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Inner product and hermitic product

  1. Sep 21, 2012 #1
    1. The problem statement, all variables and given/known data

    prove the triangle inequality ||v+w|| ≤ ||v||+||w||
    if (ℂ, V, + , [ , ] ) is a Hermitic space where [ , ] is the Hermitic product.


    2. Relevant equations

    |[v,w]| ≤ ||v|| . ||w||

    [itex]\overline{a+b i}[/itex]= a-b i (where i is the imaginary number, a+bi a complex number)

    3. The attempt at a solution

    ||v+w||² = [v+w,v+w]. Since [v+w,v+w] = [itex]\overline{[v+w,v+w]}[/itex], [v+w,v+w] must be a real number. this means that the same rules will apply as in the inner product on the innerproductspace (ℝ, V, +, [ , ] ) ( true ?). so [v+w,v+w] = [v,v] + [w,w] + 2[v,w] ≤ ||v||²+||w||²+2||v|| ||w|| = (||v|| + ||w|| ) ²
    this means ||v+w|| ≤ ||v|| + ||w||
     
    Last edited: Sep 21, 2012
  2. jcsd
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