# ch09 - CHAPTER 9 0 1 1 With A = 0 0 2 0 0 0 0 0 3 0 0 0 0 0...

• Notes
• 9

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CHAPTER 9 1. With A + = 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 0 we have ( A + ) 2 = 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 0 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 0 = 0 0 0 00 0 0 0 00 2 0 0 00 0 6 0 00 0 0 12 00 It follows that ( A + ) 3 = 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 0 0 0 0 00 0 0 0 00 2 0 0 00 0 6 0 00 0 0 12 00 = 0 0 000 0 0 000 0 0 000 3 .2 .1 0 000 0 4 .3 .2 000 The next step is obvious: In the 5 x 5 format, there is only one entry in the bottom left- most corner, and it is 4 .3 .2 .1 . 2. [The reference should be to Eq. (6-36) instead of Eq. (6-4) x = h 2 m ϖ ( A + A + ) = h 2 m ϖ 0 1 0 0 0 1 0 2 0 0 0 2 0 3 0 0 0 3 0 4 0 0 0 4 0 from which it follows that

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x 2 = h 2 m ϖ 1 0 2 .1 0 0 0 3 0 3 .2 0 2 .1 0 5 0 4 .3 0 3 ,2 0 7 0 0 0 4 .3 0 9 3. The procedure here is exactly the same. We have p = i m h ϖ 2 ( A + - A ) = i m h ϖ 2 0 - 1 0 0 0 1 0 - 2 0 0 0 2 0 - 3 0 0 0 3 0 - 4 0 0 0 4 0 from which it follows that p 2 = m h ϖ 2 1 0 - 2 .1 0 0 0 3 0 - 3 .2 0 - 2 .1 0 5 0 - 4 .3 0 - 3 .2 0 7 0 0 0 - 4 .3 0 9 4. We have u 1 = A + u 0 = 0 0 0 0 0 1 0 0 0 0 0 2 0 0 0 0 0 3 0 0 0 0 0 4 0 1 0 0 0 0 = 0 1 0 0 0
5. We write u 2 = 1 2! ( A + ) 2 u 0 = 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 6 0 0 0 0 0 12 0 0 1 0 0 0 0 = 0 0 1 0 0 Similarly u 3 = 1 3 !

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• Spring '05
• mokhtari
• Matrices, 0 m, 1m, Unitary matrix, Hermitian matrix

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