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| [[File:Linearly independent vectors in R3.svg|thumb|right|Linearly independent vectors in '''R'''<sup>3</sup>.]]
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| [[File:Linearly dependent vectors in R3.svg|thumb|right|Linearly dependent vectors in a plane in '''R'''<sup>3</sup>.]]
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| In [[linear algebra]], two slightly different notions of linear independence are used: the linear independence of a ''family'' of vectors, and the linear independence of a ''set'' of vectors.
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| * An [[indexed family]] of [[vector space|vector]]s is a '''linearly independent family''' if none of them can be written as a [[linear combination]] of finitely many other vectors in the family. A family of vectors which is not linearly independent is called '''linearly dependent'''.
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| * A [[set]] of vectors is a '''linearly independent set''' if the set (regarded as a family indexed by itself) is a linearly independent family.
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| These two notions aren't equivalent: the difference being that in a family, we allow repeated elements, while in a set we don't. For example if <math>V</math> is a vector space, then the family <math>F : \{1,2\} \to V</math> such that <math>f(1) = v</math> and <math>f(2) = v</math> is a ''linearly dependent family'', but the singleton set of the images of that family is <math>\{v\}</math> which is a ''linearly independent set''.
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| Both notions are important and used in common, and sometimes even confused in the literature.
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| For instance, in the three-dimensional [[real vector space]] <math>\mathbb{R}^3</math> we have the following example:
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| :<math>
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| \begin{matrix}
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| \mbox{independent}\qquad\\
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| \underbrace{
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| \overbrace{
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| \begin{bmatrix}0\\0\\1\end{bmatrix},
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| \begin{bmatrix}0\\2\\-2\end{bmatrix},
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| \begin{bmatrix}1\\-2\\1\end{bmatrix}
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| },
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| \begin{bmatrix}4\\2\\3\end{bmatrix}
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| }\\
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| \mbox{dependent}\\
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| \end{matrix}
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| </math><!-- weights 9, 5, 4 -->
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| Here the first three vectors are linearly independent; but the fourth vector equals 9 times the first plus 5 times the second plus 4 times the third, so the four vectors together are linearly dependent. Linear dependence is a property of the family, not of any particular vector; for example in this case we could just as well write the first vector as a linear combination of the last three.
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| :<math>\bold{v}_1 = \left(-\frac{5}{9}\right) \bold{v}_2 + \left(-\frac{4}{9}\right) \bold{v}_3 + \frac{1}{9} \bold{v}_4 . </math>
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| In [[probability theory]] and [[statistics]] there is an unrelated measure of linear dependence between [[random variable]]s.
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| == Definition ==
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| A [[subset]] '''S''' of a [[vector space]] '''V''' is called''''' linearly dependent''''' if there exist a '''finite''' number of '''distinct''' vectors ''v'' <sub>1</sub>, ''v'' <sub>2</sub>,..., ''v<sub>n</sub>'' in S and scalars''''' '''''<nowiki/>''a<sub>1</sub>, a<sub>2</sub>,..., a<sub>n</sub>'', not all zero, such that
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| :<math> a_1 \mathbf{v}_1 + a_2 \mathbf{v}_2 + \cdots + a_n \mathbf{v}_n = \mathbf{0}. </math>
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| Note that the zero on the right is the [[Null vector (vector space)|zero vector]], not the number zero.
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| For any vectors ''u'' <sub>1</sub>, ''u'' <sub>2</sub>,..., ''u''<sub>n</sub> we have that
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| :<math> 0 \mathbf{u}_1 + 0 \mathbf{u}_2 + \cdots + 0 \mathbf{u}_n = \mathbf{0}, </math>
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| This is called the [[Triviality (mathematics)|trivial]] representation of ''0'' as a linear combination of ''u'' <sub>1</sub>, ''u'' <sub>2</sub>,..., ''u''<sub>n</sub>, this motivates a very simple definition of both linear independence and linear dependence, for a set to be linearly dependent, there must exist a non-trivial representation of ''0'' as a linear combination of vectors in the set.
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| A subset '''S''' of a vector space''' V''' is then said to be '''''linearly independent''''' if it is not linearly dependent, in other words, a set is linearly independent if the only representation of ''0'' as a linear combination its vectors are trivial representations.<ref>{{cite book|last=Friedberg, Insel, Spence|first=Stephen, Arnold, Lawrence|title=Linear Algebra|publisher=Pearson, 4th Edition|isbn=0130084514|pages=48-49}}</ref>
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| Note that in both definitions we also say that the vectors in the subset '''S''' are linearly dependent or linearly independent.
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| More generally, let '''V''' be a vector space over a [[field (mathematics)|field]] '''K''', and let {'''v'''<sub>''i''</sub> | ''i''∈''I''} be a [[indexed family|family]] of elements of '''V'''. The family is ''linearly dependent'' over '''''K''''' if there exists a family {''a''<sub>''j''</sub> | ''j''∈''J''} of elements of''' ''K''''', not all zero, such that
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| :<math> \sum_{j \in J} a_j \mathbf{v}_j = \mathbf{0} \,</math>
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| where the index set ''J'' is a nonempty, finite subset of ''I''.
