# Triangle inequality implies nonnegative scalar multiple

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

## Answers and Replies

andrewkirk
Science Advisor
Homework Helper
Gold Member
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?).

fresh_42
Mentor
(ie can there be more than one inner product function?).
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.

andrewkirk
Science Advisor
Homework Helper
Gold Member
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.

• fresh_42