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Math Help - Normalizing complex vector

  1. #1
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    Normalizing complex vector

    Hi.
    I'm trying to normalize the vector v=(i,i,0), but I'm getting the wrong answer.

    here's what I did:

    \hat{v}=\frac{v}{\left \| v \right \|}=\frac{(i,i,0)}{\sqrt{i^2+i^2+0}}=\frac{(i,i,0)  }{\sqrt{-2}}=\frac{(i,i,0)}{2i}=(\frac{i}{2i},\frac{i}{2i},  0)=(\frac{1}{2},\frac{1}{2},0)

    but \hat{v} length is not 1, since: \left \| \hat{v} \right \|=\sqrt{(\frac{1}{2})^2+(\frac{1}{2})^2+0}=\frac{  1}{\sqrt{2}}\neq 1

    thanks in advanced!
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  2. #2
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    Re: Normalizing complex vector

    Quote Originally Posted by Stormey View Post
    Hi.
    I'm trying to normalize the vector v=(i,i,0), but I'm getting the wrong answer.

    here's what I did:

    \hat{v}=\frac{v}{\left \| v \right \|}=\frac{(i,i,0)}{\sqrt{i^2+i^2+0}}=\frac{(i,i,0)  }{\sqrt{-2}}=\frac{(i,i,0)}{2i}=(\frac{i}{2i},\frac{i}{2i},  0)=(\frac{1}{2},\frac{1}{2},0)

    but \hat{v} length is not 1, since: \left \| \hat{v} \right \|=\sqrt{(\frac{1}{2})^2+(\frac{1}{2})^2+0}=\frac{  1}{\sqrt{2}}\neq 1

    thanks in advanced!
    I'm not familiar with the norm you have. I'm pretty sure \| v\| = \sqrt{v^T\overline v} = \sqrt{2}.

    Ignoring that for now, you went from \sqrt{-2} to 2i in the denominator instead of i\sqrt{2}.
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  3. #3
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    Re: Normalizing complex vector

    I'm not familiar with the norm you have. I'm pretty sure \| v\| = \sqrt{v^T\overline v} = \sqrt{2}.

    Quote Originally Posted by Gusbob View Post
    Ignoring that for now, you went from \sqrt{-2} to 2i in the denominator instead of i\sqrt{2}.

    I'm always doing this stupid mistakes...

    so:

    \hat{v}=\frac{v}{\left \| v \right \|}=\frac{(i,i,0)}{\sqrt{i^2+i^2+0}}=\frac{(i,i,0)  }{\sqrt{-2}}=\frac{(i,i,0)}{\sqrt{2}i}=(\frac{i}{\sqrt{2}i}  ,\frac{i}{\sqrt{2}i},0)=(\frac{1}{\sqrt{2}},\frac{  1}{\sqrt{2}},0)

    and then:

    \left \| \hat{v} \right \|=\sqrt{(\frac{1}{\sqrt{2}})^2+(\frac{1}{\sqrt{2}  })^2+0}=1



    Quote Originally Posted by Gusbob View Post
    I'm not familiar with the norm you have. I'm pretty sure \| v\| = \sqrt{v^T\overline v} = \sqrt{2}.
    What do you mean?
    Isn't the norm is defined to be:
    \left \| x \right \|=\sqrt{|x_1|^2+|x_2|^2+...+|x_n|^2}
    ?
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  4. #4
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    Re: Normalizing complex vector

    Quote Originally Posted by Stormey View Post
    What do you mean?
    Isn't the norm is defined to be:
    \left \| x \right \|=\sqrt{|x_1|^2+|x_2|^2+...+|x_n|^2}
    ?
    Yes, so in your case, |x_1|=|i|=1,|x_2|=|i|=1,|x_3|=|0|=0 and \|x\| = \sqrt{1^2+1^2+0^2}=\sqrt{2}
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  5. #5
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    Re: Normalizing complex vector

    Quote Originally Posted by Gusbob View Post
    Yes, so in your case, |x_1|=|i|=1,|x_2|=|i|=1,|x_3|=|0|=0 and \|x\| = \sqrt{1^2+1^2+0^2}=\sqrt{2}
    isn't |x_1|,|x_2|,|x_3| supposed to be squared?

    plus if \left \| v \right \|=\sqrt{2}, then:

     \hat{v}=\frac{(i,i,0)}{\sqrt{2}}=(\frac{i}{\sqrt{2  }},\frac{i}{\sqrt{2}},0)

    and then: \left \| \hat{v} \right \|\neq 1

    which means that \left \| v \right \| can't be equal to \sqrt{2}
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  6. #6
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    Re: Normalizing complex vector

    Yes, and they are squared! 1^2= 1 and 0^2= 0.

    And the length of \left(\frac{i}{\sqrt{2}},\frac{i}{\sqrt{2}}, 0\right) is 1:
    \sqrt{\left(\frac{i}{\sqrt{2}}\right)\left(\frac{-i}{\sqrt{2}}\right)+ \left(\frac{i}{\sqrt{2}}\right)\left(\frac{-i}{\sqrt{2}}\right)+ (0)(0)= \sqrt{\frac{1}{2}+ \frac{1}{2}+ 0}= \sqrt{1}= 1
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  7. #7
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    Re: Normalizing complex vector

    Quote Originally Posted by Stormey View Post
    . . . then:

     \hat{v}=\frac{(i,i,0)}{\sqrt{2}}=(\frac{i}{\sqrt{2  }},\frac{i}{\sqrt{2}},0)
    That looks right to me.

    And \left \| \hat{v} \right \| does equal 1.

    And \left \| v \right \| does equal \sqrt{2}.
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  8. #8
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    Re: Normalizing complex vector

    Oh... I forgot about the conjugate.

    Thank you all for the help.
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