MA353_HW10_sol

MA353_HW10_sol - Purdue University MA 353: Linear Algebra...

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Purdue University MA 353: Linear Algebra II with Applications Homework 10, due Apr. 2, Some Solutions (I) Sec. 1.6#34: (a) Prove that if W 1 is any subspace of a finite-dimensional vector space V , then there exists a subspace W 2 of V such that V = W 1 W 2 . Proof. Let γ 1 = { v 1 ,v 2 ,...,v k } be a basis of W 1 . Then extend it to a basis γ = { v 1 ,v 2 ,...,v k ,v k +1 ,...,v n } of V . Let γ 2 = { v k +1 ,...,v n } and W 2 = span γ . Then V = W 1 W 2 . This can be shown as follows. First, one can see that V = W 1 + W 2 because W 1 + W 2 = span γ 1 + span γ 2 = span γ 1 γ 2 = span γ = V . Next we have to show that W 1 W 2 = { 0 } . Let x W 1 W 2 . Then since x W 1 , x = a 1 v 1 + a 2 v 2 + ··· + a k v k , and since x W 2 , x = a k +1 v k +1 + a k +2 v k +1 + ··· + a n v n . Hence a 1 v 1 + a 2 v 2 + ··· + a k v k = a k +1 v k +1 + a k +2 v k +1 + ··· + a n v n which gives a 1 v 1 + a 2 v 2 + ··· + a k v k - a k +1 v k +1 - a k +2 v k +1 - ··· - a n v n = 0 . Now since { v 1 ,v 2 ,...,v k ,v k +1 ,...,v n } is linearly independent, we have a 1 = a 2 = ··· = a k = a k +1 = ··· = a n = 0 , namely x = 0. This completes the proof. ± (b) Let V = R 2 and W 1 = { ( a 1 , 0) : a 1 R } . Give examples of two differen subspaces W 2 and W 0 2 such that V = W 1 W 2 and V = W 1 W 0 2 . Answer. Let W 2 = { (0 ,a 2 ) : a 2 R } and W 0 2 = { ( a,a ) : a R } . Then one can verify that those are desired examples. ± 1
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Sec. 2.3#16: Let V be a finite-dimensional vector space, and let T : V V be linear. (a) If rank( T ) = rank( T 2 ), prove that R ( T ) N ( T ) = { 0 } and deduce that V = R ( T ) N ( T ). Proof. Assume rank( T ) = rank( T 2 ). Then by Rank-Nullity Theo- rem, this implies dim N ( T ) = dim N ( T 2 ). But notice that N ( T ) N ( T 2 ) because if x N ( T ), then T ( x ) = 0 and hence
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MA353_HW10_sol - Purdue University MA 353: Linear Algebra...

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