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A a 32 based on prob ax x x ax let x be an eigenvector

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Unformatted text preview: hat is not upper triangular. If it were upper triangular, backsubstitution would imply that x 0 . Hence a linear combination of all the rows results in a row containing only zeros. That is, there are scalars, , one for each row and not all zero, such that for each for column , . Now consider A for each , . By definition, , or It follows that . Let y . We therefore have nonzero vector, y , such that A y 0. 26. By definition, Ax y Let , so that Ax Now interchanging the order or summation, Ax y Now note that ________________________________________________________________________ page 359 —————————————————————————— CHAPTER 7. —— Ay . Therefore Ax y Ay x Ay . 28. By linearity, Ax 29. Let Ax b0 b A . By the hypothesis, there is a nonzero vector, y , such that , . Taking the conjugate of both sides, and interchanging the indices, we have This implies that a linear combination of each row of A is equal to zero. Now consider the augmented...
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