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Linear independence

Linear independence is a fundamental concept in linear algebra that characterizes sets or families of vectors within a vector space. A set of vectors \{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k \} is linearly independent if the equation c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \dots + c_k \mathbf{v}_k = \mathbf{0} holds only when all scalars c_1 = c_2 = \dots = c_k = 0, meaning no vector in the set can be expressed as a nontrivial linear combination of the others.^[1] This property distinguishes linearly independent collections from linearly dependent ones, where at least one vector is redundant as a linear combination of the rest.^[2] Linear independence is essential for defining the structure of vector spaces, particularly in relation to span, basis, and dimension. A basis is a linearly independent set of vectors that spans the entire vector space, providing a minimal generating set for all elements in the space.^[3] The dimension of a vector space is the cardinality of any such basis, which remains consistent regardless of the choice of basis, and it determines the maximum size of a linearly independent subset.^[4] For instance, in \mathbb{R}^n, the standard basis vectors form a linearly independent set of size n, confirming the dimension is n.^[5] Beyond theoretical foundations, linear independence has practical implications in matrix theory and applications. The columns (or rows) of a matrix are linearly independent if and only if the associated homogeneous system has only the trivial solution, which relates directly to the matrix's rank and invertibility.^[6] This concept extends to solving systems of linear equations,^[7] optimizing in machine learning,^[8] and analyzing data structures in fields like physics and computer science, where it ensures non-redundant representations.^[7]

Definitions

Finite-dimensional vector spaces

In the context of finite-dimensional vector spaces, the concepts of vector spaces, linear combinations, and the zero vector are foundational prerequisites. A vector space V over a field F consists of vectors that can be added and scaled by elements of F, with the zero vector \mathbf{0} serving as the additive identity. Linear combinations involve sums of the form \sum a_i \mathbf{v}_i, where a_i \in F and \mathbf{v}_i \in V.^[9] A finite set of vectors \{\mathbf{v}_1, \dots, \mathbf{v}_n\} in a vector space V over a field F is linearly independent if the only solution in F to the equation a_1 \mathbf{v}_1 + \dots + a_n \mathbf{v}_n = \mathbf{0} is a_1 = \dots = a_n = 0.^[10] This condition ensures that no vector in the set can be expressed as a linear combination of the others. Formally, the set is linearly independent if

\sum_{i=1}^n a_i \mathbf{v}_i = \mathbf{0} \quad \implies \quad a_i = 0 \quad \forall i = 1, \dots, n.

^[9] The negation of this property defines linear dependence: a set is linearly dependent if there exist scalars a_1, \dots, a_n \in F, not all zero, such that \sum_{i=1}^n a_i \mathbf{v}_i = \mathbf{0}.^[10] The concept of linear independence was formalized by Giuseppe Peano in his 1888 work Calcolo geometrico secondo l'Ausdehnungslehre di H. Grassmann, where he provided the first axiomatic treatment of vector spaces over the reals, building on earlier ideas from mathematicians like Grassmann and Möbius.^[11] Although detailed discussions of span and bases appear later in the theory, linear independence is essential for identifying sets that form bases in finite-dimensional spaces.^[9]

Infinite-dimensional vector spaces

In infinite-dimensional vector spaces, the notion of linear independence extends to possibly infinite families of vectors indexed by a set I, which may be countably or uncountably infinite. A family \{v_i \mid i \in I\} is linearly independent if, for every finite subset J \subseteq I, the only solution to the equation

\sum_{j \in J} a_j v_j = 0

is a_j = 0 for all j \in J.^[12] This condition ensures that no nontrivial finite linear combination of the vectors vanishes, mirroring the finite-dimensional case but restricting nontrivial relations to finite subcollections.^[12] A key structure in this context is the Hamel basis, defined as a linearly independent set that spans the vector space algebraically, meaning every vector in the space can be expressed as a finite linear combination of basis elements.^[13] By Zorn's lemma (an equivalent of the axiom of choice), every vector space possesses a Hamel basis, though in infinite dimensions, this basis is typically uncountable and its explicit construction is impossible without additional assumptions.^[13] For instance, in the Hilbert space \ell^2 of square-summable sequences, any Hamel basis must be uncountable, as the space has cardinality $2^{\aleph_0} and cannot be spanned algebraically by a countable set using only finite combinations.^[14] The standard orthonormal basis \{e_n\}_{n=1}^\infty in \ell^2, where e_n has a 1 in the nth position and zeros elsewhere, provides an example of a countably infinite linearly independent set.^[15] However, this set does not form a Hamel basis, as its algebraic span consists only of sequences with finite support, which is a proper subspace of \ell^2. In contrast, Schauder bases in topological vector spaces like \ell^2 permit spanning via infinite convergent linear combinations, highlighting a distinction from the purely algebraic Hamel framework.^[16] A fundamental subtlety in infinite dimensions is that linear independence governs only finite combinations, so spanning sets like Hamel bases require potentially uncountably many elements to cover the space without infinite sums, unlike finite-dimensional cases where independence directly ties to dimension.^[16] This algebraic restriction often renders Hamel bases impractical for analysis in spaces equipped with topology, such as Banach or Hilbert spaces.^[14]

