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Unformatted text preview: Linear Algebra &amp; Geometry: Solutions to sheet 5 1. We have to show that if k X i =1 i v i = 0 then 1 = 2 = = k = 0. Let us take the dot product of the sum with v j , then since v j v i = 0 if i 6 = j we find v j k X i =1 i v i = k X i =1 i v j v i = i v j v j , but v j 6 = 0 and hence we must have j = 0 for j = 1 , 2 , ,k . 2. (a) Since ij = 0 if i 6 = j there is only one term in the sum n i =1 b i ij which is not 0, namely the one with i = j , and this term is b j jj = b j . By the same argument in the sum n j =1 b j ij only the term with j = i is non zero, and this term is b i ii = b i . (b) Let us treat first the case that i = k , then we have to show that n j =1 ij ji = 1. But since ij ij = ij we can apply the result from part (a) with b i = 1. Now for the case i 6 = k we have ij jk = 0 for all j , because by the first factor ij the product could only be nonzero for j = i , but since...
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This note was uploaded on 02/25/2010 for the course MATHEMATIC 11007 taught by Professor Spyros during the Spring '10 term at Bristol Community College.
 Spring '10
 Spyros
 Linear Algebra, Algebra, Geometry, Dot Product

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