6.4 - Math 2940 Solutions, Fall 2011 Section 6.4 2) We have...

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Math 2940 Solutions, Fall 2011 Section 6 . 4 2 ) We have ( A T CA ) T = A T C T ( A T ) T = A T CA . When A is 6 × 3 we have A T is 3 × 6. So C must be 6 × 6 and A T CA is 3 × 3. 5 ) We need to find the unit eigenvectors of A . They will be our columns. The characteristic polynomial of A = 1 0 2 0 - 1 - 2 2 - 2 0 is (1 - λ )( - 1 - λ )( - λ ) - 4(1 - λ ) + 4(1 + λ ) = λ (9 - λ 2 ). The eigenvalues are 0 , 3 , - 3 wih eigenvectors (2 , 2 , 1), (2 , - 1 , 2) and (1 , - 2 , - 2) . So our matrix is 2 / 3 2 / 3 1 / 3 2 / 3 - 1 / 3 - 2 / 3 1 / 3 - 2 / 3 - 2 / 3 . 7 ) (a) ± 1 b b 1 ² has characteristic polynomial λ 2 - 2 λ + 1 - b 2 which has roots 1 ± b . So as long as | b | > 1 one eigenvalue will be negative. (b) The pivots have the same signs as the eigenvalues. (c) The sum of the eigenvalues is the trace of the matrix, 2. So both eigenvalues cannot be negative. 15 ) ± i 1 1 - i ² has characteristic polynomial λ 2 which has 0 as a double root. If there were independent vectors with 0 eigenvalue, our matrix would be the zero matrix! 23
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This note was uploaded on 01/02/2012 for the course MATH 2940 at Cornell.

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