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$\Bbb{E}^n$ is an Euclidean space with dimension $n$. $v_0,v_1,\cdots,v_m\in \Bbb{E}^n,m\le n $ , and $(v_i,v_j)\lt0$ for $0\le i\ne j\le m$. Prove $v_1,v_2,\cdots,v_m $ are linear independent.

My try: I tried to make an argument like this, if $v_1,v_2$ are linearly dependent, then there exists $a_1$ such that $v_2=a_1v_1$, then $(v_2,v_2)=a_1(v_2,v_1)$. At first I thought that the left hand $\ge 0$, the right hand $\lt0$. But I notice that the constant $a_1$ could also be negative. So this contradiction failed.
Any hints would be helpful.

Jaqen Chou
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    Does $(v_i,v_j)$ refer to the standard inner product? – peek-a-boo Jun 17 '19 at 14:12
  • @ peek-a-boo Yes – Jaqen Chou Jun 17 '19 at 14:13
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    The claim is false. Consider $n=m =2$, $v_1 = (1,0)$, and $v_2 = (-1,0)$. Then, $(v_1,v_2) = -1 < 0$, but $v_1$ and $v_2$ are linearly dependent. – peek-a-boo Jun 17 '19 at 14:14
  • And what is false there? The claim is, that they are linearly independent. – kolobokish Jun 17 '19 at 14:15
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    @kolobokish but $v_1$ and $v_2$ are linearly dependent because $v_2 = -v_1$ – peek-a-boo Jun 17 '19 at 14:16
  • Note that $v_1, v_2$ linearly dependent doesn't necessarily mean that $v_2 = av_1$. In particular because $v_1$ could be the zero vector. Handling this correctly becomes more important as you include more vectors. The correct (and more easily generalizable) way to set it up is that $v_1, v_2$ lineary dependent means there are $a_1, a_2$ that are not both $0$ such that $a_1v_1 + a_2v_2 = 0$. – Arthur Jun 17 '19 at 14:16
  • Do you really want $m+1$ $v_i$? – ancient mathematician Jun 17 '19 at 14:17
  • Ah, sory my mistake. – kolobokish Jun 17 '19 at 14:17
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    @peek-a-boo I think there need $v_0$ in your case and must satisfy $(v_i,v_j)\lt 0$. – Jaqen Chou Jun 17 '19 at 15:35
  • For a slightly larger counterexample with $m=n=3$, consider $v_1=(2,0,0)$, $v_2=(-1,\sqrt{3},0)$, $v_3=(-1,-\sqrt{3},0)$. – Daniel Schepler Jun 17 '19 at 18:42
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    Actually, with the question as stated, with $v_0$ included in the negative inner product condition but not included in the linearly independent set it looks like a true statement - and I have a proof which convinces me. Not sure how I could give a hint to the proof short of giving the full answer, though... – Daniel Schepler Jun 17 '19 at 19:04
  • my mistake, I read the question hastily, and misread it, so you're right, my "counterexample" wasn't actually a counterexample. – peek-a-boo Jun 17 '19 at 23:34

2 Answers2

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Some hints:

Suppose we apply the Gram-Schmidt process to $v_1, \ldots, v_m$ to get a sequence of orthogonal vectors $w_1 = v_1$, $w_2 = v_2 - \frac{\langle v_2, w_1 \rangle}{\langle w_1, w_1 \rangle} w_1$, $\ldots$. Then we can write $w_j = c_{1j} v_1 + c_{2j} v_2 + \cdots + c_{jj} v_j$ for some scalars $c_{ij}$. Can you prove that each $c_{ij} > 0$?

Now, if $v_1, \ldots, v_m$ is linearly dependent, then we will have $w_j = 0$ for some $j$; thus, a linear combination of $v_1, \ldots, v_j$ with strictly positive coefficients gives 0. From here, can you see a way to derive a contradiction?

Daniel Schepler
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You're claiming that the Gram-Matrix of the vectors is invertible. But all we know about that symmetric matrix is that all entries are negative except the diagonal ones, which are positive. That's not enough to be invertible.

Michael Hoppe
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