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Unformatted text preview: Review Problems for Exam Three THEORY 1. Use the Law of Cosines to prove that x T y =  x  y  cos ( ) for x,y R 3 . By the Law of Cosines, k x y k 2 = k x k 2 + k y k 2 2 k x kk y k cos ( ), where is the angle between x and y . By definition, k x y k 2 = h x y,x y i , which equals h x,x i + h y,y i  2 h x,y i by properties of inner products. Thus 2 h x,y i = 2 k x kk y k cos ( ), so that x T y = h x,y i = k x kk y k cos ( ). 2. Let A be an m n matrix. Identify Rowspace ( A ) , Colspace ( A ) and N ( A ) in terms of the row, column and null spaces of A and/or A t . rs ( A ) = ns ( A ), cs ( A ) = ns ( A T ) and ns ( A ) = rs ( A ) 3. Prove that Rowspace ( A ) = Nullspace ( A ). Let A have rows A T 1 ,...,A T m . Suppose first that x rs ( A ) , so that y T x = 0 for all y rs ( A ). Then in particular A T i x = 0 for each i . It follows that Ax = A T 1 . . . A T m x = 0 , so that x ns ( A ). Suppose next that x ns ( A ), so that Ax = 0, so that A i x = 0 for each i . Then for any y = c 1 A 1 + + c m A m rs ( A ), y T x = ( c 1 A 1 + ... + c m A m ) T x = c 1 A T 1 x + + c m A T m x = c 1 0 + + c m 0 = 0, so that x rs ( A ) . 4. (a) Prove that N ( A ) = N ( A T A ) for any m n matrix A . (b) Show that rk ( A T A ) = rk ( A T ). (c) Show that A T A x = A T b has a solution for any b R m . (a) Suppose first that x ns ( A ), so that Ax = 0. Then ( A T A ) x = A T ( Ax ) = A T 0 = 0, so that x ns ( A T A ). Suppose next that x ns ( A T A ), so that A T Ax = 0. Then Ax ns ( A T ). Also, Ax cs ( A ). But ns ( A T ) = rs ( A T ) = cs ( A ), so that Ax ns ( A T ) . Thus k Ax k = ( Ax ) T Ax = 0, so that Ax = 0. This means that x ns ( A ). (b) It follows from (a) that A and A T A have the same nullity. A T A is an n n matrix, so that, by the rank plus nullity theorem, rk ( A T A ) = n null ( A T A ) = n null ( A ) = rk ( A ). (c) It follows from (a) that rs ( A T A ) = ns ( A T A ) = ns ( A ) = rs ( A ), so that cs ( A T A ) = rs ( A T A ) = rs ( A ) = cs ( A T ). Now A T b cs ( A T ), so that A T b cs ( A T A ), which means that A T b = A T Ax for some x . 5. Show that for any matrix A , the transformation L A is an isomorphism from the Row Space of A onto the Column Space of A . Let A be m n , so that L A maps R n into R m . L A ( v ) = Av is in colspace(A) for any v , so L A maps rowspace(A) into colspace(A). To show that this map is onto, take any y R ( A ) and let y = Ax for some x R m . Now let x = u + v where u rowspace ( A ) and v N ( A ), by Theorem 5.2.3. Then Ax = Au + Av = Au , since Av = 0. Therefore y = Au with u in rowspace(A). To show that this map is onetoone, suppose that Ax = Au for some x and u in rowspace(A). Then A ( x u ) = 0, so that x u N ( A ). But x u rowspace ( A ), which is N ( A )...
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This note was uploaded on 12/27/2011 for the course MAS 4105 taught by Professor Rudyak during the Spring '09 term at University of Florida.
 Spring '09
 RUDYAK

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