Equivalent characterizations

A finite set of vectors \{v_1, \dots, v_n\} in a vector space V over a field F is linearly independent if and only if the dimension of its span is n, meaning the vectors form a basis for \span\{v_1, \dots, v_n\}.^[17] This equivalence holds because linear independence ensures no redundancies, allowing the set to achieve the maximal dimension equal to its cardinality within the subspace it generates.^[18] To see this, consider a proof by induction on n. For n=1, the set \{v_1\} is linearly independent if v_1 \neq 0, in which case \dim \span\{v_1\} = 1. Assume the statement holds for sets of size k-1. For a set of size k, the subset \{v_1, \dots, v_{k-1}\} is linearly independent by the definition, so \dim \span\{v_1, \dots, v_{k-1}\} = k-1 by the induction hypothesis. The full set is linearly independent if and only if v_k \notin \span\{v_1, \dots, v_{k-1}\}, which increases the dimension to k.^[19] Equivalently, the set \{v_1, \dots, v_n\} is linearly independent if and only if every vector in \span\{v_1, \dots, v_n\} has a unique representation as a linear combination of the vectors in the set.^[19] This uniqueness follows directly from the triviality of the kernel of the coordinate map associating coefficients to linear combinations. Another characterization uses linear maps: the vectors \{v_1, \dots, v_n\} are linearly independent if and only if the linear map T: F^n \to V defined by T(e_i) = v_i, where \{e_1, \dots, e_n\} is the standard basis of F^n, is injective.^[18] Injectivity means the kernel is trivial, which corresponds precisely to the only solution of \sum a_i v_i = 0 being a_i = 0 for all i. A set S is linearly independent if and only if it can be extended to a basis of the ambient vector space V (assuming V is finite-dimensional).^[20] In finite dimensions, starting from a linearly independent S, one can iteratively add vectors from a spanning set until spanning V, preserving independence at each step; the converse follows from subsets of bases being independent. Thus, the vectors \{v_1, \dots, v_n\} form a basis for their span if and only if they are linearly independent, as they trivially span \span\{v_1, \dots, v_n\}.^[19]

Geometric and Visual Interpretations

In two-dimensional space

In two-dimensional Euclidean space \mathbb{R}^2, the concept of linear independence for vectors gains an intuitive geometric interpretation. A set consisting of two vectors is linearly independent if they are not collinear, meaning neither vector is a scalar multiple of the other. Geometrically, such vectors point in different directions and form the sides of a parallelogram with positive area, allowing them to span the entire plane. In contrast, collinear vectors lie along the same line and span only that line, forming a degenerate parallelogram with zero area.^[21] For example, the standard basis vectors \mathbf{e}_1 = (1, 0) and \mathbf{e}_2 = (0, 1) are linearly independent, as they are perpendicular and together span all of \mathbb{R}^2. On the other hand, the vectors (1, 0) and (2, 0) are linearly dependent, since (2, 0) = 2 \cdot (1, 0), and they both lie along the x-axis. Similarly, (1, 2) and (2, 4) are dependent because (2, 4) = 2 \cdot (1, 2). A single nonzero vector in \mathbb{R}^2, such as (1, 1), is linearly independent, as the equation c \cdot (1, 1) = (0, 0) implies c = 0. However, the zero vector (0, 0) by itself is linearly dependent, since $1 \cdot (0, 0) = (0, 0) with a nonzero scalar.^[22]^[21]^[23] Any set of three or more vectors in \mathbb{R}^2 is always linearly dependent, as the space has dimension 2 and cannot be spanned by more than two linearly independent vectors; at least one vector must lie in the span of the others, akin to the pigeonhole principle in geometry. Non-collinear pairs span the full plane, while collinear ones are confined to a one-dimensional line. To check dependence for two vectors \mathbf{u} = (u_1, u_2) and \mathbf{v} = (v_1, v_2), compute the determinant u_1 v_2 - u_2 v_1; the vectors are dependent if this equals zero, which measures the signed area of the parallelogram they form.^[24]^[21]

In higher-dimensional spaces

In three-dimensional space \mathbb{R}^3, three vectors are linearly independent if they span the entire space without being coplanar, forming a parallelepiped with nonzero volume that corresponds to the tetrahedron they define with the origin having positive volume.^[25] For instance, the standard basis vectors \mathbf{e}_1 = (1,0,0), \mathbf{e}_2 = (0,1,0), and \mathbf{e}_3 = (0,0,1) are linearly independent, as they align along mutually orthogonal axes and collectively span \mathbb{R}^3.^[26] In contrast, any set including the zero vector or where one vector is a scalar multiple of another fails to add a new dimension and is thus dependent.^[25] This geometric intuition generalizes to \mathbb{R}^n for n > 3, where a set of k vectors (with k \leq n) is linearly independent if their span forms a full k-dimensional subspace without dimensional collapse, meaning each successive vector extends the span by one dimension.^[25] However, in \mathbb{R}^n, any collection of n+1 vectors must be linearly dependent, as they can occupy at most an n-dimensional space and thus cannot all contribute unique directions.^[27] Visually, linear independence in higher dimensions preserves a "full rank" orientation, where the vectors maintain their maximal possible spread; dependence, conversely, causes a flattening into a lower-dimensional subspace, such as vectors collapsing onto a hyperplane.^[25] Fundamentally, a set of vectors is linearly independent if and only if they do not all lie within any proper subspace of dimension less than the size of the set.^[25]

