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It is true that every orthogonal set of non-null vectors are linearly independent. Is the converse true? That is, if $\{v_1,\cdots,v_n\}$ is a linearly independent set of vectors, then there is an inner product such that $\langle v_i,v_j\rangle=0$ for every $i\neq j$.

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Yes, you can complete $v_1,..,v_n$ to a basis $v_1,..,v_n,..,v_m$ define the scalar product, $\langle v_i,v_j\rangle =\delta_{ij}$. If $x=\sum_{i=1}^{i=m}x_iv_i$ and $y=\sum_{i=1}^{i=m}y_iv_i$, then $\langle x,y\rangle =\sum_{i=1}^{i=m} x_iy_i$.

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  • $\begingroup$ What if $n$ is less than the dimension of the space? Isn't your answer assuming that the $\{v_1, ..., v_n\}$ is a basis? What if the space is infinite dimensional? $\endgroup$ – Tom Aug 11 '16 at 15:30
  • $\begingroup$ An infinite dimensional space also has a basis (this requires the Axiom of Choice). $\endgroup$ – Robert Israel Aug 11 '16 at 15:39

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