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Math Help - diagonalizable

  1. #1
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    diagonalizable

    in my book, it states that the 2x2 matrix,

    (1 2i
    2 1) is unitarily diagonalizable. i thought it has to be normal to be unitarily diagonalizable? in this case, the matrix doesnt seem to be normal.

    may i know what is wrong with my reasoning?
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    in my book, it states that the 2x2 matrix,

    (1 2i
    2 1) is unitarily diagonalizable. i thought it has to be normal to be unitarily diagonalizable? in this case, the matrix doesnt seem to be normal.

    may i know what is wrong with my reasoning?
    It is normal...(you should remember that a matrix is normal if and only if it is unitarily diagonalisable).

    It is normal as if

    (1 2i)
    (2 1)=A

    A=\left( \begin{array}{cc}<br />
1 & 2i \\<br />
2 & 1 \end{array} \right)

    and

    (1 2)
    (-2i 1)=B

    B=\left( \begin{array}{cc}<br />
1 & 2 \\<br />
-2i & 1 \end{array} \right)

    then A and B commute. As B is the conjugate transpose thing of A, you are done...

    (I've left the LaTeX in in the hope that the compiler will start working again...)
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  3. #3
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    shouldnt
    (1 2
    -2i 1) =B?
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    shouldnt
    (1 2
    -2i 1) =B?
    Yes, sorry, the minus sign went walkabout between the LaTeX and the Ascii stuff. What did you get when you calculated AB and BA?
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  5. #5
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    i got that

    (5 2+2i
    2-2i 5) = AB

    (5 2i+2
    2i+2 5)=BA

    thus dont commute but it should commute
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by alexandrabel90 View Post
    i got that

    (5 2+2i
    2-2i 5) = AB

    (5 2i+2
    2i+2 5)=BA

    thus dont commute but it should commute
    You BA is incorrect, but your AB is correct. Looks like a simple case of a wandering minus sign to me...
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