253Solutions5

Linear Algebra with Applications (3rd Edition)

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ma253 ,Fa l l 2007 — Problem Set 5 Solutions 1. Problems from the textbook: a. Section 3 . 1 , problems * 18 ,* 22 38 48 . *18. Row-reducing A gives rref ( A )= ± 14 00 ² so the image is spanned by the Frst row, which is the only one that has a pivot. So Im ( A Span ³± 1 3 ²´ . (It’s fairly easy to see this directly, of course.) This is a line in R 2 . *22. Row-reducing A gives rref ( A 10 2 01 - 1 00 0 so the pivots are in the Frst two columns. Hence, Im ( A Span 2 3 6 , 1 4 5 , which is a plane through the origin in R 3 . *38. We discussed part (a) in class, and I talked about it with lots of you too. If A ~ x = ~ 0 , then certainly A 2 ~ x = A ( A ~ x ~ 0 .So ker ( A ) ker ( A 2 ) .The rei sno reason to expect they’d be equal. ±or a dramatic example, suppose A = ± 10 ² . Then A 2 = 0 ,so ker ( A Span ³± 0 1 ²´ while ker ( A 2 R 2 . So, in general, ker ( A ) ker ( A 2 ) ker ( A 3 ) ... with equality being possible but not required. To speak algebraically, Mr. M. is execrable, but Mr. G. is ( x + 1 ) -ecrable. – Edgar Allen Poe
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Part (b) is similar, except that the containments are reversed: if ~ y Im ( A ) ,then ~ y = A ~ x for some ~ x ;if ~ y Im ( A 2 ) ~ y = A 2 ~ w = A ( A ~ w )= A ~ x for ~ x = A ~ w . So Im ( A 2 ) Im ( A ) . In general, Im ( A ) Im ( A 2 ) Im ( A 3 ) ... and once again equality is possible but not required. *48. This is basically the same as one of the problems in the midterm. Part (a) wasexact lythesameasoneo ftheprob lemsinthem idterm :i f ~ w Im ( A ) A ~ w = ~ w . See the midterm solutions for details. For part (b), note ±rst that the only matrix with rank zero is the zero matrix. If, on the other hand, A has rank two, it is invertible (since it’s a 2 × 2 matrix). Multiplying A 2 = A by A - 1 on both sides gives A = I .Sotheon lyranktwomatr ix that satis±es A 2 = A is the identity. (Another way to see this: if A has rank two, then the image of A is all of R 2 . By part (a), it follows that A ~ w = ~ w for every ~ w , so A is the identity.) Finally, suppose A has rank one. Then both the kernel and the image of A are spanned by one vector, i.e., they are lines in R 2 .G i v e n ~ x R 2 ,le t ~ w = A ~ x , ~ v = ~ x - A ~ x .Then ~ w Im ( A ) , ~ v ker ( A ) , ~ x = ~ w + ~ v ,and A ~ x = ~ w . This is what problem 2.2.33 de±nes as “the projection on the image along the kernel.” b. Section 3 . 2 , problems 1 ,* 2 , 3 4 6 , 8 , 10 , 12 14 16 20 32 34 , * 36 37 . 1. Not a subspace, since 0 0 0 is not in W . *2. Not a subspace. The easiest way to see this is to note that if x<y<z then x y z W but (- 1 ) x y z / W . (Multiplying by - 1 reverses inequalities!) 3. In this case, W = Span 1 4 7 , 2 5 8 , 3 6 9 , so it is a subspace. *4. Yes, as we checked in class. If you don’t see how to do this, come ask. *6. The intersection is always a subspace, but the union hardly ever is. Few men speak humbly of humility, chastely of chastity, skeptically of skepticism. – Blaise Pascal
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For the frst, just test the three conditions and use the ±act that both V and W are subspaces: Since ~ 0 V and ~ 0 W , ~ 0 V W .
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253Solutions5 - ma253, Fall 2007 - Problem Set 5 Solutions...

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