In this case following the same procedure we get zj

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Unformatted text preview: ￿ ￿ 1 cn with the unitary U = [z1 z2 · · · zn ] \\ Now, consider the general case in which A may not be invertible. In this case, following the same procedure we get {zj }n=1 which possibly contain zero vectors. And, if zj = 0, then cj > 0 is j arbitrary. Then, we get ˜ A=U 1 c1 ￿ 1 c2 ￿ ￿ ... ￿ ￿ ￿ 1 cn ˜ with U containing zero columns so possibly it is not unitary. We can fix this by inserting unit ˜ vectors into the zero columns of U , complementing the other column vectors to an orthonormal bas...
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This note was uploaded on 01/16/2014 for the course MATH 6304 taught by Professor Bernhardbodmann during the Fall '12 term at University of Houston.

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