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| A set '''''X''''' of elements of '''''V''''' is ''linearly independent'' if the corresponding family {'''x'''}<sub>'''x'''∈''X''</sub> is linearly independent.
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| Equivalently, a family is dependent if a member is in the [[linear span]] of the rest of the family, i.e., a member is a [[linear combination]] of the rest of the family.
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| The trivial case of the empty family must be regarded as linearly independent for theorems to apply.
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| A set of vectors which is linearly independent and [[linear span|spans]] some vector space, forms a [[basis (linear algebra)|basis]] for that vector space. For example, the vector space of all polynomials in ''x'' over the reals has the (infinite) subset {1, ''x'', ''x''<sup>2</sup>, ...} as a basis.
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| == Geometric meaning ==
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| A geographic example may help to clarify the concept of linear independence. 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." Although this last statement is ''true'', it is not necessary.
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| 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.
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| Also note that if altitude is not ignored, it becomes necessary to add a third vector to the linearly independent set. In general, ''n'' linearly independent vectors are required to describe any location in ''n''-dimensional space.
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| == Example I ==
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| The vectors (1, 1) and (−3, 2) in <math>\mathbb{R}^2</math> are linearly independent.
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| === Proof ===
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| Let λ<sub>1</sub> and λ<sub>2</sub> be two [[real number]]s such that
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| :<math> (1, 1) \lambda_1 + (-3, 2) \lambda_2 = (0, 0) . \,\! </math>
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| Taking each coordinate alone, this means
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| :<math> \begin{align}
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| \lambda_1 - 3 \lambda_2 &{}= 0 , \\
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| \lambda_1 + 2 \lambda_2 &{}= 0 .
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| \end{align} </math>
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| Solving for λ<sub>1</sub> and λ<sub>2</sub>, we find that λ<sub>1</sub> = 0 and λ<sub>2</sub> = 0.
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| === Alternative method using determinants ===
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| An alternative method uses the fact that ''n'' vectors in <math>\mathbb{R}^n</math> are linearly '''independent''' [[if and only if]] the [[determinant]] of the [[matrix (mathematics)|matrix]] formed by taking the vectors as its columns is non-zero.
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| In this case, the matrix formed by the vectors is
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| :<math>A = \begin{bmatrix}1&-3\\1&2\end{bmatrix} . \,\!</math>
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| We may write a linear combination of the columns as
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| :<math> A \Lambda = \begin{bmatrix}1&-3\\1&2\end{bmatrix} \begin{bmatrix}\lambda_1 \\ \lambda_2 \end{bmatrix} . \,\!</math>
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| We are interested in whether ''A''Λ = '''0''' for some nonzero vector Λ. This depends on the determinant of ''A'', which is
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| :<math> \det A = 1\cdot2 - 1\cdot(-3) = 5 \ne 0 . \,\!</math>
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| Since the [[determinant]] is non-zero, the vectors (1, 1) and (−3, 2) are linearly independent.
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| Otherwise, suppose we have ''m'' vectors of ''n'' coordinates, with ''m'' < ''n''. Then ''A'' is an ''n''×''m'' 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 〈''i''<sub>1</sub>,...,''i''<sub>''m''</sub>〉 is any list of ''m'' rows, then the equation must be true for those rows.
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| :<math> A_{{\lang i_1,\dots,i_m} \rang} \Lambda = \bold{0} . \,\!</math>
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| Furthermore, the reverse is true. That is, we can test whether the ''m'' vectors are linearly dependent by testing whether
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| :<math> \det A_{{\lang i_1,\dots,i_m} \rang} = 0 \,\!</math>
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| 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.
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| == Example II ==
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| Let ''V'' = '''R'''<sup>''n''</sup> and consider the following elements in ''V'':
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| :<math>\begin{matrix}
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| \mathbf{e}_1 & = & (1,0,0,\ldots,0) \\
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| \mathbf{e}_2 & = & (0,1,0,\ldots,0) \\
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| & \vdots \\
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| \mathbf{e}_n & = & (0,0,0,\ldots,1).\end{matrix}</math>
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| Then '''e'''<sub>1</sub>, '''e'''<sub>2</sub>, ..., '''e<sub>n</sub>''' are linearly independent.
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| === Proof ===
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| Suppose that ''a''<sub>1</sub>, ''a''<sub>2</sub>, ..., ''a<sub>n</sub>'' are elements of '''R''' such that
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| :<math> a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + \cdots + a_n \mathbf{e}_n = 0 . \,\!</math>
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| Since
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| :<math> a_1 \mathbf{e}_1 + a_2 \mathbf{e}_2 + \cdots + a_n \mathbf{e}_n = (a_1 ,a_2 ,\ldots, a_n) , \,\!</math>
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| then ''a<sub>i</sub>'' = 0 for all ''i'' in {1, ..., ''n''}.
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| == Example III ==
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| Let ''V'' be the [[vector space]] of all [[function (mathematics)|function]]s of a real variable ''t''. Then the functions ''e<sup>t</sup>'' and ''e''<sup>2''t''</sup> in ''V'' are linearly independent.
