Let $V=\mathbb{R}^3$ with the dot product, find an orthonormal basis $\{e_1,e_2,e_3\}$ such that:

$\textrm{Span}\left\{e_1\right\}=\textrm{Span}\left\{\left(\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right)\right\}$

and $\textrm{Span}\left\{e_1,e_2\right\}=\left(\textrm{Span}\left\{\left(\begin{array}{c} 0 \\ -\frac{2}{5} \\ \frac{1}{5} \end{array}\right)\right\}\right)^\perp$

How should I approach this question?


Given the span of $e_1$, you can choose to let \begin{equation} e_1 = \left(\begin{matrix} 1\\ 0\\ 0\end{matrix}\right). \end{equation} Then we're told that the span of $e_1$ and $e_2$ is orthogonal to (the span of) a particular vector. But the span of $e_1$ and $e_2$ must also be orthogonal to the span of $e_3$, so why not let $e_3$ be a unit length vector pointing in the direction of \begin{equation} \left(\begin{matrix} 0\\ -\frac{2}{5}\\ \frac{1}{5}\end{matrix}\right)? \end{equation} In fact, this vector is unit length, so just set \begin{equation} e_3=\left(\begin{matrix} 0\\ -\frac{2}{5}\\ \frac{1}{5}\end{matrix}\right). \end{equation} Finally, $e_2$ needs to be orthogonal to both $e_1$ and $e_3$, and it needs to have unit length. Do you know how to find a vector which is orthogonal to two given vectors? Once you have such a vector, can you scale it to make it unit length?

  • $\begingroup$ So we need to look for $<e_1,(x,y,z)>=0$ and $<e_3,(x,y,z)>=0$? $\endgroup$ – newhere Aug 31 '16 at 18:12
  • 1
    $\begingroup$ Yes. Are you familiar with the cross product of two vectors? $\endgroup$ – 211792 Aug 31 '16 at 18:16

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