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Unitary matrix proof

  1. Dec 5, 2013 #1
    Let u be a unitary matrix in M2(ℝ).
    Prove that if {b1, b2} is an orthonormal basis of ℝ2, then u(b2) is determined up to a negative sign by u(b1).

    Can anyone provide some intuition that will help me understand the question (don't really understand it)? Any tips/hints appreciated.


    Thanks.
     
  2. jcsd
  3. Dec 5, 2013 #2

    ShayanJ

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    Gold Member

    u is unitary so [itex] u u^{\dagger}=I [/itex]
    [itex] \{b_1,b_2\} [/itex] is an orthonormal basis of [itex] \mathbb{R}^2 [/itex] so [itex] b_i \cdot b_j =\delta_{ij} [/itex]
    The point is:
    [itex] (ub_1)\cdot(ub_2)=(ub_1)^{\dagger}(ub_2)=b_1^{\dagger}u^{\dagger}ub_2=b_1^{\dagger}b_2=b_1\cdot b_2=\delta_{ij}[/itex]
    So because [itex] \{b_1,b_2\} [/itex] is an orthonormal basis,so is [itex]\{ub_1,ub_2\} [/itex] which means [itex]ub_2[/itex] is orthogonal to [itex]ub_1[/itex] which means once [itex] ub_1[/itex] is determined,[itex]ub_2[/itex] is determined up to a negative sign!
     
  4. Dec 5, 2013 #3
    Thank you! I'm confused by the meaning of "up to a negative sign"...what does that meant exactly?
     
  5. Dec 5, 2013 #4

    HallsofIvy

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    That knowing what u(b1) is will tell you what the absolute value of u(b2) is but not whether it is positive or negative.
     
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