# Triangle inequality implies nonnegative scalar multiple

1. Nov 12, 2015

### Bipolarity

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. Nov 12, 2015

### andrewkirk

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?).

3. Nov 12, 2015

### Staff: Mentor

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.

4. Nov 12, 2015

### andrewkirk

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.