Determination Methods

For two or three vectors

For two vectors \mathbf{u} and \mathbf{v} in a vector space over a field, the set \{\mathbf{u}, \mathbf{v}\} is linearly independent if and only if neither vector is the zero vector and \mathbf{v} is not a scalar multiple of \mathbf{u}.^[1] This condition ensures that the only solution to the equation a \mathbf{u} + b \mathbf{v} = \mathbf{0} is the trivial solution a = b = 0.^[18] To verify linear independence for two nonzero vectors, one can check whether \mathbf{v} lies in the span of \{\mathbf{u}\}, which occurs if and only if there exists a scalar c such that \mathbf{v} = c \mathbf{u}.^[18] If no such scalar exists, the vectors are linearly independent. Geometrically, in the plane, this corresponds to the vectors not being collinear.^[18] For three vectors \mathbf{u}, \mathbf{v}, and \mathbf{w} in \mathbb{R}^3, the set \{\mathbf{u}, \mathbf{v}, \mathbf{w}\} is linearly independent if the equation a \mathbf{u} + b \mathbf{v} + c \mathbf{w} = \mathbf{0} has only the trivial solution a = b = c = 0.^[28] The vectors are linearly dependent if there exists a nontrivial solution, meaning at least one coefficient is nonzero.^[28] For three vectors in \mathbb{R}^3, they can be checked by forming the matrix with them as columns and computing its determinant; independence holds if and only if \det \neq 0. Geometrically, this means the vectors are not coplanar.^[29] Consider the vectors (1,1), (1,2), and (2,3) in \mathbb{R}^2: these are linearly dependent because (2,3) = 1 \cdot (1,1) + 1 \cdot (1,2).^[30] Any set containing the zero vector is linearly dependent, as $1 \cdot \mathbf{0} + 0 \cdot \mathbf{u} + 0 \cdot \mathbf{v} = \mathbf{0} provides a nontrivial linear combination yielding zero.^[18] In two dimensions, a step-by-step check for two vectors \mathbf{u} = (u_1, u_2) and \mathbf{v} = (v_1, v_2) involves computing the 2D cross product analog, given by the determinant u_1 v_2 - u_2 v_1; the vectors are linearly independent if and only if this value is nonzero.^[29]

Matrix-based approaches

One effective way to determine the linear independence of a set of k vectors in \mathbb{R}^n is to form an n \times k matrix A whose columns are these vectors. The set is linearly independent if and only if A has full column rank, meaning \operatorname{rank}(A) = k.^[18] This condition ensures that the columns span a k-dimensional subspace without redundancy.^[18] Equivalently, the columns of A are linearly independent if and only if the homogeneous equation A \mathbf{x} = \mathbf{0} has only the trivial solution \mathbf{x} = \mathbf{0}, indicating that the kernel (null space) of A is trivial.^[6] This kernel characterization directly ties linear independence to the invertibility properties of the linear transformation represented by A.^[6] To compute the rank and verify full column rank, Gaussian elimination can be applied to row-reduce A to row echelon form. The vectors are linearly independent if the reduced form has k pivot positions (one in each column) with no zero rows appearing before the k-th pivot.^[31] This method systematically identifies dependencies by revealing the number of independent columns through the pivot count.^[31] For the special case where k = n (a square matrix), the vectors are linearly independent if and only if \det(A) \neq 0.^[32] A nonzero determinant confirms that A is invertible, implying full rank and thus independence of the columns.^[32] Consider an example in \mathbb{R}^3 with vectors \mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, \mathbf{v}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, and \mathbf{v}_3 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}. Form the matrix

A = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix}.

Row reduction yields

\begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{pmatrix},

with only two pivots, so \operatorname{rank}(A) = 2 < 3, confirming linear dependence.^[18] In contrast, replacing \mathbf{v}_3 with \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} gives full rank 3 via reduction to the identity matrix.^[18] The general algorithm involves placing the vectors as columns (or rows, equivalently, since row rank equals column rank) in a matrix, performing Gaussian elimination to echelon form, and counting the pivots: independence holds if this count equals k.^[31] This approach scales efficiently for computational verification in larger dimensions.^[31]

Dimension and spanning set relations

In a finite-dimensional vector space V over a field F with \dim V = n, the maximum size of a linearly independent set is n, and any linearly independent set of exactly n vectors forms a basis for V.^[33] This result establishes that the dimension n precisely measures the "size" of the space in terms of independent directions, ensuring all bases share the same cardinality.^[5] A linearly independent set \{v_1, \dots, v_k\} in V spans a subspace of dimension exactly k, as the vectors can be extended to a basis for that subspace while preserving independence.^[34] Conversely, if a set spans a subspace but contains redundant vectors, removing them yields a basis whose size equals the subspace dimension. This bidirectional relation underscores how linear independence determines the minimal number of vectors needed to generate a given span.^[33] The Steinitz exchange lemma provides a key mechanism for relating different bases: if \{u_1, \dots, u_m\} and \{v_1, \dots, v_n\} are bases for V, then m = n. Moreover, if \mathcal{B} is a basis and w \notin \operatorname{span} \mathcal{B}, then there exists some u_i \in \mathcal{B} such that \mathcal{B} \setminus \{u_i\} \cup \{w\} remains a basis.^[35] This exchange property allows iterative replacement of basis vectors without altering the spanning or independence properties, proving the invariance of basis size across all bases.^[35] If S spans V and T is a linearly independent subset of V, then |T| \leq n = \dim V, with equality holding if and only if T is a basis for V.^[5] This bound follows directly from the Steinitz exchange lemma applied to extend T within the span of S, showing that no independent set can exceed the dimension without redundancy.^[35] Consequently, any set of more than n vectors in V must be linearly dependent, as it surpasses the maximum possible independence cardinality.^[36] To see why sets larger than the dimension are dependent, consider a set \{v_1, \dots, v_{n+1}\} in V. Define the linear map \phi: F^{n+1} \to V by \phi(e_i) = v_i, where e_i are the standard basis vectors. The image of \phi has dimension at most n, so by the rank-nullity theorem, \dim \ker \phi = (n+1) - \operatorname{rank} \phi \geq 1. A nonzero vector in the kernel yields nontrivial coefficients showing \sum c_i v_i = 0 with not all c_i = 0, confirming dependence.^[37]

