DOC PREVIEW
Berkeley MATH 54 - Spans, Linear Independence and Bases

This preview shows page 1 out of 3 pages.

Save
View full document
View full document
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience
Premium Document
Do you want full access? Go Premium and unlock all 3 pages.
Access to all documents
Download any document
Ad free experience

Unformatted text preview:

Spans, Linear Independence and BasesSeptember 17, 2007Definition 1 Let (v·) := (v1, . . . , vm) be a sequence of vectors in Rn. Thena vector v ∈ Rnis said to b e a linear combination of v·if there exists asequence of numbers (c·) := (c1, . . . , cm) such thatv = c1v1+ c2v2+ · · · cmvm.The set of all such v is called the span of (v·).Note that if A is the n × m matrix whose columns are the vectors vi, thenv is in the span of (v·) if and only if there exists a column vector X in Rmsuch that v = AX, i.e., if and only if v is in the image of TA. Thus the spanof (v·) is always a linear subspace of Rn.The span of the empty sequence is defined to be the set consisting of justthe zero vector.Definition 2 Let (v·) := (v1, . . . , vm) be a sequence of vectors. Then for1 ≤ i ≤ m, the ith vector viis said to be redundant in (v·) if it belongs tothe span of the sequence (v1, . . . , vi−1). A sequence (v·) is said to be linearlydependent if it contains a redundant vector; otherwise it is said to be linearlyindependent.Lemma 3 If (v1, . . . vm) is a sequence of vectors, then for each i, the follow-ing are equivalent.1. viis redundant.2. span(v1, . . . , vi−1) = span(v1, . . . , vi).1It follows from the lemma that we can omit all the redundant vectors from asequence (v·) without changing the span. It is pretty clear that the sequencewe obtain in this way has no redundant vectors, i.e., is linearly independent.Proposition 4 Let (v·) := (v1, . . . , vm) be a sequence of vectors. Then thefollowing conditions are equivalent:1. (v·) is linearly independent.2. If (c·) := (c1, . . . cm) is a sequence of numbers such that c1v1+· · · vmvm=0, each ci= 0.3. If (c·) and (c0·) are two sequences of numbers such that c1v1+· · · cmvm=c01+ · · · c0mvm= 0, then ci= c0ifor all i.This can be translated into matrix terms for computational purposes.Arrange the vectors viinto the columns of an n × m matrix A. Consider thecolumn vector X whose coordinates are the ciappearing in the propositionabove and recall thatAX = x1C1(A) + x2C2(A) + · · · xmCm(A) = c1v1+ · · · cmvm.Then condition (2) says that the kernel of TAis {0}, and condition (3) saysthat TAis injective.Corollary 5 Let A be an n × m matrix. Then the following conditions areequivalent.1. The columns of A are linearly independent.2. The kernel of TA(the nullspace of A) is {0}.3. TAis injective.4. The rank of A is m.Recall that the rank of a matrix is always less than or equal to the the lengthof its columns.Corollary 6 If v1, . . . vmis an independent sequence of vectors in Rn, thenm ≤ n.2Definition 7 Let W be a linear subspace of Rn. A sequence (w·) of vectorsin W is a basis for W if it is linearly independent and spans W .Thus if (w·) is a basis f or W , ever vector w in W can be expressed uniquelyas a linear combination c1w1+ · · · +


View Full Document

Berkeley MATH 54 - Spans, Linear Independence and Bases

Download Spans, Linear Independence and Bases
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Spans, Linear Independence and Bases and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Spans, Linear Independence and Bases 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?