Linear Dependence and Spans | KEVOS® Mathematics

Overview

Linear independence is defined negatively: a set is independent when the only way to combine its vectors into the zero vector is the trivial way, with every scalar zero. That definition is precise but indirect, and it says nothing obvious about what dependence means. The result on this page closes that gap. A set is dependent exactly when at least one of its members can be written as a linear combination of the others — which is where the word "dependent" comes from.

The consequence for spanning sets is immediate and practical. If a set $S$ is dependent, some $\vec{u}_t$ in it is expressible from the rest. Anywhere $\vec{u}_t$ appears inside a linear combination it can be substituted away, so every vector reachable from $S$ is already reachable from $S$ with $\vec{u}_t$ deleted. The span is unchanged and the generating set is one element smaller.

Repeating that step is the natural route to an economical description of a subspace. Each pass removes one redundant generator; the process halts when the remaining set is independent, at which point nothing more can be removed without losing part of the span. Independent spanning sets are therefore the minimal descriptions, and a dependent one is always carrying weight it does not need.

There is a subtlety that catches practitioners out. Dependence guarantees that some vector can be removed, not that any vector can. If a particular vector receives a zero coefficient in every relation of linear dependence on the set, then no relation can be solved for it and it is essential to the span. Choosing which vector to discard is therefore a decision that must be justified by an explicit relation, not made arbitrarily.

Definition

Relation of Linear Dependence on Column Vectors

RLDCV

For a set $S = \{\vec{u}_1, \vec{u}_2, \dots, \vec{u}_n\}$ of vectors from $\mathbb{C}^m$ , a relation of linear dependence is an equation

$\alpha_1\vec{u}_1 + \alpha_2\vec{u}_2 + \cdots + \alpha_n\vec{u}_n = \vec{0}$

with complex scalars

\alpha_i

. The relation is trivial when every

\alpha_i = 0

and non-trivial when at least one

\alpha_i \ne 0

The trivial relation holds for every set whatsoever, so its existence carries no information. Only non-trivial relations distinguish one set from another.

Linearly Dependent Set of Column Vectors

LICV

A set $S$ of column vectors is linearly dependent when it admits at least one non-trivial relation of linear dependence. It is linearly independent when it does not — that is, when $\alpha_1\vec{u}_1 + \cdots + \alpha_n\vec{u}_n = \vec{0}$ forces $\alpha_1 = \cdots = \alpha_n = 0$ .

Span of a Set of Vectors

SSCV

The span of $S = \{\vec{u}_1, \dots, \vec{u}_n\}$ is the set of all linear combinations of its members:

$\operatorname{span}(S) = \{\, \alpha_1\vec{u}_1 + \cdots + \alpha_n\vec{u}_n \mid \alpha_i \in \mathbb{C} \,\}$

Two different sets may generate exactly the same span; the whole point of this page is to replace one generating set by a smaller one that does so.

By convention the span of the empty set is $\{\vec{0}\}$ .

Concepts

The equivalence, and both directions of its proof

The central theorem states that a set $S = \{\vec{u}_1, \dots, \vec{u}_n\}$ is linearly dependent if and only if there is an index $t$ with $1 \le t \le n$ such that $\vec{u}_t$ is a linear combination of the remaining vectors of $S$ . Both directions are short and both are constructive.

Forward: dependence gives a non-trivial relation, so some coefficient $\alpha_t$ is non-zero. Move every other term to the right-hand side and divide by $-\alpha_t$ , which is legitimate precisely because that coefficient is non-zero. The result expresses $\vec{u}_t$ with coefficients $-\alpha_i / \alpha_t$ . Backward: if $\vec{u}_t = \sum_{i \ne t} \beta_i \vec{u}_i$ , subtract $\vec{u}_t$ from both sides. The resulting relation has coefficient $-1$ on $\vec{u}_t$ , which is non-zero, so the relation is non-trivial and the set is dependent.

Substitution is what makes the span survive

Knowing $\vec{u}_t$ is redundant is not the same as proving the span is unchanged; that claim is an equality of two infinite sets and needs both inclusions. One direction is free: every combination of the smaller set is also a combination of the larger set, with zero on the missing vector, so the smaller span sits inside the larger.

