In the theory of

_{''i''} exist such that $\backslash mathbf\_3\; =\; (2,\; 4)$ can be defined in terms of $\backslash mathbf\_1\; =\; (1,\; 1)$ and $\backslash mathbf\_2\; =\; (-3,\; 2).$ Thus, the three vectors are linearly dependent.
Two vectors: Now consider the linear dependence of the two vectors $\backslash mathbf\_1\; =\; (1,\; 1)$ and $\backslash mathbf\_2\; =\; (-3,\; 2),$ and check,
:$a\_1\; \backslash begin\; 1\backslash \backslash 1\backslash end\; +\; a\_2\; \backslash begin\; -3\backslash \backslash 2\backslash end\; =\backslash begin\; 0\backslash \backslash 0\backslash end,$
or
:$\backslash begin\; 1\; \&\; -3\; \backslash \backslash \; 1\; \&\; 2\; \backslash end\backslash begin\; a\_1\backslash \backslash \; a\_2\; \backslash end=\; \backslash begin\; 0\backslash \backslash 0\backslash end.$
The same row reduction presented above yields,
:$\backslash begin\; 1\; \&\; 0\; \backslash \backslash \; 0\; \&\; 1\; \backslash end\backslash begin\; a\_1\backslash \backslash \; a\_2\; \backslash end=\; \backslash begin\; 0\backslash \backslash 0\backslash end.$
This shows that $a\_i\; =\; 0,$ which means that the vectors ''v''_{1} = (1, 1) and ''v''_{2} = (−3, 2) are linearly independent.

_{3} and obtain,
:$\backslash begin\; 1\&\; 7\; \backslash \backslash \; 0\&\; -18\; \backslash end\; \backslash begin\; a\_1\backslash \backslash \; a\_2\; \backslash end\; =\; -a\_3\backslash begin-2\backslash \backslash 9\backslash end.$
This equation is easily solved to define non-zero ''a''_{i},
:$a\_1\; =\; -3\; a\_3\; /2,\; a\_2\; =\; a\_3/2,$
where $a\_3$ can be chosen arbitrarily. Thus, the vectors $\backslash mathbf\_1,\; \backslash mathbf\_2,$ and $\backslash mathbf\_3$ are linearly dependent.

Linearly Dependent Functions

at WolframMathWorld.

on Linear Independence.

Introduction to Linear Independence

at KhanAcademy. {{Matrix classes Abstract algebra Linear algebra Articles containing proofs

vector space
In mathematics
Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in ...

s, a set of vectors is said to be if there is a nontrivial linear combination of the vectors that equals the zero vector. If no such linear combination exists, then the vectors are said to be . These concepts are central to the definition of dimension
In physics and mathematics, the dimension of a Space (mathematics), mathematical space (or object) is informally defined as the minimum number of coordinates needed to specify any Point (geometry), point within it. Thus, a Line (geometry), lin ...

.
A vector space can be of finite dimension or infinite dimension depending on the maximum number of linearly independent vectors. The definition of linear dependence and the ability to determine whether a subset of vectors in a vector space is linearly dependent are central to determining the dimension of a vector space.
Definition

A sequence of vectors $\backslash mathbf\_1,\; \backslash mathbf\_2,\; \backslash dots,\; \backslash mathbf\_k$ from avector space
In mathematics
Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in ...