Key Properties and Special Cases

Involvement of the zero vector

The zero vector occupies a distinctive position in the theory of linear independence, as its presence in any set of vectors guarantees linear dependence. Specifically, if a set S = \{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n\} in a vector space contains the zero vector \mathbf{0}, then S is linearly dependent. This holds because there exists a nontrivial linear combination equaling the zero vector: assign the coefficient 1 to \mathbf{0} and 0 to all other vectors, yielding

$1 \cdot \mathbf{0} + 0 \cdot \mathbf{v}_2 + \cdots + 0 \cdot \mathbf{v}_n = \mathbf{0},

which is nontrivial since not all coefficients are zero.^[6]^[18] This theorem has direct implications for the structure of linearly independent sets and bases. Since bases must be linearly independent and span the vector space, they cannot include the zero vector; thus, every linearly independent set consists exclusively of nonzero vectors. For instance, the singleton set \{ \mathbf{v} \} is linearly independent if and only if \mathbf{v} \neq \mathbf{0}, as the equation c \mathbf{v} = \mathbf{0} implies c = 0 precisely when \mathbf{v} is nonzero. Additionally, the empty set is conventionally regarded as linearly independent, as the only linear combination (with no vectors) is the trivial one equaling \mathbf{0}; it spans the trivial subspace \{ \mathbf{0} \} and serves as a basis for the zero vector space, which has dimension 0.^[38]^[23]^[39] A practical consequence arises when dealing with dependent sets: if linear dependence stems from the inclusion of the zero vector, removing \mathbf{0} from the set may yield a linearly independent subset that spans the same subspace as the original set minus \mathbf{0}. This removal process preserves the span while eliminating the dependence introduced by the zero vector, facilitating the extraction of maximal independent subsets.^[40]^[24]

Standard basis vectors

In the vector space F^n, where F is a field, the standard basis is the set of vectors \{ e_1, e_2, \dots, e_n \}, with e_i having a 1 in the i-th position and 0 elsewhere for i = 1, \dots, n. These vectors are linearly independent by construction, as any nontrivial linear combination equaling the zero vector would require all coefficients to be zero.^[41] The standard basis vectors span F^n and are linearly independent, thereby forming a basis for the space. To verify their linear independence, consider the n \times n matrix whose columns are these vectors; it is the identity matrix I_n, which has determinant 1 (nonzero), confirming that the only solution to I_n \mathbf{c} = \mathbf{0} is the zero vector \mathbf{c} = \mathbf{0}.^[39] As a basis, they provide unique coordinates for any vector in F^n, meaning every \mathbf{v} \in F^n can be expressed uniquely as \mathbf{v} = c_1 e_1 + \dots + c_n e_n, where the c_i are the components of \mathbf{v}.^[42] This concept generalizes to other finite-dimensional spaces, such as the space of polynomials of degree at most n-1, denoted P_{n-1}, where the standard basis is the set of monomials \{ 1, x, x^2, \dots, x^{n-1} \}. These monomials are linearly independent and span P_{n-1}, analogous to the coordinate basis in F^n.^[43] For example, in \mathbb{R}^2, the vectors e_1 = (1, 0) and e_2 = (0, 1) are linearly independent and form a basis; adding any third vector, such as (1, 1), results in a linearly dependent set, as (1, 1) = e_1 + e_2. The nonzero nature of these standard basis vectors ensures their independence, in contrast to including the zero vector, which would introduce dependence.^[44]

