Linear independence#Definition

A sequence of vectors $\backslash mathbf\{v\}\_1,\; \backslash mathbf\{v\}\_2,\; \backslash dots,\; \backslash mathbf\{v\}\_k$ from a vector space {{mvar|V}} is said to be linearly dependent, if there exist scalars $a\_1,\; a\_2,\; \backslash dots,\; a\_k,$ not all zero, such that

:$a\_1\backslash mathbf\{v\}\_1\; +\; a\_2\backslash mathbf\{v\}\_2\; +\; \backslash cdots\; +\; a\_k\backslash mathbf\{v\}\_k\; =\; \backslash mathbf\{0\},$

where $\backslash mathbf\{0\}$ 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

:$\backslash mathbf\{v\}\_1\; =\; \backslash frac\{-a\_2\}\{a\_1\}\backslash mathbf\{v\}\_2\; +\; \backslash cdots\; +\; \backslash frac\{-a\_k\}\{a\_1\}\; \backslash mathbf\{v\}\_k,$

if $k>1,$ and $\backslash mathbf\{v\}\_1\; =\; \backslash mathbf\{0\}$ 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 $\backslash mathbf\{v\}\_1,\; \backslash mathbf\{v\}\_2,\; \backslash dots,\; \backslash mathbf\{v\}\_n$ is said to be linearly independent if it is not linearly dependent, that is, if the equation

:$a\_1\backslash mathbf\{v\}\_1\; +\; a\_2\; \backslash mathbf\{v\}\_2\; +\; \backslash cdots\; +\; a\_n\backslash mathbf\{v\}\_n\; =\; \backslash mathbf\{0\},$

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.{{cite book|last1=Friedberg |last2=Insel |last3=Spence|first1=Stephen |first2=Arnold |first3=Lawrence|title=Linear Algebra|year=2003|publisher=Pearson, 4th Edition|isbn=0130084514|pages=48–49}} 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.

An infinite set of vectors is linearly independent if every finite subset is linearly independent. This definition applies also to finite sets of vectors, since a finite set is a finite subset of itself, and every subset of a linearly independent set is also 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 to be 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 polynomials in {{mvar|x}} over the reals has the (infinite) subset {{math|1={1, x, x², ...} }} as a basis.

Let $V$ be a vector space. A set $X\; \backslash subseteq\; V$ is linearly independent if and only if $X$ is a minimal element of

:$\backslash \{Y\; \backslash subseteq\; V\; \backslash mid\; X\; \backslash subseteq\; \backslash operatorname\{Span\}(Y)\backslash \}$

by the inclusion order. In contrast, $X$ is linearly dependent if it has a proper subset whose span is a superset of $X$.

right

• $\backslash vec\; u$ 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 $\backslash vec\; o\; =\; 0\; \backslash vec\; k$.

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, {{mvar|n}} linearly independent vectors are required to describe all locations in {{mvar|n}}-dimensional space.

If one or more vectors from a given sequence of vectors $\backslash mathbf\{v\}\_1,\; \backslash dots,\; \backslash mathbf\{v\}\_k$ is the zero vector $\backslash mathbf\{0\}$ then the vectors $\backslash mathbf\{v\}\_1,\; \backslash dots,\; \backslash mathbf\{v\}\_k$ 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 $\backslash \{\; 1,\; \backslash ldots,\; k\; \backslash \}$) such that $\backslash mathbf\{v\}\_i\; =\; \backslash mathbf\{0\}.$ Then let $a\_\{i\}\; :=\; 1$ (alternatively, letting $a\_\{i\}$ be equal to 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 $a\_\{j\}\; :=\; 0$ so that consequently $a\_\{j\}\; \backslash mathbf\{v\}\_j\; =\; 0\; \backslash mathbf\{v\}\_j\; =\; \backslash mathbf\{0\}$).

