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	PlanetMath: linear independence
		In the case of two vectors, linear independence means that one of these vectors is not a scalar multiple of the other.
		As an alternate characterization of dependence, we have that a set of of vectors is linearly dependent if and only if some vector in the set lies in the linear span of the other vectors in the set.
		This is version 22 of linear independence, born on 2001-11-14, modified 2003-02-03.
	planetmath.org /encyclopedia/LinearlyIndependent.html (143 words)

	Linear independence - Wikipedia, the free encyclopedia
		In linear algebra, a family of vectors is linearly independent if none of them can be written as a linear combination of finitely many other vectors in the collection.
		, the three vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) are linearly independent, while (2, −1, 1), (1, 0, 1) and (3, −1, 2) are not (since the third vector is the sum of the first two).
		The concept of linear independence is important because a set of vectors which is linearly independent and spans some vector space, forms a basis for that vector space.
	en.wikipedia.org /wiki/Linearly_independent (488 words)

	[gov1000-list] Re: collinear and linearly independent (Site not responding. Last check: 2007-10-20)
		When a vector is linearly dependent, it can be expressed as a linear function of another vector or set of vectors.
		Two vectors are linearly independent if they do not overlap or radiate out in opposite directions from the origin, which only happens when they are multiples of one another.
		This means that y falls within the 3-D space generated by x1, x2, and x3 (I am assuming that x1,x2, and x3 are linearly independent, i.e., none of these is a linear function of the other).
	lists.fas.harvard.edu /pipermail/gov1000-list/2004-March/000875.html (455 words)

	5 (Site not responding. Last check: 2007-10-20)
		Linearly dependent iff at least one of the vectors in S is expressible as a linear combination of the other vectors in S.
		Linearly independent iff no vector in S is expressible as a linear combination of the other vectors in S.
		A set with exactly two vectors is linearly independent iff neither vector is a scalar multiple of the other.
	www.apsu.edu /vandergriffj/spring99/3450/503.html (158 words)

	Linearly Independent Sets of Vectors
		The span of two independent vectors is a plane containing the origin.
		This can be verified directly in individual cases; to show it in general requires methods of the next section.
		The importance of the independence is that none of the matrices can be written in terms of the others; so the study of electron spin by Pauli matrices cannot, in general, be conducted with a proper subset of
	distance-ed.math.tamu.edu /Math640/chapter3/node9.html (877 words)

	Lindemann–Weierstrass theorem - Wikipedia, the free encyclopedia
		are algebraic numbers which are linearly independent over the rational numbers, then
		are algebraic numbers, linearly independent over the rationals (and therefore necessarily distinct), then all the different monomial products
		The transcendence of e and π are direct corollaries of this theorem.
	en.wikipedia.org /wiki/Lindemann-Weierstrass_theorem (322 words)

	Linear Dependence of Vectors. Basis of the Vector Space.
		is called a maximal linearly independent subset if V is linearly independent and it is not a proper subset of any linearly independent subset of the set U.
		Hence V is not the maximal linearly independent subset.
		is a basis of the space iff it is the maximal linearly independent subset.
	www.cs.ut.ee /~toomas_l/linalg/lin1/node7.html (377 words)

	Linear independence : Linearly independent (Site not responding. Last check: 2007-10-20)
		In linear algebra, a set of elements of a vector space is called linearly independent if none of the vectors in the set can be written as a linear combination of finitely many other vectors in the set.
		, the three vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) are linearly independent, while (2, -1, 1), (1, 0, 1) and (3, -1, 2) are not (since the third vector is the sum of the first two).
		An infinite subset of V is said to linearly independent if all its finite subsets are.
	www.city-search.org /li/linearly-independent.html (710 words)

	Math Forum - Ask Dr. Math (Site not responding. Last check: 2007-10-20)
		Date: 01/24/2001 at 19:37:24 From: sanaz golban Subject: Prove set is linearly independent I have not been able to figure this one out.
		Date: 01/24/2001 at 19:48:12 From: Doctor Schwa Subject: Re: Prove set is linearly independent To prove something is linearly independent, you want to prove that if a * (first vector) + b * (second vector) = 0, then a and b are both zero.
		I saw some complicated words, "linearly independent." Not only that, but those words were used twice, so I wrote down the definition, and wrote down what the "if-then" part of the definition said, and then found that I was already almost done with the proof.
	mathforum.org /library/drmath/view/52449.html (344 words)

	sciforums.com - Linear Transformations
		Each vector in R is linearly independent and I claim that no matter what the transformation T is (so long as it is linear) and for any set of linearly independent vectors S, the resulting set R will always contain linearly independent vectors.
		We assume that the vectors v1, v2..., vn are linearly independent.
		suppose that the v's are not linearly independent in R^n.
	www.sciforums.com /showthread.php?t=29179 (2492 words)

	[No title]
		Since {x1,x2} are linearly independent, it follows that Span{x1,x2} and thus Span{x1,x2,x3} are both 2-dimensional by the definition on p.158.
		We separate the parameters: S = {a(1,1,0,0)+b(1,-1,1,0)+c(0,2,0,1): a,b,c in R} = Span{(1,1,0,0), (1,-1,1,0), (0,2,0,1)} The three vectors {(1,1,0,0), (1,-1,1,0), (0,2,0,1)} are linearly independent, since (0,0,0,0)=a(1,1,0,0)+b(1,-1,1,0)+c(0,2,0,1) implies b=c=0 from the last two coefficients and thus a=0 from the first coefficient.
		But {x^2+2, x+3} is a linearly independent set in P3 by Theorem 3.3.3, since the Wronskian is det/ x^2+2 x+3 \ = -2x^2-6x+2 != 0 at x=0, \ 2x 1 / so S has dimension 2 by the definition on p.158.
	www.math.wustl.edu /~victor/classes/ma309/s05.txt (737 words)

