- Linear independence
In

linear algebra , a family of vectors is**linearly independent**if none of them can be written as alinear combination of finitely many other vectors in the collection. A family of vectors which is not linearly independent is called**linearly dependent**. For instance, in the three-dimensionalreal vector space **R**^{3}we have the following example. :$egin\{matrix\}mbox\{independent\}qquad\backslash underbrace\{\; overbrace\{\; egin\{bmatrix\}0\backslash 0\backslash 1end\{bmatrix\},\; egin\{bmatrix\}0\backslash 2\backslash -2end\{bmatrix\},\; egin\{bmatrix\}1\backslash -2\backslash 1end\{bmatrix\}\; \},\; egin\{bmatrix\}4\backslash 2\backslash 3end\{bmatrix\backslash mbox\{dependent\}\backslash end\{matrix\}$ 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; here we could just as well write the first vector as a linear combination of the last three.:$old\{v\}\_1\; =\; left(-frac\{5\}\{9\}\; ight)\; old\{v\}\_2\; +\; left(-frac\{4\}\{9\}\; ight)\; old\{v\}\_3\; +\; frac\{1\}\{9\}\; old\{v\}\_4\; .$In

probability theory andstatistics there is an unrelated measure of linear dependence betweenrandom variable s.**Formal definition**A subset "S" of a vector space "V" is called "linearly dependent" if there exists a finite number of distinct vectors

**v**_{1},**v**_{2}, ...,**v**_{"n"}in "S" and scalars "a"_{1}, "a"_{2}, ..., "a"_{"n"}, not all zero, such that:$a\_1\; mathbf\{v\}\_1\; +\; a\_2\; mathbf\{v\}\_2\; +\; cdots\; +\; a\_n\; mathbf\{v\}\_n\; =\; mathbf\{0\}.$

Note that the zero on the right is the zero vector, not the number zero.

If such scalars do not exist, then the vectors are said to be "linearly independent". This condition can be reformulated as follows: Whenever "a"

_{1}, "a"_{2}, ..., "a"_{"n"}are scalars such that:$a\_1\; mathbf\{v\}\_1\; +\; a\_2\; mathbf\{v\}\_2\; +\; cdots\; +\; a\_n\; mathbf\{v\}\_n\; =\; mathbf\{0\},$we have "a"_{"i"}= 0 for "i" = 1, 2, ..., "n", i.e. "only" the trivial solution exists.A set is linearly independent if and only if the only representations of the zero vector as linear combinations of its elements are trivial solutions.

More generally, let "V" be a vector space over a field "K", and let {

**v**_{"i"}}_{"i"∈"I"}be a family of elements of "V". The family is "linearly dependent" over "K" if there exists a family {"a"_{"j"}}_{"j"∈"J"}of elements of "K", not all zero, such that:$sum\_\{j\; in\; J\}\; a\_j\; mathbf\{v\}\_j\; =\; mathbf\{0\}\; ,$

where the index set "J" is a nonempty, finite subset of "I".

A set "X" of elements of "V" is "linearly independent" if the corresponding family {

**x**}_{x∈"X"}is linearly independent.Equivalently, a family is dependent if a member is in the

linear span of the rest of the family, i.e., a member is alinear combination of the rest of the family.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 "x" over the reals has the (infinite) basis {1, "x", "x"

^{2}, ...}.**Geometric meaning**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 5 miles north and 6 miles east of here." This is sufficient information to describe the location, because the geographic coordinate system may be considered a 2-dimensional vector space (ignoring altitude). The person might add, "The place is 7.81 miles northeast of here." Although this last statement is "true", it is not necessary.

In this example the "5 miles north" vector and the "6 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 "7.81 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.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.

**Example I**The vectors (1, 1) and (−3, 2) in

**R**^{2}are linearly independent.**Proof**Let λ

_{1}and λ_{2}be tworeal number s such that:$(1,\; 1)\; lambda\_1\; +\; (-3,\; 2)\; lambda\_2\; =\; (0,\; 0)\; .\; ,!$

Taking each coordinate alone, this means

:$egin\{align\}\; lambda\_1\; -\; 3\; lambda\_2\; \{\}=\; 0\; ,\; \backslash \; lambda\_1\; +\; 2\; lambda\_2\; \{\}=\; 0\; .\; end\{align\}$

Solving for λ

_{1}and λ_{2}, we find that λ_{1}= 0 and λ_{2}= 0.**Alternative method using determinants**An alternative method uses the fact that "n" vectors in

**R**^{"n"}are linearly dependentif and only if thedeterminant of the matrix formed by taking the vectors as its columns is zero.In this case, the matrix formed by the vectors is:$A\; =\; egin\{bmatrix\}1-3\backslash 12end\{bmatrix\}\; .\; ,!$We may write a linear combination of the columns as:$A\; Lambda\; =\; egin\{bmatrix\}1-3\backslash 12end\{bmatrix\}\; egin\{bmatrix\}lambda\_1\; \backslash \; lambda\_2\; end\{bmatrix\}\; .\; ,!$We are interested in whether "A"Λ =

**0**for some nonzero vector Λ. This depends on the determinant of "A", which is:$det\; A\; =\; 1cdot2\; -\; 1cdot(-3)\; =\; 5\; e\; 0\; .\; ,!$Since thedeterminant is non-zero, the vectors (1, 1) and (−3, 2) are linearly independent.When the number of vectors equals the dimension of the vectors, the matrix is square and hence the determinant is defined.