Simplifying $a\_1\; \backslash mathbf\{v\}\_1\; +\; \backslash cdots\; +\; a\_k\backslash mathbf\{v\}\_k$ gives:

:$a\_1\; \backslash mathbf\{v\}\_1\; +\; \backslash cdots\; +\; a\_k\backslash mathbf\{v\}\_k\; =\; \backslash mathbf\{0\}\; +\; \backslash cdots\; +\; \backslash mathbf\{0\}\; +\; a\_i\; \backslash mathbf\{v\}\_i\; +\; \backslash mathbf\{0\}\; +\; \backslash cdots\; +\; \backslash mathbf\{0\}\; =\; a\_i\; \backslash mathbf\{v\}\_i\; =\; a\_i\; \backslash mathbf\{0\}\; =\; \backslash mathbf\{0\}.$

Because not all scalars are zero (in particular, $a\_\{i\}\; \backslash neq\; 0$), this proves that the vectors $\backslash mathbf\{v\}\_1,\; \backslash dots,\; \backslash mathbf\{v\}\_k$ are linearly dependent.

As a consequence, the zero vector can not possibly belong to any collection of vectors that is linearly independent.

Now consider the special case where the sequence of $\backslash mathbf\{v\}\_1,\; \backslash dots,\; \backslash mathbf\{v\}\_k$ 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 $\backslash mathbf\{v\}\_1$ is any vector then the sequence $\backslash mathbf\{v\}\_1$ (which is a sequence of length $1$) is linearly dependent if and only if {{nowrap|$\backslash mathbf\{v\}\_1\; =\; \backslash mathbf\{0\}$;}} alternatively, the collection $\backslash mathbf\{v\}\_1$ is linearly independent if and only if $\backslash mathbf\{v\}\_1\; \backslash neq\; \backslash mathbf\{0\}.$

This example considers the special case where there are exactly two vector $\backslash mathbf\{u\}$ and $\backslash mathbf\{v\}$ from some real or complex vector space.

The vectors $\backslash mathbf\{u\}$ and $\backslash mathbf\{v\}$ are linearly dependent if and only if at least one of the following is true:

1. $\backslash mathbf\{u\}$ is a scalar multiple of $\backslash mathbf\{v\}$ (explicitly, this means that there exists a scalar $c$ such that $\backslash mathbf\{u\}\; =\; c\; \backslash mathbf\{v\}$) or
2. $\backslash mathbf\{v\}$ is a scalar multiple of $\backslash mathbf\{u\}$ (explicitly, this means that there exists a scalar $c$ such that $\backslash mathbf\{v\}\; =\; c\; \backslash mathbf\{u\}$).

If $\backslash mathbf\{u\}\; =\; \backslash mathbf\{0\}$ then by setting $c\; :=\; 0$ we have $c\; \backslash mathbf\{v\}\; =\; 0\; \backslash mathbf\{v\}\; =\; \backslash mathbf\{0\}\; =\; \backslash mathbf\{u\}$ (this equality holds no matter what the value of $\backslash mathbf\{v\}$ is), which shows that (1) is true in this particular case. Similarly, if $\backslash mathbf\{v\}\; =\; \backslash mathbf\{0\}$ then (2) is true because $\backslash mathbf\{v\}\; =\; 0\; \backslash mathbf\{u\}.$

If $\backslash mathbf\{u\}\; =\; \backslash mathbf\{v\}$ (for instance, if they are both equal to the zero vector $\backslash mathbf\{0\}$) then both (1) and (2) are true (by using $c\; :=\; 1$ for both).

If $\backslash mathbf\{u\}\; =\; c\; \backslash mathbf\{v\}$ then $\backslash mathbf\{u\}\; \backslash neq\; \backslash mathbf\{0\}$ is only possible if $c\; \backslash neq\; 0$ and $\backslash mathbf\{v\}\; \backslash neq\; \backslash mathbf\{0\}$; in this case, it is possible to multiply both sides by $\backslash frac\{1\}\{c\}$ to conclude $\backslash mathbf\{v\}\; =\; \backslash frac\{1\}\{c\}\; \backslash mathbf\{u\}.$

This shows that if $\backslash mathbf\{u\}\; \backslash neq\; \backslash mathbf\{0\}$ and $\backslash mathbf\{v\}\; \backslash neq\; \backslash mathbf\{0\}$ 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 independent).

