Linear Algebra with Applications (3rd Edition)

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ma253 , Fall 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 first that it clearly spans V + W , so the hard part is to show it is independent. For 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 } , so in fact we get 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 } , we have 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 . Now do the same 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 first span. Therefore, we get rank ( A + B ) rank ( A ) + rank ( B ) . (We could use problem 47 to get a better estimate.) *63. For any v , ABv = A ( Bv ) , so Im ( AB ) Im ( A ) , so 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 find an element of one that is not an element of the other. Looking at matrix A we see that 4 times the first column plus 5 times the second plus 6 times the fourth gives the fifth. This tells us that the vector x = 4 5 0 6 - 1 0 is in the kernel of A . But Bx = 0 0 - 1 0 = 0, so x is not in the kernel of B . That shows the kernels are different.
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