The other direction is the substitution. Take any $\vec{v}$ in the larger span, write it as $\sum_i \alpha_i \vec{u}_i$ , and replace the single term $\alpha_t \vec{u}_t$ by $\alpha_t$ times the expression for $\vec{u}_t$ . Collecting like terms leaves a linear combination of the remaining vectors only, so $\vec{v}$ lies in the smaller span too. Both inclusions hold, so the spans are equal.

Not every vector is removable

The theorem promises an index $t$ ; it does not let you choose $t$ . If a vector $\vec{u}_k$ carries coefficient zero in every relation of linear dependence on $S$ , then no relation can ever be rearranged to isolate it, and $\vec{u}_k$ genuinely contributes a direction that the other vectors cannot reproduce.

This is detectable mechanically. Form the matrix whose columns are the vectors of $S$ and reduce it. The relations of linear dependence are exactly the null space vectors of that matrix, so $\vec{u}_k$ is essential precisely when every null space vector has a zero in position $k$ — equivalently, when column $k$ is a pivot column whose row in the reduced form has zeros across all free columns. Removing such a vector would strictly shrink the span.

Relations of linear dependence are null space vectors

Assembling the vectors of $S$ as the columns of a matrix $A$ turns the question into a homogeneous system. A vector $\vec{x}$ of scalars satisfies $A\vec{x} = \vec{0}$ if and only if $[\vec{x}]_1 \vec{u}_1 + \cdots + [\vec{x}]_n \vec{u}_n = \vec{0}$ — that is, if and only if it encodes a relation of linear dependence on $S$ .

This translation is the reason one row reduction answers everything. The null space is trivial exactly when $S$ is independent. When it is non-trivial, each of its vectors is an explicit relation ready to be solved for any vector whose coefficient is non-zero. The pivot pattern simultaneously tells you which vectors are essential and which are candidates for removal.

Why independent generating sets are the useful ones

Building a span from a dependent set is wasteful in a measurable sense: the representation of a vector in the span is not unique, because any non-trivial relation can be added to a set of coefficients without changing the vector produced. Design parameters that overlap in this way are indistinguishable from data and make downstream fitting or control problems ill-posed.

From an independent generating set the coefficients are unique, so "which combination produced this vector" has exactly one answer. The reduction process on this page is the bridge from an arbitrary generating set to an independent one, and it never changes the object being described.

Special cases worth recognising on sight

Some dependencies need no computation. Any set containing the zero vector is dependent, since $1 \cdot \vec{0}$ plus zero coefficients elsewhere is a non-trivial relation; note the zero vector is then the removable one, and removing it never changes the span. Any set containing the same vector twice, or two scalar multiples of one another, is dependent for the same kind of reason.

A counting argument covers another class: any set of more than $m$ vectors from $\mathbb{C}^m$ is dependent, because the associated homogeneous system has more unknowns than equations and therefore has a non-trivial solution. Recognising these cases avoids reducing matrices whose answer is already determined.

Procedure: shrink a spanning set without changing the span

Assemble the columnsForm the matrix

A

whose columns are the vectors of

S

, in the order you intend to keep.

Reduce and inspectRow-reduce

A

. If every column is a pivot column,

S

is independent and cannot be shrunk. Stop.

Extract one relationChoose a free column, set its variable to

1

and the other free variables to

0

, and read off a non-trivial relation of linear dependence.

Choose a vector with non-zero coefficientOnly vectors carrying a non-zero coefficient in this relation may be removed. A vector with coefficient zero here may still be removable using a different relation — or may be essential.

Solve and substituteRearrange the relation to express the chosen vector in terms of the others, and delete it from the set. The span is unchanged.

Repeat or stopTest the reduced set again. When it becomes independent, no further removal is possible without losing part of the span.

Equations

Non-trivial relation of linear dependence

EQ-LDS-01

Non-trivial relation of linear dependence

The defining feature of a dependent set. The existence of one non-zero coefficient is what makes the rearrangement below legal.