If $\backslash mathbf\{u\}\; =\; c\; \backslash mathbf\{v\}$ but instead $\backslash mathbf\{u\}\; =\; \backslash mathbf\{0\}$ then at least one of $c$ and $\backslash mathbf\{v\}$ must be zero.

Moreover, if exactly one of $\backslash mathbf\{u\}$ and $\backslash mathbf\{v\}$ is $\backslash mathbf\{0\}$ (while the other is non-zero) then exactly one of (1) and (2) is true (with the other being false).

The vectors $\backslash mathbf\{u\}$ and $\backslash mathbf\{v\}$ are linearly independent if and only if $\backslash mathbf\{u\}$ is not a scalar multiple of $\backslash mathbf\{v\}$ and $\backslash mathbf\{v\}$ is not a scalar multiple of $\backslash mathbf\{u\}$.

Three vectors: Consider the set of vectors $\backslash mathbf\{v\}\_1\; =\; (1,\; 1),$ $\backslash mathbf\{v\}\_2\; =\; (-3,\; 2),$ and $\backslash mathbf\{v\}\_3\; =\; (2,\; 4),$ then the condition for linear dependence seeks a set of non-zero scalars, such that

:$a\_1\; \backslash begin\{bmatrix\}\; 1\backslash \backslash 1\backslash end\{bmatrix\}\; +\; a\_2\; \backslash begin\{bmatrix\}\; -3\backslash \backslash 2\backslash end\{bmatrix\}\; +\; a\_3\; \backslash begin\{bmatrix\}\; 2\backslash \backslash 4\backslash end\{bmatrix\}\; =\backslash begin\{bmatrix\}\; 0\backslash \backslash 0\backslash end\{bmatrix\},$

or

:

Row reduce this matrix equation by subtracting the first row from the second to obtain,

:

Continue the row reduction by (i) dividing the second row by 5, and then (ii) multiplying by 3 and adding to the first row, that is

:

Rearranging this equation allows us to obtain

:

which shows that non-zero a_i exist such that $\backslash mathbf\{v\}\_3\; =\; (2,\; 4)$ can be defined in terms of $\backslash mathbf\{v\}\_1\; =\; (1,\; 1)$ and $\backslash mathbf\{v\}\_2\; =\; (-3,\; 2).$ Thus, the three vectors are linearly dependent.

Two vectors: Now consider the linear dependence of the two vectors $\backslash mathbf\{v\}\_1\; =\; (1,\; 1)$ and $\backslash mathbf\{v\}\_2\; =\; (-3,\; 2),$ and check,

:$a\_1\; \backslash begin\{bmatrix\}\; 1\backslash \backslash 1\backslash end\{bmatrix\}\; +\; a\_2\; \backslash begin\{bmatrix\}\; -3\backslash \backslash 2\backslash end\{bmatrix\}\; =\backslash begin\{bmatrix\}\; 0\backslash \backslash 0\backslash end\{bmatrix\},$

or

:

The same row reduction presented above yields,

:

This shows that $a\_i\; =\; 0,$ which means that the vectors $\backslash mathbf\{v\}\_1\; =\; (1,\; 1)$ and $\backslash mathbf\{v\}\_2\; =\; (-3,\; 2)$ are linearly independent.

