# every orthonormal set is linearly independent

Proof. We denote by $\langle\cdot,\cdot\rangle$ the inner product of $L$. Let $S$ be an orthonormal set of vectors. Let us first consider the case when $S$ is finite, i.e., $S=\{e_{1},\ldots,e_{n}\}$ for some $n$. Suppose

 $\lambda_{1}e_{1}+\cdots+\lambda_{n}e_{n}=0$

for some scalars $\lambda_{i}$ (belonging to the field on the underlying vector space of $L$). For a fixed $k$ in $1,\ldots,n$, we then have

 $0=\langle e_{k},0\rangle=\langle e_{k},\lambda_{1}e_{1}+\cdots+\lambda_{n}e_{n% }\rangle=\lambda_{1}\langle e_{k},e_{1}\rangle+\cdots+\lambda_{n}\langle e_{k}% ,e_{n}\rangle=\lambda_{k},$

so $\lambda_{k}=0$, and $S$ is linearly independent. Next, suppose $S$ is infinite (countable or uncountable). To prove that $S$ is linearly independent, we need to show that all finite subsets of $S$ are linearly independent. Since any subset of an orthonormal set is also orthonormal, the infinite case follows from the finite case. $\Box$

Title every orthonormal set is linearly independent EveryOrthonormalSetIsLinearlyIndependent 2013-03-22 13:33:48 2013-03-22 13:33:48 mathcam (2727) mathcam (2727) 14 mathcam (2727) Theorem msc 15A63