Independence in function spaces

In function spaces, such as the vector space of continuous functions on an interval or polynomials over a field, a set of functions \{f_1, \dots, f_n\} is linearly independent if whenever \sum_{i=1}^n a_i f_i(x) = 0 for all x in the domain, it follows that a_1 = \dots = a_n = 0.^[45] This definition mirrors that in finite-dimensional spaces but applies to the pointwise addition and scalar multiplication of functions.^[45] A classic example occurs in the space of polynomials of degree at most 2, denoted \mathbb{P}_2, over the reals. The set \{1, x, x^2\} is linearly independent because if a + b x + c x^2 = 0 for all x, equating coefficients yields a = b = c = 0.^[46] However, adjoining x^3 to form \{1, x, x^2, x^3\} in \mathbb{P}_2 results in linear dependence, as the dimension of \mathbb{P}_2 is 3, so any four elements must satisfy a nontrivial linear relation.^[46] For exponential functions, the set \{e^{\lambda_1 t}, \dots, e^{\lambda_k t}\} with distinct \lambda_i \in \mathbb{C} is linearly independent over the reals or complexes.^[47] This follows from the Wronskian determinant W(e^{\lambda_1 t}, \dots, e^{\lambda_k t})(t) = \det \begin{pmatrix} e^{\lambda_1 t} & \cdots & e^{\lambda_k t} \\ \lambda_1 e^{\lambda_1 t} & \cdots & \lambda_k e^{\lambda_k t} \\ \vdots & \ddots & \vdots \\ \lambda_1^{k-1} e^{\lambda_1 t} & \cdots & \lambda_k^{k-1} e^{\lambda_k t} \end{pmatrix} = e^{(\lambda_1 + \dots + \lambda_k) t} \prod_{1 \leq i < j \leq k} (\lambda_j - \lambda_i), which is nonzero for all t when the \lambda_i are distinct.^[48] More generally, for a set of n sufficiently differentiable functions f_1, \dots, f_n, linear independence holds if and only if their Wronskian W(f_1, \dots, f_n)(t) \neq 0 at some point t in the domain.^[49] To see this, suppose \sum_{i=1}^n a_i f_i(t_0) = 0 at some t_0 with not all a_i = 0; without loss of generality, assume a_n \neq 0. Then f_n(t_0) = -\sum_{i=1}^{n-1} (a_i / a_n) f_i(t_0). Differentiating the relation repeatedly and evaluating at t_0 yields a system whose determinant is the Wronskian at t_0, leading to a contradiction if it is nonzero.^[48] Conversely, if the functions are dependent, the Wronskian vanishes identically.^[49] In the infinite-dimensional case, the monomials \{x^n \mid n = 0, 1, 2, \dots \} form a linearly independent set in the vector space of formal power series over a field, as any finite linear combination equaling zero implies all coefficients are zero by uniqueness of series representations.^[50]

Advanced Concepts and Generalizations

Linear dependence relations

In the context of a finite set of vectors \{v_1, \dots, v_n\} in a vector space over a field F, the linear dependence relations are the nontrivial tuples (a_1, \dots, a_n) \in F^n satisfying \sum_{i=1}^n a_i v_i = 0.^[51] These relations correspond precisely to the elements of the kernel of the linear map defined by the matrix A with columns v_1, \dots, v_n, where the kernel consists of all solutions to A \mathbf{a} = \mathbf{0}.^[51] The set of all such coefficient tuples forms a subspace of F^n, known as the dependence space, and its dimension equals n - r, where r is the rank of A (equivalently, the dimension of the span of \{v_1, \dots, v_n\}).^[52] This follows directly from the rank-nullity theorem applied to the linear map from F^n to the ambient space induced by A.^[52] For example, consider two collinear vectors in \mathbb{R}^2, such as v_1 = (1, 0) and v_2 = (2, 0). The span has dimension 1, so the dependence space has dimension $2 - 1 = 1, yielding essentially one relation up to scalar multiple: $2 v_1 - v_2 = 0.^[51] A key property is that minimal linearly dependent sets—those where the full set is dependent but every proper subset is independent—have dependence space of dimension 1, meaning exactly one relation up to scalar multiplication.^[52] In more algebraic terms, the dependence relations form a module over the field F (specifically, a vector subspace of F^n), and in module-theoretic contexts, these are studied as syzygies among the vectors.^[53] The vectors \{v_1, \dots, v_n\} are linearly independent if and only if the dependence space is the trivial subspace \{\mathbf{0}\}.^[51]

Affine independence

Affine independence generalizes the concept of linear independence to points in an affine space, focusing on affine combinations rather than linear ones. A finite set of points \{p_0, p_1, \dots, p_k\} in a vector space over \mathbb{R} (or more generally, over a field) is affinely independent if the associated difference vectors \{p_1 - p_0, p_2 - p_0, \dots, p_k - p_0\} form a linearly independent set.^[54] This condition ensures that the points do not lie in a lower-dimensional affine subspace than expected from their count.^[55] Equivalently, the points p_0, \dots, p_k are affinely independent if there is no nontrivial affine relation, meaning no scalars \lambda_0, \lambda_1, \dots, \lambda_k, not all zero, satisfying both \sum_{i=0}^k \lambda_i p_i = 0 and \sum_{i=0}^k \lambda_i = 0.^[56] This formulation captures the idea that no point lies in the affine hyperplane defined by a nontrivial combination of the others with weights summing to zero.^[57] Affine independence thus provides a coordinate-free perspective, independent of the choice of origin, unlike linear independence which is tied to the vector space structure.^[54] Geometrically, in Euclidean space \mathbb{R}^n, any set of affinely independent points with cardinality at most [n+1](/page/N+1) spans an affine subspace of dimension equal to the number of points minus one, forming the vertices of a simplex.^[57] The maximum size of an affinely independent set in \mathbb{R}^n is [n+1](/page/N+1); for instance, in \mathbb{R}^2, up to three non-collinear points can be affinely independent, as they form a triangle, while any four points are necessarily affinely dependent since they cannot all avoid lying on a common line or plane without redundancy.^[56]^[57] Affine independence relates directly to linear independence through translation: if points are affinely independent, then the differences p_i - p_0 for i = 1, \dots, k are linearly independent, and the affine span of the points has dimension equal to the linear dimension of this difference set.^[54] A key theorem states that the affine dimension of a set S is the dimension of the linear span of \{x - y \mid x, y \in S\}, which is one less than the maximum number of affinely independent points in S.^[58] This equivalence underscores the role of affine independence in defining the intrinsic geometry of point sets without reference to a fixed origin.^[55]