Solving a relation for one vector

EQ-LDS-02

{\vec{u}}_{t} = \frac{- α_{1}}{α_{t}} {\vec{u}}_{1} + \dots + \frac{- α_{t - 1}}{α_{t}} {\vec{u}}_{t - 1} + \frac{- α_{t + 1}}{α_{t}} {\vec{u}}_{t + 1} + \dots + \frac{- α_{n}}{α_{t}} {\vec{u}}_{n}

Division by $\alpha_t$ requires $\alpha_t \ne 0$ , which is exactly why a vector with zero coefficient cannot be isolated from this relation.

Converse: a combination becomes a relation

EQ-LDS-03

β_{1} {\vec{u}}_{1} + \dots + β_{t - 1} {\vec{u}}_{t - 1} + (- 1) {\vec{u}}_{t} + β_{t + 1} {\vec{u}}_{t + 1} + \dots + β_{n} {\vec{u}}_{n} = \vec{0}

Starting from $\vec{u}_t$ written as a combination of the others and moving it across. The coefficient $-1$ is non-zero, so the relation is non-trivial.

Span is preserved by deleting a redundant vector

EQ-LDS-04

span ({{\vec{u}}_{1}, \dots, {\vec{u}}_{n}}) = span ({{\vec{u}}_{1}, \dots, {\vec{u}}_{t - 1}, {\vec{u}}_{t + 1}, \dots, {\vec{u}}_{n}})

Valid whenever $\vec{u}_t$ is a linear combination of the remaining vectors. The set on the right has one fewer element and generates the same subspace.

The substitution step of the set-equality proof

EQ-LDS-05

\vec{v} = \sum_{i}^{=} α_{i} {\vec{u}}_{i} = \sum_{i \neq t} α_{i} {\vec{u}}_{i} + α_{t} \sum_{i \neq t} γ_{i} {\vec{u}}_{i} = \sum_{i \neq t} (α_{i} + α_{t} γ_{i}) {\vec{u}}_{i}

Here $\gamma_i$ are the coefficients expressing $\vec{u}_t$ from the others. Every vector of the larger span is exhibited as a combination of the smaller set.

Relations of linear dependence as a null space

EQ-LDS-06

A = [{\vec{u}}_{1} ∣ {\vec{u}}_{2} ∣ \dots ∣ {\vec{u}}_{n}], N (A) = {relations of linear dependence on S}

The bridge that makes the whole question computational: one row reduction of $A$ enumerates every relation available.

Variable Definitions

Symbols used on this page
Symbol	Name	Meaning	Domain / type
$S$	Generating set	The finite set of column vectors whose span is under study	finite subset of C^m
${\vec{u}}_{i}$	Member vector	The $i$ -th vector of $S$	vector in C^m
$α_{i}$	Relation coefficient	Scalar multiplying $\vec{u}_i$ in a relation of linear dependence	complex scalar
$t$	Removable index	Index of a vector carrying a non-zero coefficient, so it can be isolated and deleted	integer 1 to n
$n$	Set size	Number of vectors in the generating set	positive integer
$m$	Vector size	Number of entries in each vector, so the ambient space is $\mathbb{C}^m$	positive integer
$A$	Matrix of columns	The $m \times n$ matrix formed by using the vectors of $S$ as columns	m x n matrix over C
$span (S)$	Span	The set of all linear combinations of the vectors in $S$	subspace of C^m
$N (A)$	Null space	All coefficient vectors giving a relation of linear dependence on $S$	subspace of C^n

Worked Numerical Example

Problem statement

A vibration laboratory has four recorded excitation shapes for a four-degree-of-freedom rig. The set of achievable excitations is the span of the four shapes. Determine whether any recording is redundant, remove it while preserving the achievable set, and identify any shape that cannot be dispensed with.

