Lecture7

# Lecture7 - 42 Lecture 7 Gram-Schmidt Orthogonalization The...

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Lecture 7 Gram-Schmidt Orthogonalization The ”best” basis we can have for a vector space is an orthogonal basis. That is because we can most easily ﬁnd the coeﬃcients that are needed to express a vector as a linear combination of the basis vectors v 1 ,...,v n : v = ( v,v 1 ) k v 1 k 2 v 1 + ... + ( v,v n ) k v n k 2 v n . But usually we are not given an orthogonal basis. In this section we will show how to ﬁnd an orthogonal basis starting from an arbitrary basis. 7.1 Procedure Let us start with two linear independent vectors v 1 and v 2 (i.e. not on the same line through zero). Let u 1 = v 1 . How can we ﬁnd a vector u 2 which is perpendicular to u 1 and that the span of u 1 and u 2 is the same as the span of v 1 and v 2 ? We try to ﬁnd a number a R such that: u 2 = au 1 + v 2 , u 2 u 1 Take the inner product with u 1 to get: 0 = ( u 2 ,u 1 ) = a ( u 1 ,u 1 ) + ( v 2 ,u 1 ) = a k u 1 k 2 + ( v 2 ,u 1 ) 43
44 LECTURE 7. GRAM-SCHMIDT ORTHOGONALIZATION or a = - ( v 2 ,u

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## This note was uploaded on 11/23/2011 for the course MATH 2025 taught by Professor Staff during the Spring '08 term at LSU.

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Lecture7 - 42 Lecture 7 Gram-Schmidt Orthogonalization The...

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