Question

Question: In Exercises 3-6, verify that\(\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\)is an orthogonal set, and then find the orthogonal projection of y onto\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\).

3.\[y = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{4}}\\{\bf{3}}\end{aligned}} \right]\],\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{\bf{0}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{0}}\end{aligned}} \right]\)

Short answer

The orthogonal projection is \(\left[ {\begin{aligned}{ - 1}\\4\\0\end{aligned}} \right]\).

Step-by-step solution

Perform the orthogonality test for \({{\bf{u}}_1}\) and \({{\bf{u}}_{\bf{2}}}\)

The orthogonality test for vectors\({{\bf{u}}_1}\)and\({{\bf{u}}_2}\)can be performed as follows:

\(\begin{aligned}{{\bf{u}}_1} \cdot {{\bf{u}}_2} = {\bf{u}}_1^T{{\bf{u}}_2}\\ &= \left[ {\begin{aligned}1&1&0\end{aligned}} \right]\left[ {\begin{aligned}{ - 1}\\1\\0\end{aligned}} \right]\\ &= \left( 1 \right)\left( { - 1} \right) + \left( 1 \right)\left( 1 \right) + \left( 0 \right)\left( 0 \right)\\ &= - 1 + 1\\ &= 0\end{aligned}\)

Thus, the vectors \({{\bf{u}}_1}\) and \({{\bf{u}}_2}\) are orthogonal.

Find the orthogonal component

The orthogonal component in the\({\rm{span}}\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\)is:

\(\begin{aligned}{\bf{\hat y}} &= \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \frac{{{\bf{y}} \cdot {{\bf{u}}_{\bf{2}}}}}{{{{\bf{u}}_{\bf{2}}} \cdot {{\bf{u}}_{\bf{2}}}}}{{\bf{u}}_2}\\ &= \frac{{{{\bf{y}}^T} \cdot {{\bf{u}}_1}}}{{{\bf{u}}_{\bf{1}}^T \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \frac{{{{\bf{y}}^T} \cdot {{\bf{u}}_{\bf{2}}}}}{{{\bf{u}}_2^T \cdot {{\bf{u}}_{\bf{2}}}}}{{\bf{u}}_2}\\ &= \frac{{\left[ {\begin{aligned}{ - 1}&4&3\end{aligned}} \right]\left[ {\begin{aligned}1\\1\\0\end{aligned}} \right]}}{{\left[ {\begin{aligned}1&1&0\end{aligned}} \right]\left[ {\begin{aligned}1\\1\\0\end{aligned}} \right]}}{{\bf{u}}_1} + \frac{{\left[ {\begin{aligned}{ - 1}&4&3\end{aligned}} \right]\left[ {\begin{aligned}{ - 1}\\1\\0\end{aligned}} \right]}}{{\left[ {\begin{aligned}{ - 1}&1&0\end{aligned}} \right]\left[ {\begin{aligned}{ - 1}\\1\\0\end{aligned}} \right]}}{{\bf{u}}_2}\\ &= \frac{{ - 1 + 4}}{{1 + 1}}{{\bf{u}}_1} + \frac{{1 + 4}}{{1 + 1}}{{\bf{u}}_2}\\ &= \frac{3}{2}\left[ {\begin{aligned}1\\1\\0\end{aligned}} \right] + \frac{5}{2}\left[ {\begin{aligned}{ - 1}\\1\\0\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{ - 1}\\4\\0\end{aligned}} \right]\end{aligned}\)

Thus, the orthogonal projection is\(\left[ {\begin{aligned}{ - 1}\\4\\0\end{aligned}} \right]\).