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Mathematics LibreTexts

14.2: Orthogonal and Orthonormal Bases

  • Page ID
    2086
  • [ "article:topic", "authortag:waldron", "authorname:waldron" ]

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    There are many other bases that  behave in the same way as the standard basis.  As such, we will study:

    1.    \(\textit{Orthogonal bases}\) \(\{v_{1}, \ldots, v_{n} \}\):
    \[
    v_{i}\cdot v_{j}=0 \textit{ if } i\neq j\, .
    \]
    In other words, all vectors in the basis are perpendicular.

    2.    \(\textit{Orthonormal bases}\) \(\{u_{1}, \ldots, u_{n} \}\):
    \[
    u_{i}\cdot u_{j} = \delta_{ij}.
    \]
    In addition to being orthogonal, each vector has unit length.
     

    Suppose \(T=\{u_{1}, \ldots, u_{n} \}\) is an orthonormal basis for \(\Re^{n}\).  Because \(T\) is a basis, we can write any vector \(v\) uniquely as a linear combination of the vectors in \(T\):
    \[
    v=c^{1}u_{1}+\cdots c^{n}u_{n}.
    \]
    Since \(T\) is orthonormal, there is a very easy way to find the coefficients of this linear combination.  By taking the dot product of \(v\) with any of the vectors in \(T\), we get:
    \begin{eqnarray*}
    v\cdot u_{i} &=& c^{1}u_{1}\cdot u_{i} + \cdots + c^{i}u_{i}\cdot u_{i} + \cdots + c^{n}u_{n}\cdot u_{i} \\
    & = & c^{1}\cdot 0 + \cdots + c^{i}\cdot 1 + \cdots + c^{n}\cdot 0 \\
    & = & c^{i}, \\
    \Rightarrow\, c^{i} &=& v\cdot u_{i} \\
    \Rightarrow\  v &=& (v\cdot u_{1}) u_{1} + \cdots + (v\cdot u_{n})u_{n}\\
    &=& \sum_{i} (v\cdot u_{i})u_{i}.
    \end{eqnarray*}

    This proves the theorem:

     

    Theorem

    For an orthonormal basis \(\{u_{1}, \ldots, u_{n} \}\), any vector \(v\) can be expressed as
    \[
    v =\sum_{i} (v\cdot u_{i})u_{i}.
    \]


     

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