List the four shapes
The recorded shapes, as column vectors in $\mathbb{C}^4$ , are the set $R = \{\vec{v}_1, \vec{v}_2, \vec{v}_3, \vec{v}_4\}$ and the achievable set is $V = \operatorname{span}(R)$ .
${\vec{v}}_{1} = [\begin{matrix} 1 \\ 2 \\ 0 \\ 1 \end{matrix}], {\vec{v}}_{2} = [\begin{matrix} 0 \\ 1 \\ 1 \\ 2 \end{matrix}], {\vec{v}}_{3} = [\begin{matrix} 2 \\ 7 \\ 3 \\ 8 \end{matrix}], {\vec{v}}_{4} = [\begin{matrix} 1 \\ 0 \\ 2 \\ 1 \end{matrix}]$
Assemble and reduce
Place the four shapes as columns of a matrix $A$ and reduce. Pivot columns are 1, 2 and 4, so column 3 is the only free column and $x_3$ is the only free variable.
$A = [\begin{matrix} 1 & 0 & 2 & 1 \\ 2 & 1 & 7 & 0 \\ 0 & 1 & 3 & 2 \\ 1 & 2 & 8 & 1 \end{matrix}] \to B = [\begin{matrix} 1 & 0 & 2 & 0 \\ 0 & 1 & 3 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{matrix}]$
Extract a relation of linear dependence
Set the free variable $x_3 = 1$ . The reduced rows then give $x_1 = -2$ , $x_2 = -3$ and $x_4 = 0$ . That coefficient vector is a non-trivial relation of linear dependence on $R$ , so the set is dependent.
$(- 2) {\vec{v}}_{1} + (- 3) {\vec{v}}_{2} + (1) {\vec{v}}_{3} + (0) {\vec{v}}_{4} = \vec{0}$
Identify what may be removed
Three coefficients are non-zero, on $\vec{v}_1$ , $\vec{v}_2$ and $\vec{v}_3$ , so this relation can be solved for any one of those three. The coefficient on $\vec{v}_4$ is zero, so this relation cannot be solved for $\vec{v}_4$ . Since $x_3$ is the only free variable, every relation on $R$ is a multiple of this one, and every relation therefore gives $\vec{v}_4$ a coefficient of zero. Shape 4 is essential.
Solve for the chosen vector
Choose to discard $\vec{v}_3$ . Rearranging the relation and checking the arithmetic entrywise: $2(1,2,0,1) + 3(0,1,1,2) = (2,4,0,2) + (0,3,3,6) = (2,7,3,8)$ , which is $\vec{v}_3$ exactly.
${\vec{v}}_{3} = 2 {\vec{v}}_{1} + 3 {\vec{v}}_{2}$
Confirm the span is unchanged
The inclusion $\operatorname{span}(R') \subseteq \operatorname{span}(R)$ is automatic because $R' \subset R$ . For the reverse, take any $\vec{v} = \alpha_1\vec{v}_1 + \alpha_2\vec{v}_2 + \alpha_3\vec{v}_3 + \alpha_4\vec{v}_4$ and substitute the expression for $\vec{v}_3$ .
$\vec{v} = (α_{1} + 2 α_{3}) {\vec{v}}_{1} + (α_{2} + 3 α_{3}) {\vec{v}}_{2} + α_{4} {\vec{v}}_{4}$
Verify the reduced set is independent
Reduce the matrix whose columns are $\vec{v}_1$ , $\vec{v}_2$ and $\vec{v}_4$ . All three columns are pivot columns, so the associated homogeneous system has only the trivial solution and $R' = \{\vec{v}_1, \vec{v}_2, \vec{v}_4\}$ is linearly independent. No further reduction is possible.
$[\begin{matrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 1 & 2 \\ 1 & 2 & 1 \end{matrix}] \to [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{matrix}]$

Result

Three of the four recordings suffice: every excitation the rig can produce from all four shapes can be produced from shapes 1, 2 and 4 alone, and shape 3 can be dropped from the test library with no loss of coverage. Shape 4 is irreplaceable — it carries a direction no combination of the others reproduces — so a proposal to discard it instead would silently reduce the achievable set.

Applications & Industry Use

Experimental mechanics

Pruning a load case library

Test programmes accumulate load cases over years, many of which are combinations of earlier ones. Assembling the cases as columns and reducing identifies which can be deleted without shrinking the set of loadings the programme covers, cutting test time while preserving coverage exactly.

Chemical engineering

Independent reactions in a mechanism

A proposed reaction mechanism lists many elementary steps, some of which are linear combinations of others in stoichiometric space. Removing dependent steps yields a minimal reaction set generating the same composition changes, which is what kinetic parameter estimation actually requires.

