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No it is not. It's possible to have to have a set of vectors that are linearly dependent but still Span R^3. Same holds true for reverse. Linear Independence does not guarantee Span R^3. IF both conditions are met then that set of vectors is called the Basis for R^3. So, for a set of vectors, S, to be a Basis it must be:
(1) Linearly Independent
(2) Span S = R^3.
This means that both conditions are independent.

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1 1 Check whether the following set of vectors is LD or LI?

(i) They are linearly dependent since the 2nd vector is twice the 1st vector. All 3 vectors lie in the x-z plane, so they don't span 3D space. (ii) They are linearly independent. Note that the cross-product of the first two is (-1,1,1). If the third vector is not perpendicular to the above cross-product, then the third vector does not lie in the plane defined by the first two vectors. (-1,1,1) "dot" (1,1,-1) = -1+1-1 = -1, not zero, so 3rd vector is not perpendicular to the cross product of the other two.


What is the definition of domain used in math?

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Assume that the table shows a set of values for two variables that are linearly related. If these assumptions are false, you cannot do answer the question.Suppose the first set of values is for a variable, x, and the second set is for y.Select any two [ordered] pairs data: (x1, y1) and (x2, y2). Then the slope between these two points is (y2 - y1)/(x2 - x1) provided x2 ≠ x1. If x2 = x1 then the slope is undefined.


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A set of variables or equations are said to be independent if no one of them can be expressed in terms of the others. In statistics, a variable, X, is said to be independent of another variable, Y, if changes in Y do not cause changes in X. The reverse need not be true.

Related Questions

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1 1 Check whether the following set of vectors is LD or LI?

(i) They are linearly dependent since the 2nd vector is twice the 1st vector. All 3 vectors lie in the x-z plane, so they don't span 3D space. (ii) They are linearly independent. Note that the cross-product of the first two is (-1,1,1). If the third vector is not perpendicular to the above cross-product, then the third vector does not lie in the plane defined by the first two vectors. (-1,1,1) "dot" (1,1,-1) = -1+1-1 = -1, not zero, so 3rd vector is not perpendicular to the cross product of the other two.


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