Independence of subspaces

In linear algebra, a family of subspaces \{U_1, \dots, U_k\} of a vector space V over a field F is said to be linearly independent if, for each i = 1, \dots, k, the intersection U_i \cap \left( \sum_{j \neq i} U_j \right) = \{0\}. This condition ensures that no nonzero element of U_i can be expressed as a linear combination of elements from the other subspaces. Equivalently, the family is linearly independent if the sum \sum_{i=1}^k U_i is a direct sum, meaning every element in the sum admits a unique representation as \sum_{i=1}^k u_i with u_i \in U_i for each i.^[59] The direct sum notation V = \bigoplus_{i=1}^k U_i is used when the family \{U_1, \dots, U_k\} is linearly independent and their sum spans the entire space V, i.e., V = \sum_{i=1}^k U_i. In this case, every vector in V decomposes uniquely into components from each U_i, providing a canonical decomposition of the space. This structure is fundamental in decomposing vector spaces into simpler components, such as in the study of invariant subspaces under linear transformations.^[60] A concrete example occurs in the Euclidean space \mathbb{R}^n, where the standard coordinate subspaces—such as the x-axis spanned by ([1](/page/1),0,\dots,0) and the y-axis spanned by (0,[1](/page/1),0,\dots,0) in \mathbb{R}^n for n \geq 2—form a linearly independent family. Here, the intersection of one axis with the sum of the others is trivially \{[0](/page/0)\}, and their direct sum yields the full space \mathbb{R}^n. This illustrates how orthogonal directions contribute independently to the overall structure.^[61] One characterization of linear independence for such a family is that the natural inclusion map \iota: \bigoplus_{i=1}^k U_i \to V, which sends (u_1, \dots, u_k) \mapsto \sum_{i=1}^k u_i, is an isomorphism onto its image \sum_{i=1}^k U_i. This isomorphism property highlights the absence of relations between the subspaces beyond their trivial overlaps at the zero vector.^[60] A key consequence of linear independence is the additivity of dimensions: if \{U_1, \dots, U_k\} is linearly independent, then \dim\left( \sum_{i=1}^k U_i \right) = \sum_{i=1}^k \dim U_i. This equality holds because bases of the individual subspaces can be concatenated to form a basis for the sum, without redundancy. Conversely, if the dimensions add up in this way for a sum of subspaces, the family must be linearly independent.^[61] While the primary focus here is on vector spaces, the notion of linear independence extends analogously to modules over a ring, where a family of submodules is independent if each intersects the sum of the others trivially, leading to a direct sum decomposition. This generalization appears in the study of module theory, preserving the core ideas of unique decompositions and dimension-like invariants where applicable.^[62]