In order to determine if the three vectors in $\backslash mathbb\{R\}^4,$

:$\backslash mathbf\{v\}\_1=\; \backslash begin\{bmatrix\}1\backslash \backslash 4\backslash \backslash 2\backslash \backslash -3\backslash end\{bmatrix\},\; \backslash mathbf\{v\}\_2=\backslash begin\{bmatrix\}7\backslash \backslash 10\backslash \backslash -4\backslash \backslash -1\backslash end\{bmatrix\},\; \backslash mathbf\{v\}\_3=\backslash begin\{bmatrix\}-2\backslash \backslash 1\backslash \backslash 5\backslash \backslash -4\backslash end\{bmatrix\}.$

are linearly dependent, form the matrix equation,

:

Row reduce this equation to obtain,

:

Rearrange to solve for v₃ and obtain,

:

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 $\backslash mathbf\{v\}\_1,\; \backslash mathbf\{v\}\_2,$ and $\backslash mathbf\{v\}\_3$ are linearly dependent.

An alternative method relies on the fact that $n$ vectors in $\backslash mathbb\{R\}^n$ are linearly independent if and only if the determinant of the matrix formed by taking the vectors as its columns is non-zero.

In this case, the matrix formed by the vectors is

:

We may write a linear combination of the columns as

:

We are interested in whether {{math|1=AΛ = 0}} 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 is non-zero, the vectors $(1,\; 1)$ and $(-3,\; 2)$ are linearly independent.

Otherwise, suppose we have $m$ vectors of $n$ coordinates, with Then A is an n×m matrix and Λ is a column vector with $m$ entries, and we are again interested in AΛ = 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 {{math|⟨i₁,...,i_m⟩}} is any list of $m$ rows, then the equation must be true for those rows.

:$A\_\{\backslash lang\; i\_1,\backslash dots,i\_m\; \backslash rang\}\; \backslash Lambda\; =\; \backslash mathbf\{0\}\; .$

Furthermore, the reverse is true. That is, we can test whether the $m$ vectors are linearly dependent by testing whether

:$\backslash det\; A\_\{\backslash lang\; i\_1,\backslash dots,i\_m\; \backslash rang\}\; =\; 0$

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.

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

Let $V\; =\; \backslash R^n$ and consider the following elements in $V$, known as the natural basis vectors:

:$\backslash begin\{matrix\}$

\mathbf{e}_1 & = & (1,0,0,\ldots,0) \\

\mathbf{e}_2 & = & (0,1,0,\ldots,0) \\

& \vdots \\

\mathbf{e}_n & = & (0,0,0,\ldots,1).\end{matrix}

Then $\backslash mathbf\{e\}\_1,\; \backslash mathbf\{e\}\_2,\; \backslash ldots,\; \backslash mathbf\{e\}\_n$ are linearly independent.

{{math proof|

Suppose that $a\_1,\; a\_2,\; \backslash ldots,\; a\_n$ are real numbers such that

:$a\_1\; \backslash mathbf\{e\}\_1\; +\; a\_2\; \backslash mathbf\{e\}\_2\; +\; \backslash cdots\; +\; a\_n\; \backslash mathbf\{e\}\_n\; =\; \backslash mathbf\{0\}.$

Since

:$a\_1\; \backslash mathbf\{e\}\_1\; +\; a\_2\; \backslash mathbf\{e\}\_2\; +\; \backslash cdots\; +\; a\_n\; \backslash mathbf\{e\}\_n\; =\; \backslash left(\; a\_1\; ,a\_2\; ,\backslash ldots,\; a\_n\; \backslash right),$

then $a\_i\; =\; 0$ for all $i\; =\; 1,\; \backslash ldots,\; n.$

}}

Let $V$ be the vector space of all differentiable functions of a real variable $t$. Then the functions $e^t$ and $e^\{2t\}$ in $V$ are linearly independent.

Suppose $a$ and $b$ are two real numbers such that

:$ae\; ^\; t\; +\; be\; ^\; \{2t\}\; =\; 0$

Take the first derivative of the above equation:

:$ae\; ^\; t\; +\; 2be\; ^\; \{2t\}\; =\; 0$

for {{em|all}} 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 $be^\{2t\}\; =\; 0$. Since $e^\{2t\}$ is not zero for some $t$, $b=0.$ It follows that $a\; =\; 0$ too. Therefore, according to the definition of linear independence, $e^\{t\}$ and $e^\{2t\}$ are linearly independent.