Statistics & machine learning

Removing collinear predictors

An exactly collinear predictor is a linear combination of the others and adds nothing to the column space of the design matrix. Detecting and dropping it restores identifiability, and the theorem here is the guarantee that the fitted subspace — and hence the predictions — are unaffected.

Signal processing

Trimming an over-complete dictionary

Dictionaries used for sparse representation are deliberately over-complete, which makes them dependent by construction. When a compact exactly-equivalent dictionary is needed, this reduction identifies atoms that can be removed without changing the set of representable signals.

Robotics & kinematics

Redundant actuator directions

The columns of an actuation matrix are the wrench directions individual actuators can apply. A dependent column means one actuator's effect is reproducible by the others; identifying which ones informs both fault-tolerance analysis and decisions about removing hardware.

Electrical power systems

Redundant measurement channels

In state estimation, measurement rows that are linear combinations of others add no new observability. Reducing the measurement matrix distinguishes genuinely redundant channels, which may be dropped, from critical channels whose loss makes the state unobservable.

Design Considerations

Justify the removal, do not assume it

The theorem provides an index, not a licence to delete anything convenient. Before removing a vector, produce the explicit relation in which its coefficient is non-zero and record it. That record is also the substitution needed if anyone later asks how the removed vector is recovered from the survivors.

Order the vectors deliberately

When a matrix of columns is reduced, the earlier columns tend to become pivot columns and the later ones free. If certain vectors are physically meaningful and should be retained, place them first. The subspace does not change, but which vectors survive the reduction does.

Removing one vector can change what is removable next

Reduction is iterative, not simultaneous. After deleting one vector, the relations available on the remaining set are those old relations that had a zero coefficient on the deleted vector — a strictly smaller collection. Recompute after each removal rather than deleting several vectors on the strength of a single reduction.

Exact dependence is rare in measured data

Real recordings are almost never exactly dependent; they are nearly dependent. A tolerance-free reduction will then report an independent set and remove nothing, even when one vector is redundant to within measurement error. Decide in advance whether the question is algebraic (exact dependence) or numerical (near dependence), and use a rank tolerance or singular values for the latter.

Recognise the free cases before computing

More than $m$ vectors in $\mathbb{C}^m$ is automatically dependent; a set containing $\vec{0}$ or a repeated vector is automatically dependent. Screening for these costs nothing and sometimes removes the need to reduce a large matrix at all.

Deleting a vector does not preserve every property

The span is preserved. Orthogonality, norm balance, physical interpretability and numerical conditioning are not. A reduced set can be badly conditioned even when the original was benign, so if the generating set will be used for projection or least squares, follow the reduction with an orthogonalisation step.

Standards & Codes

Notation, interchange and numerical standards that govern how this material is written down, stored and computed in production systems.

Applicable standards, conventions and reference implementations
Reference	Title	Relevance to this topic
`ISO 80000-2`	Quantities and units — Part 2: Mathematics	Prescribes the set notation, subset symbols and italic scalar conventions used throughout the statements and proofs on this page.
`BLAS Level 2`	Basic Linear Algebra Subprograms, matrix-vector operations	Provides `xGEMV`, the kernel used to verify a claimed relation of linear dependence by forming $A\vec{x}$ and checking it against zero.
`LAPACK / BLAS reference`	Linear Algebra PACKage reference implementation	Supplies `xGEQP3`, the column-pivoted QR that identifies redundant columns numerically and returns the permutation naming which to keep.
`ISO/IEC 40314`	Mathematical Markup Language (MathML) Version 3.0	Encodes the relations, spans and matrix displays here as semantic markup so they remain machine-readable and searchable.
`W3C WCAG 2.1 AA`	Web Content Accessibility Guidelines	Requires the vector and matrix presentations on this page to carry text alternatives rather than being delivered as images.

Material Selection

For a mathematical topic, "material" is the numeric representation: the scalar field, storage format and precision the computation is built from.

