# Is it possible that 3 vectors have all negative pairwise correlations?

Given three vectors $a$, $b$, and $c$, is it possible that correlations between $a$ and $b$, $a$ and $c$, and $b$ and $c$ are all negative? I.e. is this possible?

It is possible if the size of the vector is 3 or larger. For example

a=(−1,1,1)b=(1,−9,−3)c=(2,3,−1)\begin{align} a &= (-1, 1, 1)\\ b &= (1, -9, -3)\\ c &= (2, 3, -1)\\ \end{align}

The correlations are
$$cor(a,b)=−0.80...cor(a,c)=−0.27...cor(b,c)=−0.34...$$\text{cor}(a,b) = -0.80...\\ \text{cor}(a,c) = -0.27...\\ \text{cor}(b,c) = -0.34...$$$$

We can prove that for vectors of size 2 this is not possible:
cor(a,b)<02(∑iaibi)−(∑iai)(∑ibi)<02(a1b1+a2b2)−(a1+a2)(b1b2)<02(a1b1+a2b2)−(a1+a2)(b1b2)<02(a1b1+a2b2)−a1b1+a1b2+a2b1+a2b2<0a1b1+a2b2−a1b2+a2b1<0a1(b1−b2)+a2(b2−b1)<0(a1−a2)(b1−b2)<0\begin{align} \text{cor}(a,b) &< 0\\[5pt] 2\Big(\sum_i a_i b_i\Big) - \Big(\sum_i a_i\Big)\Big(\sum_i b_i\Big) &< 0\\[5pt] 2(a_1 b_1 + a_2 b_2) - (a_1 + a_2)(b_1 b_2) &< 0\\[5pt] 2(a_1 b_1 + a_2 b_2) - (a_1 + a_2)(b_1 b_2) &< 0\\[5pt] 2(a_1 b_1 + a_2 b_2) - a_1 b_1 + a_1 b_2 + a_2 b_1 + a_2 b_2 &< 0\\[5pt] a_1 b_1 + a_2 b_2 - a_1 b_2 + a_2 b_1 &< 0\\[5pt] a_1 (b_1-b_2) + a_2 (b_2-b_1) &< 0\\[5pt] (a_1-a_2)(b_1-b_2) &< 0 \end{align}

The formula makes sense: if $$a1a_1$$ is larger than $$a2a_2$$, $$b2b_2$$ has to be larger than $$b1b_1$$ to make the correlation negative.

Similarly for correlations between (a,c) and (b,c) we get

$$(a1−a2)(c1−c2)<0(b1−b2)(c1−c2)<0$$(a_1-a_2)(c_1-c_2) < 0\\ (b_1-b_2)(c_1-c_2) < 0\\$$$$

Clearly, all of these three formulas can not hold at the same time.