is said to be ''linearly dependent'', if there exist scalars $a\_1,\; a\_2,\; \backslash dots,\; a\_k,$ not all zero, such that
:$a\_1\backslash mathbf\_1\; +\; a\_2\backslash mathbf\_2\; +\; \backslash cdots\; +\; a\_k\backslash mathbf\_k\; =\; \backslash mathbf,$
where $\backslash mathbf$ denotes the zero vector.
This implies that at least one of the scalars is nonzero, say $a\_1\backslash ne\; 0$, and the above equation is able to be written as
:$\backslash mathbf\_1\; =\; \backslash frac\backslash mathbf\_2\; +\; \backslash cdots\; +\; \backslash frac\; \backslash mathbf\_k,$
if $k>1,$ and $\backslash mathbf\_1\; =\; \backslash mathbf$ if $k=1.$
Thus, a set of vectors is linearly dependent if and only if one of them is zero or a linear combination of the others.
A sequence of vectors $\backslash mathbf\_1,\; \backslash mathbf\_2,\; \backslash dots,\; \backslash mathbf\_n$ is said to be ''linearly independent'' if it is not linearly dependent, that is, if the equation
:$a\_1\backslash mathbf\_1\; +\; a\_2\; \backslash mathbf\_2\; +\; \backslash cdots\; +\; a\_n\backslash mathbf\_n\; =\; \backslash mathbf,$
can only be satisfied by $a\_i=0$ for $i=1,\backslash dots,n.$ This implies that no vector in the sequence can be represented as a linear combination of the remaining vectors in the sequence. In other words, a sequence of vectors is linearly independent if the only representation of $\backslash mathbf\; 0$ as a linear combination of its vectors is the trivial representation in which all the scalars $a\_i$ are zero. Even more concisely, a sequence of vectors is linearly independent if and only if $\backslash mathbf\; 0$ can be represented as a linear combination of its vectors in a unique way.
If a sequence of vectors contains the same vector twice, it is necessarily dependent. The linear dependency of a sequence of vectors does not depend of the order of the terms in the sequence. This allows defining linear independence for a finite set of vectors: A finite set of vectors is ''linearly independent'' if the sequence obtained by ordering them is linearly independent. In other words, one has the following result that is often useful.
A sequence of vectors is linearly independent if and only if it does not contain the same vector twice and the set of its vectors is linearly independent.
Infinite case

An infinite set of vectors is ''linearly independent'' if every nonempty finitesubset
In mathematics, Set (mathematics), set ''A'' is a subset of a set ''B'' if all Element (mathematics), elements of ''A'' are also elements of ''B''; ''B'' is then a superset of ''A''. It is possible for ''A'' and ''B'' to be equal; if they are ...

is linearly independent. Conversely, an infinite set of vectors is ''linearly dependent'' if it contains a finite subset that is linearly dependent, or equivalently, if some vector in the set is a linear combination of other vectors in the set.
An indexed family of vectors is ''linearly independent'' if it does not contain the same vector twice, and if the set of its vectors is linearly independent. Otherwise, the family is said ''linearly dependent''.
A set of vectors which is linearly independent and spans some vector space, forms a basis for that vector space. For example, the vector space of all polynomial
In mathematics, a polynomial is an expression (mathematics), expression consisting of indeterminate (variable), indeterminates (also called variable (mathematics), variables) and coefficients, that involves only the operations of addition, subtrac ...

s in over the reals has the (infinite) subset as a basis.
Geometric examples

* $\backslash vec\; u$ and $\backslash vec\; v$ are independent and define the plane P. * $\backslash vec\; u$, $\backslash vec\; v$ and $\backslash vec\; w$ are dependent because all three are contained in the same plane. * $\backslash vec\; u$ and $\backslash vec\; j$ are dependent because they are parallel to each other. * $\backslash vec\; u$ , $\backslash vec\; v$ and $\backslash vec\; k$ are independent because $\backslash vec\; u$ and $\backslash vec\; v$ are independent of each other and $\backslash vec\; k$ is not a linear combination of them or, equivalently, because they do not belong to a common plane. The three vectors define a three-dimensional space. * The vectors $\backslash vec\; o$ (null vector, whose components are equal to zero) and $\backslash vec\; k$ are dependent since $\backslash vec\; o\; =\; 0\; \backslash vec\; k$Geographic location

A person describing the location of a certain place might say, "It is 3 miles north and 4 miles east of here." This is sufficient information to describe the location, because the geographic coordinate system may be considered as a 2-dimensional vector space (ignoring altitude and the curvature of the Earth's surface). The person might add, "The place is 5 miles northeast of here." This last statement is ''true'', but it is not necessary to find the location. In this example the "3 miles north" vector and the "4 miles east" vector are linearly independent. That is to say, the north vector cannot be described in terms of the east vector, and vice versa. The third "5 miles northeast" vector is a linear combination of the other two vectors, and it makes the set of vectors ''linearly dependent'', that is, one of the three vectors is unnecessary to define a specific location on a plane. Also note that if altitude is not ignored, it becomes necessary to add a third vector to the linearly independent set. In general, linearly independent vectors are required to describe all locations in -dimensional space.Evaluating linear independence