	Linear independence of error vectors (Site not responding. Last check: 2007-10-20)
		If we can find a member of the row space of H that produces a zero syndrome, then every position where that n-bit row space vector is a one is error-free, if the r-bit error vectors are linearly independent.
		If the error vectors are independent, the number of error locations and the number of zeroes usually will be exactly ρ.
		Take a ρ x ρ binary submatrix of H whose rows correspond to the positions of the discovered independent syndrome rows, and whose columns correspond to the error locations; invert this to solve for the error values in terms of the original syndrome values.
	www.cse.psu.edu /~cg554/Vector1.html (735 words)

	Basis and Dimension
		, it is linearly independent and the above equation implies that all the coefficients are zero.
		is not linearly independent then we can remove the redundant vectors until we arrive at a basis.
		is not linearly independent, then one of the vectors is a linear combination of the rest.
	www.ltcconline.net /greenl/courses/203/Vectors/basisDimension.htm (519 words)

	Exam #2
		Yes, they are linearly independent from the point of view of pure mathematics.
		We need to start with a set of linearly independent set of vectors to generate a set of orthogonal vectors (which are linearly independent).
		Form a nxm matrix of independent variables X, where n is the number of students and m is the number of independent variables.
	www.ee.umd.edu /~nsw/ench620/exam2-02.htm (1721 words)

	[No title]
		It is true that if S is linearly independent, then the number of elements in S is less than the dimension of V, but it is possible for S to have fewer elements than the dimension of V without S being linearly independent.
		May or may not be linearly independent, and may or may not span V. Answer.
		May or may not be linearly independent, and may or may not span V. Correct Answer.
	www.math.ucla.edu /~tao/java/MultipleChoice/vector_spaces.txt (1470 words)

	[No title]
		If instead you want to test independence for a proof of something (that is, if you need 100% accuracy) you'll need collections of particular 4-tuples of vectors which will collectively give an "iff" condition.
		The method I proposed makes it easy to demonstrate independence; to prove dependence is also in a probabilistic sense easy until the number of independent vectors is nearly the dimension of possible vectors.
		Indeed, if you have N vectors which you have already proven to be independent, and you evaluate N+1 vectors at N+1 points, then two outcomes can occur: either you get linearly independent function values, which proves the N+1 functions are still independent, or else you get linearly dependent function values.
	www.math.niu.edu /~rusin/known-math/95/polynom (6731 words)

	Physics Help and Math Help - Physics Forums - subset linearly independent?
		If the question is about linearly dependant sets of vectors then obviously any non-zero vector by itself is a linearly independant subset - so it's false.
		A subset of linearly dependent set is linearly dependent.
		The reason I posted it as 'IN'dependent is because I think it is independent, but not sure, hence, question mark.
	www.physicsforums.com /printthread.php?t=59442 (878 words)

	Quiz 6 key
		This will constitute a basis for W since it will be formed by linearly independent vectors that span W.
		Furthermore these two vectors are linearly independent hence they form a basis for the subspace they span.
		, thus the vectors are linearly independent and constitute a basis for
	math.la.asu.edu /~tracogna/MAT342/q6 (188 words)

	Bases of Vector Spaces, the Basis Problem
		is linearly dependent, then one of the vectors can be written as a linear combination of the others.
		Therefore, the vectors are linearly independent and form a basis
		The remaining rows are therefore all linear combinations of the independent vectors from the original set.
	distance-ed.math.tamu.edu /Math640/chapter3/node11.html (1642 words)

	On Almost Interpolation and Locally Linearly Independent Bases - Davydov, Sommer, Strauss (ResearchIndex) (Site not responding. Last check: 2007-10-20)
		Moreover, we show that this characterization can be significantly simplified in the case of existence of a locally linearly independent basis, so that almost interpolation sets can be constructed by taking a point in a support of each basis function.
		Some further results, including several equivalent definitions of a locally linearly independent system of functions, are...
		Locally linearly independent bases play an important role in the theory of interpolation and almost interpolation by multivariate...
	citeseer.ist.psu.edu /200387.html (566 words)

	The Ch 3 Equation
		If two vectors are linearly dependent then one is a scalar multiple of the other.
		If three vectors are linearly dependent then one is a scalar multiple of the other.
		The set of vectors, {2x-1 and 4-x and 3}, is linearly independent.
	www.fiu.edu /~hudsons/la/lahelp/maineqn.htm (564 words)

	Vector Spaces, Bases, and Dimension
		Thus the condition of linear independence is related to another condition, that of expressing other vectors as a linear combination of the given vectors.
		Linear independence, spanning, and basis are defined solely in terms of such finite linear combinations.
		Then any linearly independent subset of V can be expanded to a basis, and any subset that spans V can be contracted to a basis.
	www.uwm.edu /~adbell/Teaching/631/1999/631notes7L/node1.html (3461 words)

	Linear Algebra: The Basics (Site not responding. Last check: 2007-10-20)
		The goal of these concepts will be to provide a background for signal decomposition and to lead up to the derivation of the Fourier Series.
		Looking at these two vectors geometrically (as in figure 1), one can again prove that these vectors are not linearly independent.
		Based on the definition, this proof shows that these vectors are indeed linearly independent.
	cnx.rice.edu /content/m10734/latest (675 words)

	[No title]
		A sufficient condition for nonsingularity of a square matrix is that the rows of the matrix be linearly independent.
		B = These examples serve to illustrate that the condition of linear independence cannot necessarily be ascertained at a quick glance.
		Conversely, supposing r is the maximum number of linearly independent rows which can be found, then the matrix is said to be of rank r.
	www.lancs.ac.uk /people/ecajj/213l2.doc (3231 words)

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