A linear dependency or linear relation among vectors {{math|v₁, ..., v_n}} is a tuple {{math|(a₁, ..., a_n)}} with {{mvar|n}} scalar components such that

:$a\_1\; \backslash mathbf\{v\}\_1\; +\; \backslash cdots\; +\; a\_n\; \backslash mathbf\{v\}\_n=\; \backslash mathbf\{0\}.$

If such a linear dependence exists with at least a nonzero component, then the {{mvar|n}} vectors are linearly dependent. Linear dependencies among {{math|v₁, ..., v_n}} 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.

{{See also|Affine space}}

A set of vectors is said to be affinely dependent if at least one of the vectors in the set can be defined as an affine combination 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. Contrapositively, every linearly independent set is affinely independent. Note that an affinely independent set is not necessarily linearly independent.

Consider a set of $m$ vectors $\backslash mathbf\{v\}\_1,\; \backslash ldots,\; \backslash mathbf\{v\}\_m$ of size $n$ each, and consider the set of $m$ augmented vectors $\backslash left(\backslash left[\backslash begin\{smallmatrix\}\; 1\; \backslash \backslash \; \backslash mathbf\{v\}\_1\backslash end\{smallmatrix\}\backslash right],\; \backslash ldots,\; \backslash left[\backslash begin\{smallmatrix\}1\; \backslash \backslash \; \backslash mathbf\{v\}\_m\backslash end\{smallmatrix\}\backslash right]\backslash right)$ of size $n\; +\; 1$ each. The original vectors are affinely independent if and only if the augmented vectors are linearly independent.{{Cite Lovasz Plummer}}{{Rp|256}}

Two vector subspaces $M$ and $N$ of a vector space $X$ are said to be {{em|linearly independent}} if $M\; \backslash cap\; N\; =\; \backslash \{0\backslash \}.${{Bachman Narici Functional Analysis 2nd Edition}} pp. 3–7

More generally, a collection $M\_1,\; \backslash ldots,\; M\_d$ of subspaces of $X$ are said to be {{em|linearly independent}} if $M\_i\; \backslash cap\; \backslash sum\_\{k\; \backslash neq\; i\}\; M\_k\; =\; \backslash \{0\backslash \}$ for every index $i,$ where $\backslash sum\_\{k\; \backslash neq\; i\}\; M\_k\; =\; \backslash Big\backslash \{m\_1\; +\; \backslash cdots\; +\; m\_\{i-1\}\; +\; m\_\{i+1\}\; +\; \backslash cdots\; +\; m\_d\; :\; m\_k\; \backslash in\; M\_k\; \backslash text\{\; for\; all\; \}\; k\backslash Big\backslash \}\; =\; \backslash operatorname\{span\}\; \backslash bigcup\_\{k\; \backslash in\; \backslash \{1,\backslash ldots,i-1,i+1,\backslash ldots,d\backslash \}\}\; M\_k.$

The vector space $X$ is said to be a {{em|direct sum}} of $M\_1,\; \backslash ldots,\; M\_d$ if these subspaces are linearly independent and $M\_1\; +\; \backslash cdots\; +\; M\_d\; =\; X.$

• {{annotated link|Matroid}}

{{reflist}}

• {{springer|title=Linear independence|id=p/l059290}}
• [http://mathworld.wolfram.com/LinearlyDependentFunctions.html Linearly Dependent Functions] at WolframMathWorld.
• [http://people.revoledu.com/kardi/tutorial/LinearAlgebra/LinearlyIndependent.html Tutorial and interactive program] on Linear Independence.
• [https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/linear_independence/v/linear-algebra-introduction-to-linear-independence Introduction to Linear Independence] at KhanAcademy.

{{linear algebra}}

{{Matrix classes}}

Category:Abstract algebra

Category:Linear algebra

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