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| === Proof ===
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| Suppose ''a'' and ''b'' are two real numbers such that
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| :''ae<sup>t</sup>'' + ''be''<sup>2''t''</sup> = 0
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| for ''all'' values of ''t''. We need to show that ''a'' = 0 and ''b'' = 0. In order to do this, we divide through by ''e''<sup>''t''</sup> (which is never zero) and subtract to obtain
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| :''be<sup>t</sup>'' = −''a''.
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| In other words, the function ''be''<sup>''t''</sup> must be independent of ''t'', which only occurs when ''b'' = 0. It follows that ''a'' is also zero.
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| == Example IV ==
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| The following vectors in '''R'''<sup>4</sup> are linearly dependent.
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| :<math>
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| \begin{matrix}
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| \\
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| \begin{bmatrix}1\\4\\2\\-3\end{bmatrix},
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| \begin{bmatrix}7\\10\\-4\\-1\end{bmatrix} \mathrm{and}
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| \begin{bmatrix}-2\\1\\5\\-4\end{bmatrix}
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| \\
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| \\
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| \end{matrix}
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| </math>
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| === Proof ===
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| We need to find not-all-zero scalars <math>\lambda_1</math>, <math>\lambda_2</math> and <math>\lambda_3</math> such that
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| :<math>
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| \begin{matrix}
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| \\
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| \lambda_1 \begin{bmatrix}1\\4\\2\\-3\end{bmatrix}+
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| \lambda_2 \begin{bmatrix}7\\10\\-4\\-1\end{bmatrix}+
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| \lambda_3 \begin{bmatrix}-2\\1\\5\\-4\end{bmatrix}=
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| \begin{bmatrix}0\\0\\0\\0\end{bmatrix}.
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| \end{matrix}
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| </math>
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| Forming the [[simultaneous equation]]s:
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| :<math>
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| \begin{align}
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| \lambda_1& \;+ 7\lambda_2& &- 2\lambda_3& = 0\\
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| 4\lambda_1& \;+ 10\lambda_2& &+ \lambda_3& = 0\\
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| 2\lambda_1& \;- 4\lambda_2& &+ 5\lambda_3& = 0\\
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| -3\lambda_1& \;- \lambda_2& &- 4\lambda_3& = 0\\
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| \end{align}
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| </math>
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| we can solve (using, for example, [[Gaussian elimination]]) to obtain:
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| :<math>
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| \begin{align}
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| \lambda_1 &= -3 \lambda_3 /2 \\
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| \lambda_2 &= \lambda_3/2 \\
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| \end{align}
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| </math>
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| where <math>\lambda_3</math> can be chosen arbitrarily.
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| Since these are nontrivial results, the vectors are linearly dependent.
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| == Projective space of linear dependences ==
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| A '''linear dependence''' among vectors '''v'''<sub>1</sub>, ..., '''v'''<sub>''n''</sub> is a [[tuple]] (''a''<sub>1</sub>, ..., ''a''<sub>''n''</sub>) with ''n'' [[scalar (mathematics)|scalar]] components, not all zero, such that
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| :<math>a_1 \mathbf{v}_1 + \cdots + a_n \mathbf{v}_n=0. \,</math>
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| If such a linear dependence exists, then the ''n'' vectors are linearly dependent. It makes sense to identify two linear dependences if one arises as a non-zero multiple of the other, because in this case the two describe the same linear relationship among the vectors. Under this identification, the set of all linear dependences among '''v'''<sub>1</sub>, ...., '''v'''<sub>''n''</sub> is a [[projective space]].
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| == Linear dependence between random variables ==
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| The [[covariance]] is sometimes called a measure of "linear dependence" between two [[random variable]]s. That does not mean the same thing as in the context of [[linear algebra]]. When the covariance is normalized, one obtains the [[correlation matrix]]. From it, one can obtain the [[Pearson coefficient]], which gives us the goodness of the fit for the best possible [[linear function]] describing the relation between the variables. In this sense covariance is a linear gauge of dependence.
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| == See also ==
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| * [[Gramian matrix]]
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| * [[Matroid]]
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| * [[Orthogonality]]
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| * [[Wronskian]]
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| == References ==
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| {{reflist}}
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| == External links ==
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| * {{springer|title=Linear independence|id=p/l059290}}
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| * [http://mathworld.wolfram.com/LinearlyDependentFunctions.html Linearly Dependent Functions] at WolframMathWorld.
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| * [http://people.revoledu.com/kardi/tutorial/LinearAlgebra/LinearlyIndependent.html Tutorial and interactive program] on Linear Independence.
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| * [http://www.khanacademy.org/video/linear-algebra--introduction-to-linear-independence Introduction to Linear Independence] at KhanAcademy.
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| {{linear algebra}}
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| {{DEFAULTSORT:Linear Independence}}
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| [[Category:Abstract algebra]]
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| [[Category:Linear algebra]]
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| [[Category:Articles containing proofs]]
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