253Solutions5

# Linear Algebra with Applications (3rd Edition)

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ma253 , Fall 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 ) = 1 4 0 0 so the image is spanned by the first row, which is the only one that has a pivot. So Im ( A ) = Span 1 3 . (It’s fairly easy to see this directly, of course.) This is a line in R 2 . *22. Row-reducing A gives rref ( A ) = 1 0 2 0 1 - 1 0 0 0 so the pivots are in the first 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 Ax = 0 , then certainly A 2 x = A ( Ax ) = 0 . So ker ( A ) ker ( A 2 ) . There is no reason to expect they’d be equal. For a dramatic example, suppose A = 1 0 0 0 . Then A 2 = 0 , so ker ( A ) = Span 0 1 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 = Ax for some x ; if y Im ( A 2 ) , then y = A 2 w = A ( Aw ) = Ax for x = Aw . 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) was exactly the same as one of the problems in the midterm: if w Im ( A ) , then Aw = w . See the midterm solutions for details. For part (b), note first 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 . So the only rank two matrix that satisfies 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 Aw = 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 . Given x R 2 , let w = Ax , v = x - Ax . Then w Im ( A ) , v ker ( A ) , x = w + v , and Ax = w . This is what problem 2.2.33 defines 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
For the first, just test the three conditions and use the fact that both V and W are subspaces: Since 0 V and 0 W , 0 V W .

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