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

Info icon This preview shows pages 1–4. Sign up to view the full content.

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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

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
Image of page 2
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 !
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 4
This is the end of the preview. Sign up to access the rest of the document.
  • Spring '05
  • mokhtari
  • Matrices, 0 m, 1m, Unitary matrix, Hermitian matrix

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern