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Prove ||w + z|| <= ||w|| + ||z|| for complex vectors

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data

    Let z, w be complex vectors of C^n.

    Prove ||w + z|| <= ||w|| + ||z||

    (using the standard inner product for C^n)
    (i.e. <w,z> = w*z', where * is the dot product and ' denotes the complex conjugate)


    3. The attempt at a solution

    Well, I found that

    ||w + z||
    = sqrt( w*w' + z*z' + z*w' + w*z')
    = sqrt( w*w' + z*z' + <w,z> + <w,z>' )

    and

    ||w|| + ||z||
    = sqrt(w*w' + z*z')

    So by showing that <w,z> + <w,z>' <= 0 then I guess that will finish the proof but I am unsure of how to do this.

    Any help?
     
  2. jcsd
  3. Nov 28, 2009 #2

    Mark44

    Staff: Mentor

    The equation above isn't true. ||w|| = sqrt(w*w') and ||z|| = sqrt(z*z'), but the two square roots don't add to sqrt(w*w' + z*z'). This is like saying that the lengths of two sides of a right triangle add up to the length of the hypotenuse.
     
  4. Nov 28, 2009 #3
    Oooohhh deeaarr how embarrassing. Thank you for that insightful information. I should be able to get it now.
     
  5. Nov 28, 2009 #4
    Sorry to double post but I didn't think simply editing would bump this up.

    Alright so I played around with this some more but still could not get it to work.

    I tried to prove this using the method of squaring both sides for the regular 'scalar triangle inequality' proof like so:

    ||w + z||^2 = (sqrt<w + z,w + z>)^2
    = <w + z, w + z>
    = <w,w> + <w,z> + <z,w> + <z,z>
    = ||w||^2 + <w,z> + <w,z>' + ||z||^2


    and so if I can show that

    <w,z> + <w,z>' <= 2*||w||*||z||


    then I would have

    ||w + z||^2 <= (||w|| + ||z||)^2

    but I have been unable to do so. Any assistance would be greatly appreciated!
     
  6. Nov 28, 2009 #5

    lanedance

    User Avatar
    Homework Helper

    have you tried assuming a complex form, say w = aj + b and expanding the inner products?
     
  7. Nov 28, 2009 #6
    [tex] \|w + z\|^2 [/tex] is not equal to [tex] \langle w + z, w + z \rangle [/tex] unless you're in a real vector space. But since these are complex, then

    [tex]
    \|w + z\|^2 = \langle w + z, \overline{w + z} \rangle
    [/tex]

    Some hints:
    If [tex] w [/tex] is a complex number (or vector, or whatever), then [tex] w = a + bi [/tex], where [tex] a [/tex] and [tex] b [/tex] are real numbers (or vectors), and

    [tex]
    w + \overline{w} = a + bi + a - bi = 2a = 2\operatorname{Re}(w).
    [/tex]

    Also, [tex]\operatorname{Re}(w) \le \|w\| [/tex]

    These should hopefully get you on the right track.
     
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