References

[1]
[PDF] Math 2331 – Linear Algebra - 1.7 Linear Independence
A set of vectors is linearly independent if the equation x1v1 + x2v2 + ... + xpvp = 0 has only the trivial solution.
[2]
Linear Independence — Linear Algebra, Geometry, and Computation
A set of vectors is linearly independent if the equation c1v1+...+cpvp=0 has only the trivial solution c1=0,...,cp=0.
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Lecture 9: Independence, basis, and dimension | Linear Algebra
A basis is a set of vectors, as few as possible, whose combinations produce all vectors in the space. The number of basis vectors for a space equals the ...
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[PDF] Bases and Dimension - Purdue Math
Feb 16, 2007 · If a vector space has a basis containing n vectors, then from Theorem 4.6.4, the maximum number of vectors in any linearly independent set is n.
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Linear Independence
If you make a set of vectors by adding one vector at a time, and if the span got bigger every time you added a vector, then your set is linearly independent.
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[PDF] 1.7 Linear Independence - UC Berkeley math
The columns of matrix A are linearly independent if and only if the equation Ax = 0 has only the trivial solution.
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[PDF] Section 2.3 Linear Independence
Linear independence is an important concept in linear algebra and sets of vectors which are linearly independent allow us to guarantee useful properties ...
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[PDF] Key Linear Independence Theorems
Apr 21, 2010 · Definition. A vector space V over a field F is finite-dimensional if it has a basis which has finitely many elements. The dimension of a finite ...
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[PDF] Finite-Dimensional Vector Spaces
This situation is so important that we give it a special name—linear independence—which we now define. A list (v1,...,vm) of vectors in V is called linearly ...
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Abstract linear spaces - MacTutor History of Mathematics
The first to give an axiomatic definition of a real linear space was Peano in a book published in Torino in 1888. He credits Leibniz, Möbius's 1827 work ...
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Linear Independence of Vectors - Department of Mathematics at UTSA
Nov 17, 2021 · An infinite set of vectors is linearly independent if every nonempty finite subset is linearly independent. Conversely, an infinite set of ...
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[PDF] Economics 204 Summer/Fall 2011 Lecture 8–Wednesday
Definition 3 A Hamel basis (often just called a basis) of a vector space X is a linearly independent set of vectors in X that spans X. Example: {(1, 0), (0 ...
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Banach space with uncountable basis - MathOverflow
Nov 15, 2022 · So the Hamel dimension of ℓ2 is at least 2ℵ0, and can't be more because the cardinality of ℓ2 is 2ℵ0. Now, for every infinite-dimensional Banach ...Defects of Hamel bases for analysis in infinite dimensionsNo Hilbert space can have countable Hamel basis without using ...More results from mathoverflow.net
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LA8 for DE
An infinite-dimensional space contains an infinite collection of linearly independent elements. For example, the vector space P=⋃nPn is infinite ...
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[PDF] Bases for Infinite Dimensional Vector Spaces
By definition, a basis for a vector space V is a linearly independent set which ... A basis for an infinite dimensional vector space is also called a Hamel basis.
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[PDF] Span, linear independence, basis and dimension 1 - UT Math
Nov 19, 2012 · If T ⊆ V is linearly independent, then T must be finite and |T |≤|S|. As we now show, this implies that if a vector space has a finite basis, ...
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[PDF] SUPPLEMENT TO CHAPTER I Vector spaces
The following are equivalent: (1) S is linearly independent. (2) Every vector in span(S) has a unique expression as a linear combination of vectors in S. ( ...
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[PDF] Linear Span and Bases - UC Davis Math
Jan 23, 2007 · A basis of a finite-dimensional vector space V is a list of vectors (v1,...,vm) in V that is linearly independent and spans V . If (v1,...,vm) ...
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[PDF] MATH 308. Differential Equations
Geometric interpretation of independence: two vectors are indepen- dent if they are not in one line. Three vectors in 3-dimensional space are independent if ...<|control11|><|separator|>
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[PDF] 2 Lecture 2 - 2.1 Span, Basis and Dimensions
In R2, the span of any single vector is the line that goes through the origin and that vector.2. The span of any two vectors in R2 is generally equal to R2 ...Missing: characterization | Show results with:characterization
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Lecture 5 Linear Dependence and Independence
(1) A set consisting of a single nonzero vector is linearly independent. On the other hand, any set containing the vector 0 is linearly dependent. (2) A set ...
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[PDF] Linear Dependence and Linear Independence - Purdue Math
Feb 16, 2007 · Linearly dependent vectors have a non-zero scalar combination summing to zero. Linearly independent vectors only sum to zero with all scalars ...
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[PDF] A Geometric Interpretation of Linear (In)dependence - Ohio University
This set is linearly independent if, and only if, it is not linearly dependent. These two definitions are equivalent; they define the same concepts. Winfried ...
[24]
[PDF] Unit 4: Basis and dimension
linear independence. The assumption q<p can not be true. (ii) Because A spans X and B is linearly independent, we know that q ≤ p. Because. B spans X and A ...Missing: characterization | Show results with:characterization
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[PDF] MATH 23A SOLUTION SET #3 (PART C) Problem (7). Show that if V ...
Linear Independence. Let a1,...,an ∈ R be such that a1f1 + ... + anfn = f0 ... find a basis for V , we can take the standard basis for Rn, namely: B ...
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[PDF] Linear independence. Basis of a vector space.
Determine whether vectors v2,v2,v3 are linearly independent. We have to check if there exist r1,r2,r3 ∈ R not all zero such that r1v1 + r2v2 + r3v3 = 0 ...
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Testing linear dependence via Gaussian Elimination
I use an example to show how to test if a system of vectors is linearly dependent. The method is based on Gaussian Elimination and Row Echelon Form analysis.Missing: check | Show results with:check
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Math 21b: Determinants
The determinant of a square matrix A detects whether A is invertible: If det(A)=0 then A is not invertible (equivalently, the rows of A are linearly dependent.
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[PDF] The dimension of a vector space - Keith Conrad
If V has dimension n and W is a subspace with dimension n, then W = V . Proof. When W has dimension n, any basis for W is a linearly independent subset of V.Missing: scholarly | Show results with:scholarly
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[PDF] 19. Basis and Dimension - UC Davis Math
For a finite dimensional vector space V , any two bases for V have the same number of vectors. Proof. Let S and T be two bases for V . Then both are linearly ...