The zero vector

If one or more vectors from a given sequence of vectors $\backslash mathbf\_1,\; \backslash dots,\; \backslash mathbf\_k$ is the zero vector $\backslash mathbf$ then the vector $\backslash mathbf\_1,\; \backslash dots,\; \backslash mathbf\_k$ are necessarily linearly dependent (and consequently, they are not linearly independent). To see why, suppose that $i$ is an index (i.e. an element of $\backslash $) such that $\backslash mathbf\_i\; =\; \backslash mathbf.$ Then let $a\_\; :=\; 1$ (alternatively, letting $a\_$ be equal any other non-zero scalar will also work) and then let all other scalars be $0$ (explicitly, this means that for any index $j$ other than $i$ (i.e. for $j\; \backslash neq\; i$), let $a\_\; :=\; 0$ so that consequently $a\_\; \backslash mathbf\_j\; =\; 0\; \backslash mathbf\_j\; =\; \backslash mathbf$). Simplifying $a\_1\; \backslash mathbf\_1\; +\; \backslash cdots\; +\; a\_k\backslash mathbf\_k$ gives: :$a\_1\; \backslash mathbf\_1\; +\; \backslash cdots\; +\; a\_k\backslash mathbf\_k\; =\; \backslash mathbf\; +\; \backslash cdots\; +\; \backslash mathbf\; +\; a\_i\; \backslash mathbf\_i\; +\; \backslash mathbf\; +\; \backslash cdots\; +\; \backslash mathbf\; =\; a\_i\; \backslash mathbf\_i\; =\; a\_i\; \backslash mathbf\; =\; \backslash mathbf.$ Because not all scalars are zero (in particular, $a\_\; \backslash neq\; 0$), this proves that the vectors $\backslash mathbf\_1,\; \backslash dots,\; \backslash mathbf\_k$ are linearly dependent. As a consequence, the zero vector can not possibly belong to any collection of vectors that is linearly ''in''dependent. Now consider the special case where the sequence of $\backslash mathbf\_1,\; \backslash dots,\; \backslash mathbf\_k$ has length $1$ (i.e. the case where $k\; =\; 1$). A collection of vectors that consists of exactly one vector is linearly dependent if and only if that vector is zero. Explicitly, if $\backslash mathbf\_1$ is any vector then the sequence $\backslash mathbf\_1$ (which is a sequence of length $1$) is linearly dependent if and only if alternatively, the collection $\backslash mathbf\_1$ is linearly independent if and only if $\backslash mathbf\_1\; \backslash neq\; \backslash mathbf.$Linear dependence and independence of two vectors

This example considers the special case where there are exactly two vector $\backslash mathbf$ and $\backslash mathbf$ from some real or complex vector space. The vectors $\backslash mathbf$ and $\backslash mathbf$ are linearly dependentif and only if
In logic and related fields such as mathematics and philosophy, "if and only if" (shortened as "iff") is a biconditional logical connective between statements, where either both statements are true or both are false.
The connective is bicondi ...

at least one of the following is true:
# $\backslash mathbf$ is a scalar multiple of $\backslash mathbf$ (explicitly, this means that there exists a scalar $c$ such that $\backslash mathbf\; =\; c\; \backslash mathbf$) or
# $\backslash mathbf$ is a scalar multiple of $\backslash mathbf$ (explicitly, this means that there exists a scalar $c$ such that $\backslash mathbf\; =\; c\; \backslash mathbf$).
If $\backslash mathbf\; =\; \backslash mathbf$ then by setting $c\; :=\; 0$ we have $c\; \backslash mathbf\; =\; 0\; \backslash mathbf\; =\; \backslash mathbf\; =\; \backslash mathbf$ (this equality holds no matter what the value of $\backslash mathbf$ is), which shows that (1) is true in this particular case. Similarly, if $\backslash mathbf\; =\; \backslash mathbf$ then (2) is true because $\backslash mathbf\; =\; 0\; \backslash mathbf.$
If $\backslash mathbf\; =\; \backslash mathbf$ (for instance, if they are both equal to the zero vector $\backslash mathbf$) then ''both'' (1) and (2) are true (by using $c\; :=\; 1$ for both).
If $\backslash mathbf\; =\; c\; \backslash mathbf$ then $\backslash mathbf\; \backslash neq\; \backslash mathbf$ is only possible if $c\; \backslash neq\; 0$ ''and'' $\backslash mathbf\; \backslash neq\; \backslash mathbf$; in this case, it is possible to multiply both sides by $\backslash frac$ to conclude $\backslash mathbf\; =\; \backslash frac\; \backslash mathbf.$
This shows that if $\backslash mathbf\; \backslash neq\; \backslash mathbf$ and $\backslash mathbf\; \backslash neq\; \backslash mathbf$ then (1) is true if and only if (2) is true; that is, in this particular case either both (1) and (2) are true (and the vectors are linearly dependent) or else both (1) and (2) are false (and the vectors are linearly ''in''dependent).
If $\backslash mathbf\; =\; c\; \backslash mathbf$ but instead $\backslash mathbf\; =\; \backslash mathbf$ then at least one of $c$ and $\backslash mathbf$ must be zero.
Moreover, if exactly one of $\backslash mathbf$ and $\backslash mathbf$ is $\backslash mathbf$ (while the other is non-zero) then exactly one of (1) and (2) is true (with the other being false).
The vectors $\backslash mathbf$ and $\backslash mathbf$ are linearly ''in''dependent if and only if $\backslash mathbf$ is not a scalar multiple of $\backslash mathbf$ ''and'' $\backslash mathbf$ is not a scalar multiple of $\backslash mathbf$.
Vectors in R^{2}

