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Unformatted text preview: MA 35100 LECTURE NOTES: MONDAY, MARCH 1 Characterizations of Linear Independence We summarize all that we have discussed concerning the linear independence of a set of vectors. Theorem. Consider a list of vectors ~v 1 ,~v 2 ,...,~v m ∈ R n . Denote their span V = span( ~v 1 ,~v 2 ,...,~v m ) as a subspace of R n , and A = ~v 1 ~v 2 ··· ~v m as an n × m matrix. The following are equivalent: i. The vectors ~v 1 ,~v 2 ,...,~v m are linearly independent. ii. None of the vectors ~v 1 ,~v 2 ,...,~v m is redundant i.e., none of the vectors ~v i is a linear combination of the other vectors ~v 1 ,...,~v i 1 ,~v i +1 ,...,~v m . iii. The vectors ~v 1 ,~v 2 ,...,~v m form a basis for V . iv. The only relation c 1 ~v 1 + c 2 ~v 2 + ··· + c m ~v m = ~ 0 is the trivial relation i.e., c 1 = c 2 = ··· = c m = 0. v. ker( A ) = { ~ } . vi. rank( A ) = m ≤ n . We sketch why this is true: • (i) ⇐⇒ (ii): This follows from Definition 3.2.3 on pg. 116. • (i) ⇐⇒ (iii): This also follows from Definition 3.2.3 on pg. 116. • (i) ⇐⇒ (iv): This follows from the previous fact (Theorem 3.2.7 on pg. 118). • (iv) ⇐⇒ (v): This also follows from the previous fact (Theorem 3.2.8 on pg. 119). • (v) ⇐⇒ (vi): This follows from Theorem 3.1.7 on pg. 109. Dimension of a Subspace of R n Let V denote a subspace of R n . As previously mentioned, if there is a basis for V , this basis is not unique. Indeed, say that V is a line, spanned by a vector ~v ; then ~v is a basis for V . Then for any nonzero scalar k , the vector ~w = k~v also spans V ; then ~w is also a basis for V . Which the elements in a basis are not unique, the number of elements in a basis are....
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This note was uploaded on 10/18/2010 for the course MATH 351 taught by Professor Egoins during the Spring '10 term at Purdue UniversityWest Lafayette.
 Spring '10
 EGoins
 Linear Algebra, Algebra, Vectors, Linear Independence

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