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L2 norm for complex valued vector

  1. Jan 26, 2013 #1
    Let's say I have a vector (4+2i, 1-i), how do I take an L2 norm?
    Dont tell me I simply do sqrt(16+4+1+1)..?
     
  2. jcsd
  3. Jan 27, 2013 #2

    jbunniii

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    The short answer is yes, you simply do sqrt(16+4+1+1). Here is why:

    The L2 norm of a vector [itex]v[/itex] is [itex]\sqrt{v^* v}[/itex], where [itex]v^*[/itex] is the conjugate transpose. In your case,
    $$v = \left[\begin{matrix} 4 + 2i \\ 1 - i \end{matrix}\right]$$
    so
    $$v^* = \left[\begin{matrix} 4 - 2i & 1 + i \end{matrix}\right]$$
    and
    $$v^* v = \left[\begin{matrix} 4 - 2i & 1 + i \end{matrix}\right] \left[\begin{matrix} 4 + 2i \\ 1 - i \end{matrix}\right] = (4-2i)(4+2i) + (1+i)(1-i) = 16 + 4 + 1 + 1$$
    and therefore
    $$||v||_2 = \sqrt{v^H v} = \sqrt{16 + 4 + 1 + 1}$$
     
    Last edited: Jan 27, 2013
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