Vectors in R^{4}

Alternative method using determinants

An alternative method relies on the fact that $n$ vectors in $\backslash mathbb^n$ are linearly independentif and only if
In logic and related fields such as mathematics and philosophy, "if and only if" (shortened as "iff") is a biconditional logical connective between statements, where either both statements are true or both are false.
The connective is bicondi ...

the determinant
In mathematics, the determinant is a Scalar (mathematics), scalar value that is a function (mathematics), function of the entries of a square matrix. It characterizes some properties of the matrix and the linear map represented by the matrix. In p ...

of the matrix formed by taking the vectors as its columns is non-zero.
In this case, the matrix formed by the vectors is
:$A\; =\; \backslash begin1\&-3\backslash \backslash 1\&2\backslash end\; .$
We may write a linear combination of the columns as
:$A\; \backslash Lambda\; =\; \backslash begin1\&-3\backslash \backslash 1\&2\backslash end\; \backslash begin\backslash lambda\_1\; \backslash \backslash \; \backslash lambda\_2\; \backslash end\; .$
We are interested in whether for some nonzero vector Λ. This depends on the determinant of $A$, which is
:$\backslash det\; A\; =\; 1\backslash cdot2\; -\; 1\backslash cdot(-3)\; =\; 5\; \backslash ne\; 0.$
Since the determinant
In mathematics, the determinant is a Scalar (mathematics), scalar value that is a function (mathematics), function of the entries of a square matrix. It characterizes some properties of the matrix and the linear map represented by the matrix. In p ...

is non-zero, the vectors $(1,\; 1)$ and $(-3,\; 2)$ are linearly independent.
Otherwise, suppose we have $m$ vectors of $n$ coordinates, with $m\; <\; n.$ Then ''A'' is an ''m''×''n'' matrix and Λ is a column vector with $m$ entries, and we are again interested in ''A''Λ = 0. As we saw previously, this is equivalent to a list of $n$ equations. Consider the first $m$ rows of $A$, the first $m$ equations; any solution of the full list of equations must also be true of the reduced list. In fact, if is any list of $m$ rows, then the equation must be true for those rows.
:$A\_\; \backslash Lambda\; =\; \backslash mathbf\; .$
Furthermore, the reverse is true. That is, we can test whether the $m$ vectors are linearly dependent by testing whether
:$\backslash det\; A\_\; =\; 0$
for all possible lists of $m$ rows. (In case $m\; =\; n$, this requires only one determinant, as above. If $m\; >\; n$, then it is a theorem that the vectors must be linearly dependent.) This fact is valuable for theory; in practical calculations more efficient methods are available.
More vectors than dimensions

If there are more vectors than dimensions, the vectors are linearly dependent. This is illustrated in the example above of three vectors in $\backslash R^2.$Natural basis vectors

Let $V\; =\; \backslash R^n$ and consider the following elements in $V$, known as the natural basis vectors: :$\backslash begin\; \backslash mathbf\_1\; \&\; =\; \&\; (1,0,0,\backslash ldots,0)\; \backslash \backslash \; \backslash mathbf\_2\; \&\; =\; \&\; (0,1,0,\backslash ldots,0)\; \backslash \backslash \; \&\; \backslash vdots\; \backslash \backslash \; \backslash mathbf\_n\; \&\; =\; \&\; (0,0,0,\backslash ldots,1).\backslash end$ Then $\backslash mathbf\_1,\; \backslash mathbf\_2,\; \backslash ldots,\; \backslash mathbf\_n$ are linearly independent.Linear independence of functions