Representation and precision selection
Representation	Select when	Trade-off
Exact rational arithmetic	Symbolic reasoning where dependence is a structural claim and the answer must be certain.	Detects dependence exactly, but says nothing about near-dependence and can suffer coefficient growth on large integer inputs.
IEEE 754 binary64 with a rank tolerance	Measured vector sets such as test recordings or sensor channels.	Handles near-dependence sensibly, but the answer now depends on a threshold that must be documented alongside the result.
Column-pivoted QR factorisation	Numerically selecting which vectors to retain from a large, nearly dependent set.	Numerically stable and returns an explicit retention order, but the selected subset is not canonical and may differ from the row-reduction choice.
Singular value decomposition	Deciding how many vectors are effectively independent when the data are noisy.	Gives a defensible numerical rank from the singular value spectrum, but returns abstract directions rather than a subset of the original vectors.
Finite field arithmetic modulo a prime	Combinatorial and coding-theoretic settings, or as a fast probabilistic dependence certificate.	Cheap and exact within the field, but an unlucky prime can report dependence where none exists over the rationals.
Sparse column storage	Very large dictionaries or incidence-derived vector sets that are mostly zero.	Keeps the input compact, but elimination causes fill-in, so a sparse rank-revealing factorisation is preferable to explicit reduction.

Manufacturing Notes

Implementation notes — how the result is actually produced by hand, by algorithm and by library, including cost and numerical behaviour.

One reduction answers the whole question

Reducing the $m \times n$ matrix of columns costs $O(mn \min(m,n))$ operations and yields the dependence verdict, every available relation, and the identification of essential vectors simultaneously. Testing vectors one at a time by solving $n$ separate systems is asymptotically worse and returns strictly less information.

Hand procedure for the substitution proof

Write the relation, circle the coefficient you intend to divide by, and confirm it is non-zero before dividing — this is the single step that fails when a vector is essential. Then write the general element of the span, replace the one offending term, and collect coefficients. Keeping the surviving vectors in their original order makes the collected coefficients easy to check.

Library behaviour

SymPy's Matrix.columnspace returns the original columns indexed by pivot positions, which is exactly the surviving set from this procedure. NumPy offers no direct equivalent; the idiomatic numerical substitute is scipy.linalg.qr with pivoting=True, whose permutation vector orders columns by decreasing importance. The two select different subsets in general, and both are correct.

Verification technique

Verify a claimed removal in two steps. Recompute the removed vector from the retained ones and compare entrywise against the original, and confirm the retained set has the same rank as the original set. Equal rank plus a valid substitution is sufficient evidence that the span survived.

Guard against cascading deletions

A common implementation bug computes all relations once, then deletes every vector that appeared with a non-zero coefficient in some relation. That over-deletes and shrinks the span. Delete one vector, recompute, and loop — or use the pivot-column selection, which performs the whole cascade correctly in a single pass.

Failure Modes & Common Mistakes

Failure modes, root causes and prevention
Failure mode / mistake	Impact	Root cause	Prevention & detection
Removing a vector with a zero coefficient	high	Assuming that dependence permits any vector to be deleted, when the relation cannot be solved for that vector.	Divide only by a non-zero coefficient. If a vector has coefficient zero in every relation, it is essential and removing it shrinks the span.
Deleting several vectors from one relation	high	Treating all non-zero coefficients in a single relation as simultaneously removable.	One relation licenses one removal. Recompute the relations after each deletion, or use the pivot columns of the reduced form to make the full selection at once.
Claiming span equality without proof	medium	Asserting that the smaller set spans the same subspace without exhibiting the substitution that establishes the reverse inclusion.	Write out both inclusions once. The easy direction follows from being a subset; the other requires the explicit substitution.
Confusing dependence with all vectors being combinations	medium	Reading the theorem as saying every vector of a dependent set is a combination of the others.	The theorem asserts existence of one such index. In general only the vectors indexed by free columns are guaranteed removable.
Testing with the wrong orientation	medium	Placing the vectors as rows rather than columns, so the reduction answers a question about the row space instead of dependence among the vectors.	Relations of linear dependence among a set of vectors are null space vectors of the matrix whose columns are those vectors.
Exact zero test on floating-point coefficients	high	Declaring a coefficient non-zero because it equals $10^{-16}$ , then dividing by it and producing an enormous, meaningless expression.	Apply a scaled tolerance, and prefer a rank-revealing factorisation when the vectors come from measurement.
Overlooking the trivial dependencies	low	Reducing a large matrix to discover that the set contains the zero vector or a duplicate.	Screen for the zero vector, repeated vectors and set size exceeding $m$ before any computation.
Assuming the reduced set is unique	low	Reporting one minimal generating set as though it were the only one.	Many independent subsets generate the same span. State the selection rule used, and expect other tools to return different but equally valid subsets.

