It follows that satisfies the first alternatively

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Unformatted text preview: matrix A b . Replace the last row by We find that if b y , then the last row of the augmented matrix contains only zeros. Hence there are remaining equations. We can now set , some parameter, and solve for the other variables in terms of Therefore the system of equations Ax b has a solution. 30. If is an eigenvalue of A , then there is a nonzero vector, x , such that Ax x 0. That is, Ax 0 has a nonzero solution. This implies that the mapping defined by A is not 1-to-1, and hence not invertible. On the other hand, if A is singular, then A Thus, Ax 0 has a nonzero solution. The latter equation can be written as Ax x 31. As shown in Prob. , Ax y x A y . By definition of a Hermitian matrix, ________________________________________________________________________ page 360 —————————————————————————— CHAPTER 7. —— A A 32 . Based on Prob. , Ax x x Ax . . Let x be an eigenvector corresponding to an eigenvalue It then follows that Ax x x x and x Ax x x Based on the properties of the inner product, xx x x and x x x x . Then from Part , xx . From Part xx , xx Based on the defin...
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This note was uploaded on 03/11/2014 for the course MA 303 taught by Professor Staff during the Spring '08 term at Purdue University-West Lafayette.

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