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Given $n+2$ vectors in an $n$-dimensional space, prove that there exists a non-trivial linear combination $$a_1 v_1 + a_2v_2 + \cdots + a_{n+2}v_{n+2}=0$$ with not all $a_i = 0$, yet with $\sum\limits_{i=1}^{n+2} a_i = 0$.

I tried to take $n+1$ vectors and build a barycentric combination of them, such that they would form the $(n+2)$nd vector and $\sum\limits_{i=1}^{n+1} a_i = 1$, and if I add $(n+2)$nd vector to this combination with $a_{n+2}=-1$, it will give the needed linear combination.

But I stuck proving this barycentric combination exists. So now I think I'm moving in the wrong direction, because it seems, that there is can be no barycentric combinations formed from these vectors.

Good Boy
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Tim Creig
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  • Since this is an n dimensional space so any set of $n+2$ vectors is always linearly dependent so you get the result. I don't even know what barycentric combination is . – PNDas Jan 16 '21 at 18:26
  • @PNDas, can you elaborate a little bit, why there is a linear combination with zero-sum of coefficients? – Tim Creig Jan 16 '21 at 18:29
  • I used the definition of linear dependency(or/and linearly independency) and the fact that a n-dimension space can't have a set with more than n linearly independent vectors. You can search them on internet. – PNDas Jan 16 '21 at 18:32
  • See: https://math.stackexchange.com/questions/1387170/any-set-with-more-elements-than-the-dimension-of-vector-space-is-linearly-depend, and https://en.wikipedia.org/wiki/Linear_independence#Space_of_linear_dependencies – PNDas Jan 16 '21 at 18:39
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    Hint: embed $\Bbb R^n$ into $\Bbb R^{n+1}$ using $(x_1,\dots,x_n)\mapsto(x_1,\dots,x_n,1)$, and look for linear dependencies among the images of ${v_1,\dots,v_{n+2}}$. – Greg Martin Jan 16 '21 at 19:07

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The set $\{v_1, \ldots, v_{n+2}\}$ is linearly dependent so there is a vector which can be expressed as a linear combination of the remaining $n+1$ vectors. WLOG assume $$v_{n+2}=\sum_{i=1}^{n+1} \alpha_i v_i.$$

On the other hand, the set $\{v_1, \ldots, v_{n+1}\}$ is also linearly dependent so there are scalars $\beta_1, \ldots, \beta_{n+1}$ not all zero such that $$\sum_{i=1}^{n+1} \beta_iv_i = 0.$$

Notice that if $\sum_{i=1}^{n+1} \beta_i = 0$ we are done as $$\sum_{i=1}^{n+1} \beta_iv_i +0\cdot v_{n+2}= 0$$ would be our desired linear combination. Therefore we can assume $\sum_{i=1}^{n+1} \beta_i \ne 0$ and we have for all $\gamma \in \Bbb{R}$ that

$$\sum_{i=1}^{n+1} (\alpha_i +\gamma\beta_i )v_i - v_{n+2}=\underbrace{\sum_{i=1}^{n+1} \alpha_i v_i - v_{n+2}}_{=0} + \gamma\underbrace{\sum_{i=1}^{n+1} \beta_iv_i}_{=0} = 0$$ where the sum of these scalars is $$\sum_{i=1}^{n+1} (\alpha_i +\gamma\beta_i ) - 1 = \sum_{i=1}^{n+1}\alpha_i + \gamma \sum_{i=1}^{n+1} \beta_i - 1$$ so picking $\gamma = \frac{1-\sum_{i=1}^{n+1}\alpha_i}{\sum_{i=1}^{n+1} \beta_i}$ gives the desired linear combination.

mechanodroid
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