Solution04_121 - Problem Set 4 Solutions J Scholtz Question 1 2.1/26 Let T V V be a projection along W1 where V = W1 W2 part a T is linear Proof

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Problem Set 4 - Solutions J. Scholtz October 20, 2006 Question 1: 2.1/26 Let T : V V be a projection along W 1 , where V = W 1 W 2 . part a) T is linear. Proof: T ( x ) = T ( x 1 + x 2 ) = x 1 , where x 1 W 1 . Then T ( a + b ) = T ( a 1 + b 1 + a 2 + b 2 ) = a 1 + b 1 = T ( a )+ T ( b ). The first stap is possible because each a can be split into a 1 W 1 and a 2 W 2 , and clearly then if a + b = w 1 + w 2 , then w 1 = a 1 + b 1 W 1 and w 2 = a 2 + b 2 W 2 . Claim: W 1 = { x V : T ( x ) = x } . Proof: If w W 1 , then T ( w ) = w by definition. Hence W 1 ⊂ { x V : T ( x ) = x } . On the other hand, suppose that T ( x ) = x , but x / W 1 . Then this means T ( x ) = T ( x 1 + x 2 ) = x 1 , which implies that x 2 = 0. But this means that x = x 1 W 1 - contradiction. Therefore: W 1 = { x V : T ( x ) = x } . part b) By the definition of the map R ( T ) W 1 . However, since T ( W 1 ) = W 1 , then W 1 R ( T ). Together this means R ( T ) = W 1 . Suppose v W 2 , then T ( w ) = 0, hence W 2 N ( T ). Yet at the same time, if T ( x ) = 0, then T ( x 1 + x 2 ) = x 1 = 0, which means x 1 = 0, x = x 2 W 2 . We can conclude N ( T ) = W 2 . part c) If W 1 = V , then W 2 = 0 and also T ( x ) = x for any x , hence T = I . part d) If W 1 = 0, then N ( T ) = W 2 = V , therefore T ( x ) = 0 for all x . Question 2: 2.2/12 Let T still be a projection on W along W 0 , where V = W W 0 . Then we can pick a basis for the subspace W , β W = { w 1 ,...,w k } and extend it onto a basis 1
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for V , β V = { w 1 ,...,w k ,v k +1 ,...,v n } . Then we know that the basis for W 0 is the set { w k +1 ,...,w n } (it is certainly linearly independent as it is a part of another basis, it also must span since V = W W 0 ). Then, by the previous exercise we know that: T ( w i ) = w i and T ( w j ) = 0 for all i or j . Therefore the matrix with repsect to this basis will look like: [ T ] β = 1 . . . 1 0 . . . 0 Where the rest of entries is zeros. Question 3: 2.3/12 Let V,W,Z be vectorspaces, T : V W and U : W Z . part a) Show that if UT is one-to-one then T is one-to-one.
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This note was uploaded on 07/03/2011 for the course MATH 110 taught by Professor Gurevitch during the Spring '08 term at University of California, Berkeley.

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Solution04_121 - Problem Set 4 Solutions J Scholtz Question 1 2.1/26 Let T V V be a projection along W1 where V = W1 W2 part a T is linear Proof

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