253Solutions8

Linear Algebra with Applications (3rd Edition)

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ma253 ,Fa l l 2007 — Problem Set 8 Solutions 1. Problems from the textbook: a. Section 3 . 3 43. The point is that if we take a basis { ~ v 1 , ~ v 2 ,..., ~ v r } of V and a basis { ~ w 1 , ~ w 2 ~ w s } of W ,then { ~ v 1 , ~ v 2 ~ v r , ~ w 1 , ~ w 2 ~ w s } will be a basis of V + W . To see this, notice Frst that it clearly spans V + W ,sothe hard part is to show it is independent. ±or that, suppose c 1 ~ v 1 + ··· + c r ~ v r + b 1 ~ w 1 + + b s ~ w s = ~ 0. Then, we get c 1 ~ v 1 + + c r ~ v r =-( b 1 ~ w 1 + + b s ~ w s ) . In this equation, the expression on the left is an element of V and the expression on the right is an element of W . Since they are equal, both expressions must give (the same) element of V W .But V W = { ~ 0 } ,soinfactweget c 1 ~ v 1 + + c r ~ v r = ~ 0 and b 1 ~ w 1 + + b s ~ w s = ~ 0. Since the original sets were independent, it follows that all the c i and all the b j are zero, so the big set is still independent. The upshot is that if V W = { ~ 0 } ,wehave dim ( V + W )= dim ( V )+ dim ( W ) . 44. Since V W = { ~ 0 } , we can conclude from 43 that V + W has dimension n . Since V + W R n , it follows that V + W = R n . Hence every vector ~ x R n can be written as ~ v + ~ w as desired. To see that this representation is unique, suppose ~ v + ~ w = ~ v 1 + ~ w 1 . Then ~ v - ~ v 1 = ~ w 1 - ~ w, and the left hand side is in V while the right hand side is in W .Nowdothesame as in problem 43. You don’t have to agree with me, but it’s quicker.
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47. Hard. Basically, follow the hint, and be very careful to think clearly. *62. I guess the most direct way to see this is to let ~ v 1 , ~ v 2 ,..., ~ v m be the columns of A and ~ w 1 , ~ w 2 ~ w m be the columns of B .Then Im ( A )= Span ( ~ v 1 , ~ v 2 ~ v m ) , Im ( B Span ( ~ w 1 , ~ w 2 ~ w m ) , Im ( A + B Span ( ~ v 1 + ~ w 1 , ~ v 2 + ~ w 2 ~ v m + ~ w m ) . Now notice that Span ( ~ v 1 + ~ w 1 , ~ v 2 + ~ w 2 ~ v m + ~ w m ) Span ( ~ v 1 , ~ v 2 ~ v m , ~ w 1 , ~ w 2 ~ w m ) . The dimension of the last span is no bigger than dim ( Im ( A )) + dim ( Im ( B )) ,and the rank of A + B is the dimension of the Frst span. Therefore, we get rank ( A + B ) rank ( A )+ rank ( B ) . (We could use problem 47 to get a better estimate.) *63. ±or any ~ v , AB ~ v = A ( B ~ v ) ,so Im ( AB ) Im ( A ) rank ( AB ) rank ( A ) . On the other hand, the image of AB is the image of Im ( B ) under the transforma- tion given by A . Since the image of a transformation has dimension less than or equal to the dimension of the domain, we get, once again, rank ( AB ) rank B . *64. What we need to do to show two sets are different is to Fnd an element of one that is not an element of the other. Looking at matrix A we see that 4 times the Frst column plus 5 times the second plus 6 times the fourth gives the Ffth. This tells us that the vector ~ x = 4 5 0 6 - 1 0 is in the kernel of A .But B ~ x = 0 0 - 1 0 6 = ~ 0, so ~ x is not in the kernel of B . That shows the kernels are different.
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253Solutions8 - ma253, Fall 2007 - Problem Set 8 Solutions...

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