solexam1f11

solexam1f11 - Foundations of Computational Math I Exam 1...

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Unformatted text preview: Foundations of Computational Math I Exam 1 Take-home Exam Open Notes, Textbook, Homework Solutions Only Calculators Allowed No collaborations with anyone Due beginning of Class Wednesday, October 26, 2011 Question Points Points Possible Awarded 1. Basics 25 2. Linear operators 25 3. Floating point 25 4. Factorization 25 5. Orthogonal 25 Factorization Total 125 Points Name: Alias: to be used when posting anonymous grade list. 1 Problem 1 (25 points) Suppose A R m n and consider the matrix 2-norm k A k 2 = max k x k 2 =1 k Ax k 2 1.a (10 points) Show that k A k 2 k A 1 k 2 where A = A 1 A 2 , m = m 1 + m 2 , A 1 R m 1 n , and A 2 R m 2 n . Solution: Recall that y R m , we have y = y 1 y 2 k y k 2 2 = k y 1 k 2 2 + k y 2 k 2 2 x R n k A 1 x k 2 2 k A 1 x k 2 2 + k A 2 x k 2 2 It follows that k A k 2 2 = max k x k 2 =1 k Ax k 2 2 = max k x k 2 =1 {k A 1 x k 2 2 + k A 2 x k 2 2 } max k x k 2 =1 k A 1 x k 2 2 = k A 1 k 2 2 . We therefore have k A k 2 2 k A 1 k 2 2 . You can also prove this by exploiting the fact that there exists x 1 R n such that k A 1 x 1 k 2 = k A 1 k 2 Therefore, k A k 2 2 k Ax 1 k 2 2 = k A 1 x 1 k 2 2 + k A 1 x 1 k 2 2 k A 1 x 1 k 2 2 = k A 1 k 2 2 Note this proof does not work if you start with z R n such that k Az k 2 = k A k 2 . A more elegant proof uses the consistency of the family of matrix 2-norms. Define A = A 1 A 2 B = ( I m 1 ) Note that it is easily seen from the definition of the matrix 2-norm and B that k B k 2 = 1 We therefore have A 1 = BA k A 1 k 2 = k BA k 2 k B k 2 k A k 2 = k A k 2 as desired. 2 1.b (15 points) Let S 1 R n and S 2 R n be two subspaces of R n . (i) (5 points) Suppose x 1 S 1 , x 1 / S 1 S 2 . x 2 S 2 , and x 2 / S 1 S 2 . Show that x 1 and x 2 are linearly independent. Solution: Proof by contradiction. Assume they are dependent. If they are dependent then we must have x 1 = x 2 . x 1 = x 2 x 1 S 2 which is a contradiction. x 1 , x 2 are linearly independent. (ii) (10 points) Suppose x 1 S 1 , x 1 / S 1 S 2 . x 2 S 2 , and x 2 / S 1 S 2 . Also, suppose that x 3 S 1 S 2 and x 3 6 = 0, i.e., the intersection is not empty. Show that x 1 , x 2 and x 3 are linearly independent. (Note the result of the previous part of the problem may be useful.) Solution: Proof by contradiction. Assume they are dependent. Since we know x 1 and x 2 are linearly independent, linear dependence implies x 3 = 1 x 1 + 2 x 2 . Three cases possible: 1. 1 6 = 0 and 2 = 0 x 3 S 1 and x 3 / S 2 . Contradiction. 2. 2 6 = 0 and 1 = 0 x 3 S 2 and x 3 / S 1 . Contradiction....
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This note was uploaded on 11/27/2011 for the course MAD 5403 taught by Professor Gallivan during the Spring '09 term at FSU.

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solexam1f11 - Foundations of Computational Math I Exam 1...

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