UNIT21

# UNIT21 - Unit 21 Subspaces of n Denition 23.1. A subspace...

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Unit 21 Subspaces of < n Defnition 23.1. A subspace of < n is a nonempty set , S , of vectors from < n such that both of the following conditions are met: 1. if ~v 1 and 2 are in S then ( 1 + 2 ) S (closed under addition) 2. if 1 S and c is a scalar then c~v 1 S (closed under scalar multiplica- tion) Example 1. Istheset S of all vectors in < 4 with Frst and last components being 0 a subspace of < 4 ? Solution: Let 1 and 2 be in S ,say 1 =(0 ,a,b, 0) 2 ,c,d, 0) This is an extremely important step-choosing two arbitrary vectors from the set. We must use the deFnition of S in order to decide what form the above vectors must have. Having done this, we must show the two closure conditions are met: 1. Closure under addition: Show 1 + 2 is in S 1 + 2 =( 0 0) + (0 0) 0 ,a + c, b + d, 0) both Frst and last components are 0 and therefore 1 + 2 is in S ,and since 1 and 2 were arbitrary we have S closed under addition. 1

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2. Closure under scalar multiplication: Show c~v 1 is in S where c is a scalar 1 = c (0 ,a,b, 0) =( 0 ,ca,cb, 0) both frst and last components are 0 and thereFore 1 is in S .S ince ~v 1 was arbitrary we have S closed under scalar multiplication ThereFore, S is a subspace. Example 2. Let A be an m × n matrix and S the subset oF < n consisting oF all vectors ~x satisFying A~x = ~ 0. Show that S is a subspace oF < n Solution: Suppose ~x 1 and 2 are arbitrary vectors From S , so by the defni- tion oF S we have A~x 1 = ~ 0and 2 = ~ 0 We must show the two closure conditions are met: 1. Closure under addition: Show 1 + 2 is in S , i.e. A ( 1 + 2 )= ~ 0: A ( 1 + 2
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## This note was uploaded on 04/15/2010 for the course MATH 20C taught by Professor Lit during the Spring '10 term at UCLA.

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UNIT21 - Unit 21 Subspaces of n Denition 23.1. A subspace...

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