# every orthonormal set is linearly independent

Theorem : An orthonormal set of vectors in an inner product space^{} is linearly independent^{}.

*Proof.* We denote by $\u27e8\cdot ,\cdot \u27e9$ 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.,
$S=\{{e}_{1},\mathrm{\dots},{e}_{n}\}$ for some $n$.
Suppose

$${\lambda}_{1}{e}_{1}+\mathrm{\cdots}+{\lambda}_{n}{e}_{n}=0$$ |

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

$$0=\u27e8{e}_{k},0\u27e9=\u27e8{e}_{k},{\lambda}_{1}{e}_{1}+\mathrm{\cdots}+{\lambda}_{n}{e}_{n}\u27e9={\lambda}_{1}\u27e8{e}_{k},{e}_{1}\u27e9+\mathrm{\cdots}+{\lambda}_{n}\u27e8{e}_{k},{e}_{n}\u27e9={\lambda}_{k},$$ |

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. $\mathrm{\square}$

Title | every orthonormal set is linearly independent |
---|---|

Canonical name | EveryOrthonormalSetIsLinearlyIndependent |

Date of creation | 2013-03-22 13:33:48 |

Last modified on | 2013-03-22 13:33:48 |

Owner | mathcam (2727) |

Last modified by | mathcam (2727) |

Numerical id | 14 |

Author | mathcam (2727) |

Entry type | Theorem |

Classification | msc 15A63 |