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Worksheet 7.5.1 Worksheet

In this worksheet we will derive the process by which one may create an orthogonal basis \(\left\{\vec{u_1},\vec{u_2},\vec{u_3}\right\}\) of \(\R^3\) from a linearly independent basis \(\left\{\vec{v_1},\vec{v_2},\vec{v_3}\right\}\) of \(\R^3\text{.}\)
  1. Consider the set \(S:=\left\{\vec{v_1},\vec{v_2},\vec{v_3}\right\}=\left\{\left(\begin{array}{r}2\\8\\16 \end{array} \right), \left(\begin{array}{rrr}20\\8\\34 \end{array} \right)\left(\begin{array}{rrr}52\\-26\\-34 \end{array} \right)\right\}\text{.}\) Show that \(S\) is linearly independent in any way you like.
  2. Write \(\vec{u_1}=\vec{v_1}\text{.}\) Then subtract from \(\vec{v_2}\) its projection to span\(\{\vec{v_1}\}\text{,}\) call the result \(\vec{u_2}\text{,}\) and denote \(\displaystyle{\widehat{x}_{12}=\frac{\vec{u_1}^T\vec{v_2}}{\vec{u_1}^T\vec{u_1}}}\text{.}\) At this point you have two of the three vectors you need.
  3. Next subtract from \(\vec{v_3}\) its projections to span\(\{\vec{u_1}\}\) and to span\(\{\vec{u_2}\}\text{,}\) and call the result \(\vec{u_3}\text{,}\) writing
    \begin{equation*} \widehat{x}_{13}=\frac{\vec{u_1}^T\vec{v_3}}{\vec{u_1}^T\vec{u_1}}, \widehat{x}_{23}=\frac{\vec{u_2}^T\vec{v_3}}{\vec{u_2}^T\vec{u_2}}\text{.} \end{equation*}
  4. Verify that \(\left\{\vec{u_1},\vec{u_2},\vec{u_3}\right\}\) is orthogonal.
  5. Construct \(R=\left(\begin{array}{ccc}1\amp \widehat{x}_{12}\amp \widehat{x}_{13}\\0\amp 1\amp \widehat{x}_{23}\\0\amp 0\amp 1 \end{array} \right)\) and \(Q=\begin{pmatrix}|\amp |\amp \cdots\amp |\\\vec{u_1}\amp \vec{u_2}\amp \cdots\amp \vec{u_n}\\|\amp |\amp \cdots\amp | \end{pmatrix}\) and verify that \(A=\left(\begin{array}{rrr}2\amp 8\amp 16\\20\amp 8\amp 34\\52\amp -26\amp -34 \end{array} \right)=QR\text{.}\)
  6. Repeat your work for \(T:=\left\{\vec{v_1},\vec{v_2},\vec{v_3}\right\}=\left\{\left(\begin{array}{r}-8\\-8\\4 \end{array} \right), \left(\begin{array}{rrr}-12\\0\\12 \end{array} \right)\left(\begin{array}{rrr}20\\20\\-28 \end{array} \right)\right\}\text{.}\)
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