TITLE: Prove $v_1,v_2,\cdots,v_m $ are linear independent.
QUESTION [5 upvotes]: $\Bbb{E}^n$ 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.

REPLY [3 votes]: 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?