Let $V$ be thevector space
In mathematics
Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in ...

of all differentiable functions of a real variable $t$. Then the functions $e^t$ and $e^$ in $V$ are linearly independent.
Proof

Suppose $a$ and $b$ are two real numbers such that :$ae\; ^\; t\; +\; be\; ^\; =\; 0$ Take the first derivative of the above equation: :$ae\; ^\; t\; +\; 2be\; ^\; =\; 0$ for values of $t.$ We need to show that $a\; =\; 0$ and $b\; =\; 0.$ In order to do this, we subtract the first equation from the second, giving $be^\; =\; 0$. Since $e^$ is not zero for some $t$, $b=0.$ It follows that $a\; =\; 0$ too. Therefore, according to the definition of linear independence, $e^$ and $e^$ are linearly independent.Space of linear dependencies

A linear dependency orlinear relation
In linear algebra, a linear relation, or simply relation, between elements of a vector space or a module (mathematics), module is a linear equation that has these elements as a solution.
More precisely, if e_1,\dots,e_n are elements of a (left) m ...

among vectors is a tuple
In mathematics
Mathematics is an area of knowledge that includes the topics of numbers, formulas and related structures, shapes and the spaces in which they are contained, and quantities and their changes. These topics are represented in m ...

with scalar components such that
:$a\_1\; \backslash mathbf\_1\; +\; \backslash cdots\; +\; a\_n\; \backslash mathbf\_n=\; \backslash mathbf.$
If such a linear dependence exists with at least a nonzero component, then the vectors are linearly dependent. Linear dependencies among form a vector space.
If the vectors are expressed by their coordinates, then the linear dependencies are the solutions of a homogeneous system of linear equations, with the coordinates of the vectors as coefficients. A basis of the vector space of linear dependencies can therefore be computed by Gaussian elimination.
Generalizations

Affine independence

A set of vectors is said to be affinely dependent if at least one of the vectors in the set can be defined as anaffine combination In mathematics, an affine combination of is a linear combination
: \sum_^ = \alpha_ x_ + \alpha_ x_ + \cdots +\alpha_ x_,
such that
:\sum_^ =1.
Here, can be elements (vector (mathematics), vectors) of a vector space over a field (mathematics ...

of the others. Otherwise, the set is called affinely independent. Any affine combination is a linear combination; therefore every affinely dependent set is linearly dependent. Conversely, every linearly independent set is affinely independent.
Consider a set of $m$ vectors $\backslash mathbf\_1,\; \backslash ldots,\; \backslash mathbf\_m$ of size $n$ each, and consider the set of $m$ augmented vectors $\backslash left(\backslash left;\; href="/html/ALL/s/begin\_1\_\backslash \backslash \_\backslash mathbf\_1\backslash end\backslash right.html"\; ;"title="begin\; 1\; \backslash \backslash \; \backslash mathbf\_1\backslash end\backslash right">begin\; 1\; \backslash \backslash \; \backslash mathbf\_1\backslash end\backslash right$ of size $n\; +\; 1$ each. The original vectors are affinely independent if and only if the augmented vectors are linearly independent.
Linearly independent vector subspaces

Two vector subspaces $M$ and $N$ of a vector space $X$ are said to be if $M\; \backslash cap\; N\; =\; \backslash .$ More generally, a collection $M\_1,\; \backslash ldots,\; M\_d$ of subspaces of $X$ are said to be if $M\_i\; \backslash cap\; \backslash sum\_\; M\_k\; =\; \backslash $ for every index $i,$ where $\backslash sum\_\; M\_k\; =\; \backslash Big\backslash \; =\; \backslash operatorname\; \backslash bigcup\_\; M\_k.$ The vector space $X$ is said to be a of $M\_1,\; \backslash ldots,\; M\_d$ if these subspaces are linearly independent and $M\_1\; +\; \backslash cdots\; +\; M\_d\; =\; X.$See also

*References

*External links

*Linearly Dependent Functions

at WolframMathWorld.

on Linear Independence.

Introduction to Linear Independence

at KhanAcademy. {{Matrix classes Abstract algebra Linear algebra Articles containing proofs