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[PDF] The Theory of Finite Dimensional Vector Spaces
4.2.6 Extracting a Basis Constructively Theorem 4.6 guarantees that any spanning set of a finite dimensional vector space contains a basis. In fact, the ...
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[PDF] linear algebra: dimension and the steinitz exchange trick
We will deduce this from the following. Lemma 2. Any spanning set of a vector space is at least as large as any linearly independent set. The proof of this ...
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4.13 Finding dimensions ‣ Chapter 4 Linear algebra ‣ MATH0005 ...
1. A basis of U is a linearly independent sequence in V , so we can extend it to a basis of V . So its size is less than or equal to the size of a basis of V .Missing: scholarly | Show results with:scholarly
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1.3 Rank and Nullity
Now since { 1 , x , x 2 } is a linearly independent set, we know that rank ( T ) = 3 which means that nullity ( T ) = 1 by Theorem 1.3.1. It follows that ...
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[PDF] Contents 1 Vector Spaces - Evan Dummit
Here are a few basic properties of linear dependence and independence that follow from the definition: ◦ Any set containing the zero vector is linearly ...
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[PDF] lectures 14/15: linear independence and bases
Example. A set with only one non-zero vector is linearly independent, as if cv = 0, then we saw before that c = 0 or v = 0, and v 6= 0 by assumption. Thus, any ...Missing: single | Show results with:single
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[PDF] Span, Linear Independence, and Dimension - Penn Math
Jul 18, 2013 · If a vector space has a basis consisting of m vectors, then any set of more than m vectors is linearly dependent. Page 16. Span, Linear.Missing: scholarly | Show results with:scholarly
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[PDF] Vector spaces and bases - Columbia CS
So every standard basis vector can be obtained as a linear combinations of vectors in Fn. This implies that span(Fn) = Rn. Using mathematical induction, we ...
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Basis and Dimension
A basis of a subspace is a set of vectors that spans the subspace and is linearly independent. The dimension of a subspace is the number of vectors in any of ...
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[PDF] 6. Vector Spaces - Emory Mathematics
Vector Spaces. Example 6.3.8. Show that dim Pn = n+1 and that {1, x, x. 2, ..., xn} is a basis, called the standard basis of Pn. Solution. Each polynomial p(x) ...
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Bases - A First Course in Linear Algebra
When a vector space has a single particularly nice basis, it is sometimes called the standard basis though there is nothing precise enough about this term to ...
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[PDF] LADR4e.pdf - Linear Algebra Done Right - Sheldon Axler
... Linear Independence 28. Linear Combinations and Span 28. Linear Independence 31. Exercises 2A 37. Linear Algebra Done Right, fourth edition, by Sheldon Axler vi ...
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Linear Independence of Monomials - Branko Curgus
By the definition of a polynomial of degree less or equal n we have \mathbb{P}_n = \operatorname{Span} \bigl\{ a_0 + a_1 x + \cdots + a_n x^n \, : \, a_k \in \ ...
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[PDF] Linear dependence of exponentials - Purdue Math
Any finite set of distinct exponentials is linearly independent. Peo- ple ask me whether infinitely many exponentials exp(λkz) are linearly independent. The ...
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Proof of the theorem about Wronskian
If the Wronskian of this set of functions is not identically zero then the set of functions is linearly independent. Proof. By contradiction, suppose that ...
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[PDF] linear independence, the wronskian, and variation of parameters
Proof. We will now show that if the Wronskian of a set of functions is not zero, then the functions are linearly independent. As above suppose that {x1(t),x2 ...
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[PDF] chap. 13 standard monomials - garsia at york
surjective, it suffices to show that the standard monomials are linearly independent in the ring Z[[i1,i2,... ,in]]. This point can be achieved in several ...
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[PDF] Linear Algebra - Arizona Math
May 4, 2005 · null-space matrix columns together with unit basis vectors. ... the standard basis (the basis consisting of the columns of the identity matrix).
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[PDF] The Rank-Nullity Theorem - Purdue Math
Feb 16, 2007 · In Section 4.3, we defined the null space of a real m × n matrix A to be the set of all real solutions to the associated homogeneous linear ...
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[PDF] MA4J8 Commutative algebra II 1 Lectures 18–22. Syzygies
(Greek for “yoke” – the relations are yoked together like a pair of oxen in ploughing, or are subject to linear dependence relations like stars in.<|control11|><|separator|>
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[PDF] Integer Programming ISE 418 Lecture 4
The property of linear independence is with respect to a given origin. • Affine independence is essentially a “coordinate-free” version of linear independence.
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[PDF] Basics of Affine Geometry - UPenn CIS
2.6 Affine Independence and Affine Frames. Corresponding to the notion of linear independence in vector spaces, we have the notion of affine independence.
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[PDF] 1.2 Convex and Affine Hulls
The dimension of a convex set C is defined to be the dimension of aff(C). Definition:(Affinely Independent) x0, ..., xm ∈ Rn are affinely independent if. X λixi ...Missing: algebra | Show results with:algebra
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[PDF] Simplices Definition 1. Suppose that v 0,...,vk ∈ Rn. The convex hull ...
Three points are affinely independent if and only if they are not collinear. Rn contains at most n+1 affinely independent points. Definition 4. A k–dimensional ...
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[PDF] Lecture 5: Dimension of a polyhedron Minimal Description of a ...
Feb 9, 2021 · Linear independence implies affine independence by definition. ... Figure 5.3: Affine independence does not imply linear independence.
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[PDF] MATH251 - Algebra 2 - SUMS McGill
Definition 28 (Linearly Independent Subspaces). For V1,...,Vk ⊆ V, we say {V1,...,Vk} linearly independent if Vi ∩ Í. j≠i. Vj = {0V } and ...
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SCLA Direct Sums - A First Course in Linear Algebra
A direct sum is a short-hand way to describe the relationship between a vector space and two, or more, of its subspaces. As we will use it, it is not a way to ...
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Linear Algebra » Part 3: Vector Spaces » Direct Sums
... linearly independent, subspaces in a direct sum must be “small enough” that they only contain the zero vector in common. Another useful equivalent ...
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What is the definition of direct sum of submodules?
Mar 10, 2016 · This is usually defined as the submodules summing to the whole module, and having the property that each component intersects the sum of others trivially.Isomorphism between the direct sum of submodules and generated ...Why does the set of submodules of a module that are direct sums of ...More results from math.stackexchange.com