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Math Help - Euclidean norm of U*x when norm of x is 1!

  1. #1
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    Arrow Euclidean norm of U*x when norm of x is 1!

    Let x be a nx1 vector in Cn and y = U*x is another vector in Cn, where U is a unitary matrix and * represents the conjugate transpose.

    The Euclidean norm of y equals 1 if and only if the Euclidean norm of x equals 1:

    ||y|| = 1 if and only if ||x|| = 1.

    I can't prove this statement and don't know where to start!

    Thanks so much!
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  2. #2
    Junior Member qspeechc's Avatar
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    Quote Originally Posted by Damien View Post
    Let x be a nx1 vector in Cn and y = U*x is another vector in Cn, where U is a unitary matrix and * represents the conjugate transpose.

    The Euclidean norm of y equals 1 if and only if the Euclidean norm of x equals 1:

    ||y|| = 1 if and only if ||x|| = 1.

    I can't prove this statement and don't know where to start!

    Thanks so much!
    The unitary matrix has at least one complex eigenvalue; what do you know about the eigenvalues of a unitary matrix? U is unitary if and only if U* is unitary right? Then since U*x = ax for some eigenvalue a, take the norm on both sides and use some properties of the norm to get what you want.
    Last edited by qspeechc; October 1st 2009 at 02:43 PM.
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  3. #3
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    Quote Originally Posted by Damien View Post
    Let x be a nx1 vector in Cn and y = U*x is another vector in Cn, where U is a unitary matrix and * represents the conjugate transpose.

    The Euclidean norm of y equals 1 if and only if the Euclidean norm of x equals 1:

    ||y|| = 1 if and only if ||x|| = 1.

    I can't prove this statement and don't know where to start!

    Thanks so much!
    let e_1, \cdots , e_n be the columns of U. since U is unitary, we have e_i \cdot \bar{e_j}=\delta_{ij}, where \delta_{ij} is the Kronecker's delta. let x_i, y_i, \ \ 1 \leq i \leq n, be the entries of the vectors x,y respectively.

    now we have \sum_{i=1}^n y_ie_i = Uy=x and thus ||x||^2=||Uy||^2=\left(\sum_{i=1}^ny_ie_i \right) \cdot \left(\sum_{j=1}^n \bar{y_j} \bar{e_j} \right)=\sum_{1 \leq i,j \leq n}y_i \bar{y_j} \delta_{ij}=\sum_{i=1}^n |y_i|^2=||y||^2 and thus ||x||=||y||.
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  4. #4
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    Thank you!

    Thanks a lot guys. 2 different proofs, but both very nice.

    1.
    Every matrix has at least one eigenvalue (fundamental theorem of algebra)
    If U is unitary then U* is unitary too
    Every eigenvalue of a unitary matrix has absolute value of 1
    So: ||y|| = ||U*x|| = ||lambda x|| = |lambda| ||x|| = ||x||

    2.
    If y = U*x then x = Uy
    So: ||x||^2 = x*x = y*U*Uy = y*y = ||y||^2 <=> ||x|| = ||y||

    Thanks a lot to both of you, really appreciate it!
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