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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.

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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.

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