# linear independence

Let $n$ be a positive integer, $\lambda_{i}$ real numbers and $a_{i}$ for $1\leq i \leq n$ pairwise distinct complex numbers. Help me to prove that if $\forall z \in \mathbb{C}$ we have $\sum(\lambda_{i}|z-a_{i}|)=0$, then $\lambda_{i}=0$ for $1\leq i \leq n$.

Regards

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You are free to pick your $z$ as you like. Do so: pick about $n+1$ of them, and see what the resulting system of linear equations says about the $\lambda_i$. –  Gunnar Magnusson Apr 25 '11 at 19:35
@Gunnar : I tried in vain :( –  mathfan Apr 25 '11 at 19:37
@user10049 : It is a tricky one. I'll post a more detailed answer in a couple of minutes. (For now, a more detailed hint. If you want an answer, I'll give it, but you'll profit more from working through the hint.) –  Gunnar Magnusson Apr 25 '11 at 19:58

If $\lambda_i \neq 0$, $|z-a_i| = -1/\lambda_i \sum_{j \neq i} \lambda_j |z-a_j|$, and the RHS is differentiable at $z=a_i$, while the LHS is not.

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I concur, but I'm not sure the analytic way is the one forward here. Geometrically, the question could be interpreted as being about a weighted distance from a finite number of points. (To be sure, the weights being possibly negative is strange, but we've seen stranger things.) The result then is that always being zero distance from a finite number of points means that the weights were all zero to begin with, i.e. that the shackles were too hard to throw away. To me, this sheds more light on the problem than the property of being non-differentiable. –  Gunnar Magnusson Apr 25 '11 at 20:19
I like the solution a lot. Can you please enlighten me on how you came up with this solution? –  Thomas Rot Apr 25 '11 at 21:30

For now, let's fix $n = 2$. Then we have 2 real numbers $\lambda_1$ and $\lambda_2$, and two complex numbers $a_1$ and $a_2$. We want to see that if $\lambda_1 |z - a_1| + \lambda_2 |z - a_2| = 0$ (1) for all complex numbers $z$, then $\lambda_1 = \lambda_2 = 0$.

To do this, let's pick $z = a_1$ as a complex number. If equation (1) holds for all complex numbers, then it must hold for $z = a_1$. Thus we must have $\lambda_2 |a_1 - a_2| = 0$. As $a_1$ and $a_2$ are distinct by hypothesis, then this entails $\lambda_2 = 0$ (this would be a good place to construct a counterexample if the hypothesis fails). We now proceed either by recursion, or by picking another value for $z$, and prove that $\lambda_1 = 0$.

If you fill in the details and grok this argument, I think you'll have no trouble turning it into one that works for any given $n$.

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FYI: induction on $n$ might be the way forward. To see why, try writing down the same argument for $n = 3$. See where you have to do extra work. –  Gunnar Magnusson Apr 25 '11 at 20:26