# Prove a set of vectors $A_i$ perpendicular in pairs is linearly independent

Suppose $$A_1, ..., A_r\in F^n$$ are non-zero vectors such that $$A_i\cdot A_j=0 \space \forall i\neq j$$. Let $$c_1,...,c_r\in F$$ be scalars such that $$\sum_{i=1}^rc_iA_i=0$$ Show that $$c_i=0$$ for all $$i$$.

To prove this, appeal to Pythagoras' Theorem: $$\text{If A_1,...A_r are perpendicular, then}\space \bigg|\sum_{i=1}^rA_i\bigg|^2=\sum_{i=1}^r\big|A_r\big|^2$$

In this case, for each $$i\neq j$$ we have $$c_iA_i\cdot c_jA_j=0 \implies$$ each term in $$\sum c_iA_i$$ is perpendicular to any other term other than itself. Hence we can apply Pythagoras' Theorem: $$0=\bigg|\sum_{i=1}^rc_iA_i\bigg|^2=\sum_{i=1}^r|c_i|^2\big|A_i\big|^2$$

Since $$A_i$$ is non-zero $$\forall i$$, $$\big|A_i\big|^2\neq0 \space\forall i.$$ Furthermore, $$\sqrt{0}=\sqrt{\bigg|\sum_{i=1}^rc_iA_i\bigg|^2}=\sqrt{\sum_{i=1}^r|c_i|^2\big|A_i\big|^2}\implies$$ $$0=\sum_{I=1}^rc_iA_i=\sqrt{\sum_{i=1}^r|c_i|^2\big|A_i\big|^2}$$

But, the sum under the radical on the right is strictly positive since each term is positive via squaring. So since we're guaranteed $$A_i$$ is non-zero for all $$i$$ we know in order to enforce the sum $$=0$$ we must have $$c_i=0$$ for all $$i$$. Done.

I'm wondering if it is circular to use Pythagoras' Theorem here and whether or not this proof is correct.

Your approach is fine, provided the Pythagorean theorem has already been proved. But there's a much easier proof (without Pythagoras): Assume $$\sum_ic_iA_i=0$$. For any index $$j$$, take the dot product of this equation with $$A_j$$. All the terms where $$i\neq j$$ drop out because $$A_i\cdot A_j=0$$, so what remains is $$c_j|A_j|^2=0$$. Since $$A_j$$ is a non-zero vector, this implies $$c_j=0$$. Since $$j$$ was arbitrary, this proves linear independence.