6.5 - 1 2 3 2 d 4 3 4 5 are 1, d-4 and-4 d + 8. These are...

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Math 2940 Solutions, Fall 2011 Section 6 . 5 2 ) Only A 4 passes the test. [ a b ] ± 5 6 6 7 ²± a b ² = 5 a 2 +12 ab +7 b 2 , so [ - 6 5] ± 5 6 6 7 ²± - 6 5 ² = 180 - 360 + 175 = - 5. 6 ) [ x y ] ± 0 1 1 0 ²± x y ² = 2 xy . The eigenvalues are 1 and - 1. 7 ) ± 1 0 2 3 ²± 1 2 0 3 ² = ± 1 2 2 13 ² passes the test so it is positive definite. ± 1 1 2 1 2 1 ² 1 1 1 2 2 1 = ± 6 5 5 6 ² passes the test so it is positive definite. 1 1 1 2 2 1 ± 1 1 2 1 2 1 ² = 2 3 3 3 5 4 3 4 5 has rank at most 2 (since each factor has rank at most 2) and so has 0 as an eigenvalue. Since A T A has no negative eigenvalues for any A , 2 3 3 3 5 4 3 4 5 is positive semidefinite. 12 ) The determinants of the three upper left determinants of c 1 1 1 c 1 1 1 c are c , c 2 - 1 and ( c - 1) 2 ( c + 2). These are all positive when, simultaneously, c > 0, | c | > 1 and c > - 2, that is when c > 1. The determinants of the three upper left determinants of
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Unformatted text preview: 1 2 3 2 d 4 3 4 5 are 1, d-4 and-4 d + 8. These are positive when, simultaneously, d > 4 and 2 > d , that is never. 15 ) We know that for any ~x 6 = ~ 0 that ~x T A~x > 0 and ~x T B~x > 0. So for any ~x 6 = ~ 0 we have ~x T ( A + B ) ~x = ~x T A~x + ~x T B~x > 0. 25 ) A = C T C = 3 0 1 2 3 1 0 2 = 9 3 3 5 . For 4 8 8 25 , the rst elimination step is to add-2 times row 1 to row 2. This is multi-plying by 1 0-2 1 on the left, so 4 8 8 25 = 1 0 2 1 4 8 0 9 = 1 0 2 1 4 0 0 9 1 2 0 1 so C = 2 0 0 3 1 2 0 1 = 2 4 0 3 ....
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This note was uploaded on 01/02/2012 for the course MATH 2940 at Cornell University (Engineering School).

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