆S∪{v}S\subseteq S\cup\{v\}, we always have span⁡(S)⊆span⁡(S∪{v})\operatorname{span}(S)\subseteq\operatorname{span}(S\cup\{v\}). If v∈span⁡(S)v\in\operatorname{span}(S), every combination using vv can be rewritten using only vectors in SS, so the reverse inclusion also holds. Conversely, if the spans are equal, then vv belongs to span⁡(S∪{v})\operatorname{span}(S\cup\{v\}) and hence to span⁡(S)\operatorname{span}(S).

  1. [1]:Span
  2. [2]:Linear independence
  3. [3]:Adjoining a vector outside the span

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The homogeneous-system testIndependence and uniquenessIndependence and the span of the other vectorsAdding a vector to a span