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"_{1},…,"i"_{"m"}⟩ is any list of "m" rows, then the equation must be true for those rows.:$A\_$lang i_1,dots,i_m} ang} Lambda = old{0} . ,!Furthermore, the reverse is true. That is, we can test whether the "m" vectors are linearly dependent by testing whether:$det\; A\_$lang i_1,dots,i_m} ang} = 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.**Example II**Let "V" =

**R**^{"n"}and consider the following elements in "V"::$egin\{matrix\}mathbf\{e\}\_1\; =\; (1,0,0,ldots,0)\; \backslash mathbf\{e\}\_2\; =\; (0,1,0,ldots,0)\; \backslash \; vdots\; \backslash mathbf\{e\}\_n\; =\; (0,0,0,ldots,1).end\{matrix\}$

Then

**e**_{1},**e**_{2}, ...,**e**are linearly independent._{n}**Proof**Suppose that "a"

_{1}, "a"_{2}, ..., "a_{n}" are elements of**R**such that:$a\_1\; mathbf\{e\}\_1\; +\; a\_2\; mathbf\{e\}\_2\; +\; cdots\; +\; a\_n\; mathbf\{e\}\_n\; =\; 0\; .\; ,!$

Since :$a\_1\; mathbf\{e\}\_1\; +\; a\_2\; mathbf\{e\}\_2\; +\; cdots\; +\; a\_n\; mathbf\{e\}\_n\; =\; (a\_1\; ,a\_2\; ,ldots,\; a\_n)\; ,\; ,!$

then "a

_{i}" = 0 for all "i" in {1, ..., "n"}.**Example III**Let "V" be the

vector space of all functions of a real variable "t". Then the functions "e^{t}" and "e"^{2"t"}in "V" are linearly independent.**Proof**Suppose "a" and "b" are two real numbers such that

:"ae

^{t}" + "be"^{2"t"}= 0for "all" values of "t". We need to show that "a" = 0 and "b" = 0. In order to do this, we divide through by "e"

^{"t"}(which is never zero) and subtract to obtain:"be^{t}" = −"a"In other words, the function "be"^{"t"}must be independent of "t", which only occurs when "b" = 0. It follows that "a" is also zero.**Example IV**The following vectors in

**R**^{4}are linearly dependent.:$egin\{matrix\}\backslash \; egin\{bmatrix\}1\backslash 4\backslash 2\backslash -3end\{bmatrix\},\; egin\{bmatrix\}7\backslash 10\backslash -4\backslash -1end\{bmatrix\}\; mathrm\{and\}\; egin\{bmatrix\}-2\backslash 1\backslash 5\backslash -4end\{bmatrix\}\backslash \backslash end\{matrix\}$**Proof**We need to find scalars $lambda\_1$, $lambda\_2$ and $lambda\_3$ such that

:$egin\{matrix\}\backslash lambda\_1\; egin\{bmatrix\}1\backslash 4\backslash 2\backslash -3end\{bmatrix\}+lambda\_2\; egin\{bmatrix\}7\backslash 10\backslash -4\backslash -1end\{bmatrix\}+lambda\_3\; egin\{bmatrix\}-2\backslash 1\backslash 5\backslash -4end\{bmatrix\}=\; egin\{bmatrix\}0\backslash 0\backslash 0\backslash 0end\{bmatrix\}.end\{matrix\}$

Forming the

simultaneous equation s::$egin\{align\}\; lambda\_1\; ;+\; 7lambda\_2\; -\; 2lambda\_3\; =\; 0\backslash \; 4lambda\_1\; ;+\; 10lambda\_2\; +\; lambda\_3\; =\; 0\backslash \; 2lambda\_1\; ;-\; 4lambda\_2\; +\; 5lambda\_3\; =\; 0\backslash -3lambda\_1\; ;-\; lambda\_2\; -\; 4lambda\_3\; =\; 0\backslash end\{align\}$

we can solve (using, for example,

Gaussian elimination ) to obtain::$egin\{align\}\; lambda\_1\; =\; -3\; lambda\_3\; /2\; \backslash \; lambda\_2\; =\; lambda\_3/2\; \backslash end\{align\}$where $lambda\_3$ can be chosen arbitrarily.Since these are nontrivial results, the vectors are linearly dependent.

**The projective space of linear dependences**A

**linear dependence**among vectors**v**_{1}, ...,**v**_{"n"}is a tuple ("a"_{1}, ..., "a"_{"n"}) with "n" scalar components, not all zero, such that:$a\_1\; mathbf\{v\}\_1\; +\; cdots\; +\; a\_n\; mathbf\{v\}\_n=0.\; ,$

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**_{1}, ....,**v**_{"n"}is aprojective space .**Linear dependence between random variables**The

covariance is sometimes called a measure of "linear dependence" between tworandom variable s. That does not mean the same thing as in the context oflinear algebra . When the covariance is normalized, one obtains thecorrelation matrix . From it, one can obtain thePearson coefficient , which gives us the goodness of the fit for the best possiblelinear function describing the relation between the variables. In this sense covariance is a linear gauge of dependence.**ee also***

orthogonality

*matroid (generalization of the concept)

* Wronskian

* Gram determinant**External links*** [

*http://video.google.com/videoplay?docid=-7254479149869222300 MIT Linear Algebra Lecture on Linear Independence*] at Google Video, from MIT OpenCourseWare* [

*http://mathworld.wolfram.com/LinearlyDependentFunctions.html Linearly Dependent Functions*] at WolframMathWorld

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