FAQs

Why is the word dependent used at all?

Because of exactly this theorem. In a dependent set at least one vector depends on the others, in the sense that it is a linear combination of them and therefore carries no direction the others cannot reach. Some authors take that property as the definition and derive the relation-of-linear-dependence formulation as a theorem instead.

If a set is dependent, can I remove any vector I like?

No. You may remove a vector only if there is a relation of linear dependence in which its coefficient is non-zero, because the rearrangement divides by that coefficient. If a vector has coefficient zero in every relation, it is essential and deleting it strictly shrinks the span.

How do I find out which vectors are essential?

Form the matrix whose columns are the vectors and reduce it. The columns that are pivot columns and whose reduced rows are zero across every free column can never appear with a non-zero coefficient in any relation. More practically, the pivot columns as a whole always form a retained independent set that spans the same subspace.

Does removing a vector change the dimension of the span?

Not when the removed vector was a combination of the others — the span is literally the same set, so its dimension is unchanged. What changes is the size of the generating set, which moves closer to the dimension. When the generating set is independent, its size equals the dimension.

What is the connection between relations of linear dependence and the null space?

They are the same objects in different clothing. Assemble the vectors as the columns of $A$ ; then a vector of scalars satisfies $A\vec{x} = \vec{0}$ if and only if those scalars form a relation of linear dependence on the set. An independent set is exactly one whose associated matrix has a trivial null space.

Does this work for near-dependence in measured data?

Not directly. The theorem is exact: a coefficient is zero or it is not. Measured vectors are typically independent in exact arithmetic even when one is nearly redundant. For that situation use the singular values to define a numerical rank, or a column-pivoted QR to rank the vectors by importance, and state the tolerance.

Is the smaller generating set unique?

No. Different relations, different column orderings and different algorithms all produce different valid subsets. What is unique, once you fix the rule of keeping the pivot columns of the reduced form, is the particular subset that rule selects — because the reduced row-echelon form itself is unique.

References

Beezer, R. A. A First Course in Linear Algebra, Version 0.70. University of Puget Sound, 2006. Section LDS, Subsection LDSS. Licensed under the GNU Free Documentation License v1.2.
ISO 80000-2:2019, Quantities and units — Part 2: Mathematics. International Organization for Standardization.
Strang, G. Introduction to Linear Algebra, 5th edition. Wellesley-Cambridge Press, 2016.
Golub, G. H. and Van Loan, C. F. Matrix Computations, 4th edition. Johns Hopkins University Press, 2013.
Anderson, E. et al. LAPACK Users' Guide, 3rd edition. Society for Industrial and Applied Mathematics, 1999.

AI Suggested Questions

Construct a set of five vectors in which exactly one vector cannot be removed without shrinking the span, and show why.
If I reorder the vectors before row-reducing, which of them survive as pivot columns, and does the span ever change?
Show the full two-inclusion proof that deleting a redundant vector leaves the span unchanged, for a specific numerical set.
Compare row reduction with column-pivoted QR for choosing which of twenty nearly-dependent measurement vectors to keep.
Why does a set of more than m vectors from an m-dimensional space have to be linearly dependent?
Give an example where removing two vectors identified from a single relation of linear dependence shrinks the span.

Related Calculators

Linear Independence Checker

Test a set of column vectors and, when dependent, return an explicit non-trivial relation of linear dependence.

Spanning Set Reducer

Reduce a generating set to a linearly independent subset with the same span, showing which vectors were removed and why.

Matrix Rank & Nullity Calculator

Compute rank, nullity and the pivot column pattern for the matrix of columns, with a selectable tolerance.