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Triangle inequality implies nonnegative scalar multiple

  1. Nov 12, 2015 #1
    I'm not really sure if this is true, which is why I want your opinion. I have been trying to prove it, but it will help me a lot if someone can confirm this.

    Let ## v_{1}, v_{2} ... v_{n} ## be vectors in a complex inner product space ##V##. Suppose that ## | v_{1} + v_{2} +...+ v_{n}| = |v_{1}| + |v_{2}| +...+ |v_{n}| ##. Then is it necessarily the case that ##v_{1},v_{2}...v_{n}## are all non-negative scalar multiples of some nonzero vector ##v## ?

    It seems much easier to prove for a real inner product space, but I'm not even sure if this is true for complex inner product spaces. I'm trying an induction, first using the simple case where ## n = 2 ## but I can't seem to prove that the scaling factor must be a non-negative real number.

    BiP
     
  2. jcsd
  3. Nov 12, 2015 #2

    andrewkirk

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    I don't have a proof handy but I can confirm that it is at least true if the space is finite-dimensional.
    The reason is that ##\mathbb{C}^n## is a Hilbert Space (inner product space), every Hilbert Space is a Banach Space (normed space) using the norm derived from the inner product, and part of the definition of a norm is that it must obey the triangle inequality.

    To complete the argument, note that an n-dimensional vector space over ##\mathbb{C}## is isomorphic to ##\mathbb{C}^n##.

    Actually, now I think about it, the usual proof of that isomorphism covers only the vector space operations, not the norm, so there's still something additional to prove there: that if ##\phi## is the isomorphism then ##\|\phi(v)\|=\|v\|##. I suppose that in turn comes down to whether the inner product on a finite-dimensional complex vector space is unique (ie can there be more than one inner product function?).
     
  4. Nov 12, 2015 #3

    fresh_42

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    Every complex matrix ## A = (a_{ij}) ## defines a sesquilinear form ## <v,w> = ∑_{i,j} a_{i,j} \bar{v_i} w_j ## which is hermitian iff ##A## is.

    As far as I could see the proof for the statement in the OP uses just the properties of a sesquilinear form and the Cauchy-Schwarz inequality and does not require a finite dimensional vector space.
     
  5. Nov 12, 2015 #4

    andrewkirk

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    Picking up on fresh_42's ref to the Cauchy-Schwartz inequality, we can see from this proof of the triangle inequality for a complex inner product space, that equality holds only if the ##\leq## signs in the fourth and fifth lines become equalities. Making the first of those an equality forces ##\langle x,y\rangle## to be real and non-negative and making the second one an equality forces ##x## to be a scalar multiple of ##y##. The dimensionality of the complex vector space does not matter.
     
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