40_hwc1SolnsODDA

40_hwc1SolnsODDA - 5.19 Let A be an 3 × 3 matrix Suppose...

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and then take B = [ C v 1 , C v 2 ]. Using the given information, we then have CA = B with A = ± 1 2 2 1 ² and B = ± 2 - 1 1 1 ² . Then CA = B , we deduce C = BA - 1 . Using the formula for 2 × 2 matrix inverses, we then find C = 1 3 ± - 4 5 1 1 ² . 5.15 Are there any m × n isometries with n > m ? Give an example or explain why not. SOLUTION No. By Theorem 15, an m × n matrix A is an isometry if and only if A t is a left inverse of A . But as we have seen in Section 3.3 (see the two bullets points just before and after Example 23), matrices with more columns than rows cannot have left inverses. 5.17 Consider the matrices A = 1 2 ± 1 0 0 2 1 0 ² B = ± 0 - 1 1 2 0 0 0 1 1 ² C = 1 6 " 3 - 1 0 2 3 1 # Which, if any, are length preserving? SOLUTION To check this for A , we compute A t A and find A t A = ± 1 0 0 1 ² . Thus, by Theorem 15, A is an isometry, and hence is length preserving. Applying the same test to B and C , we find that B is not length preserving, and C is. Hence, The matrices A and C are length preserving.
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Unformatted text preview: 5.19 Let A be an 3 × 3 matrix, Suppose the third row minus the first row equals the second row. Explain why A cannot be invertible. SOLUTION Let B = A t , and write B = [ v 1 , v 2 , v 3 ]. Since, by the definition of the transpose, the columns of B are the rows of A , v 3-v 1 = v 2 , That is, v 1 + v 2-v 3 = 0. Then by V.I.F. 1 , B 1 1-1 = 0 . Therefore, by the bullet point containing (3.20), B = A t has no left inverse. Now if A has a right inverse C , then AC = I , so C t A t = I t = I , and so C t would be a left inverse of A t . But we have just seen that this is impossible. Hence A has no right inverse. It is therefore not invertible. 21 /september/ 2005; 22:09 41...
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This note was uploaded on 10/03/2010 for the course MATH 380 taught by Professor Gladue during the Fall '08 term at Roger Williams.

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