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Math Help - Interpreting a zero norm of complex vector

  1. #1
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    Post Interpreting a zero norm of complex vector

    What can we conclude about the norm of a complex vector (dimension 2) when it's equal to 0? What does it mean? What can be said about their components x and y?
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    Quote Originally Posted by janie_t View Post
    What can we conclude about the norm of a complex vector (dimension 2) when it's equal to 0? What does it mean? What can be said about their components x and y?
    z = x + iy
    |z| = 0 \Longleftrightarrow z = 0, that is x = 0 and y = 0
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    But what if x and y are not equal to 0 and the norm is 0. Basically x and y are complex vectors and their norm is 0 so what can I conclude from that?
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    Quote Originally Posted by janie_t View Post
    But what if x and y are not equal to 0 and the norm is 0. Basically x and y are complex vectors and their norm is 0 so what can I conclude from that?
    which one is it? x, y are the components or x,y are the complex vectors?

    anyway, x,y should be 0
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    Suppose you have the following vector x= (5+2i,-2+5i), then the norm ||x||=0.

    What can you say about this vector?
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  6. #6
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    Quote Originally Posted by janie_t View Post
    Suppose you have the following vector x= (5+2i,-2+5i), then the norm ||x||=0.

    What can you say about this vector?
    what is your definition for norm here?
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  7. #7
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    Let x1 and x2 be the components of the vector x, then the norm of x is SQUARE ROOT [ (x1)^2 + (x2)^2 ]
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  8. #8
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    Quote Originally Posted by janie_t View Post
    Let x1 and x2 be the components of the vector x, then the norm of x is SQUARE ROOT [ (x1)^2 + (x2)^2 ]
    That is not the norm of a 2-vector over \mathbb{C}. Assuming \bold{x} is a column vector the usual norm is:

    \|\bold{x}\|=\sqrt{\overline{\bold{x}}^T{\bold{x}}  }=\sqrt{x_1\overline{x_1}+x_2\overline{x_2}} =\sqrt{|x_1|^2+|x_2|^2}

    which 0 if and only if both x_1 and x_2 =0

    CB
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  9. #9
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    Quote Originally Posted by janie_t View Post
    Let x1 and x2 be the components of the vector x, then the norm of x is SQUARE ROOT [ (x1)^2 + (x2)^2 ]
    this is not define a "norm" because a norm must satisfy ||x|| = 0 \leftrightarrow x = 0

    in here, you have ||x|| = 0 but x \not = 0
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  10. #10
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    I am talking about the Euclidean norm.

    Suppose you have the following vector x= (5+2i,-2+5i), then how come I get ||x||=0. I don't understand why the Euclidean Norm of the vector is 0 when its components are not zero.
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  11. #11
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    Because it's not the Euclidean norm, as others have pointed out.
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  12. #12
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    x=(5+2i,-2+5i)

    Well,
    ||x||^2 = (5+2i)^2 + (-2+5i)^2
    ||x||^2 = (25 + 10i + 4i^2) + ( 4 -10i + 25i^2)
    ||x||^2 = (25+10i-4) + (4-10i-25) if we let (i^2 = -1)
    ||x||^2 = 25 + 10i -4 + 4 - 10i - 25
    ||x||^2 = 0
    ||x|| = 0

    ? Is there any mistake in my calculations? The initial components of x were not 0, yet the norm is zero.

    ?
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  13. #13
    MHF Contributor Bruno J.'s Avatar
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    I suggest you read this.
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  14. #14
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    Sorry for bothering you all! I found my mistake!
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  15. #15
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    And it was exactly what everyone had been telling you all along?
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