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\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\).

4.\(y = \left[ {\begin{aligned}{\bf{6}}\\{\bf{3}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{3}}\\{\bf{4}}\\{\bf{0}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{4}}}\\{\bf{3}}\\{\bf{0}}\end{aligned}} \right]\)

Short answer

The orthogonal projection is \(\left[ {\begin{aligned}6\\3\\0\end{aligned}} \right]\).

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:

\[\begin{aligned}{c}{{\bf{u}}_1} \cdot {{\bf{u}}_2} &= {\bf{u}}_1^T{{\bf{u}}_2}\\ &= \left[ {\begin{aligned}3&4&0\end{aligned}} \right]\left[ {\begin{aligned}{ - 4}\\3\\0\end{aligned}} \right]\\ &= - 12 + 12\\ &= 0\end{aligned}\]

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:

\(\begin{aligned}{c}{\bf{\hat y}} &= \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \frac{{{\bf{y}} \cdot {{\bf{u}}_{\bf{2}}}}}{{{{\bf{u}}_{\bf{2}}} \cdot {{\bf{u}}_{\bf{2}}}}}{{\bf{u}}_2}\\ &= \frac{{{{\bf{y}}^T} \cdot {{\bf{u}}_1}}}{{{\bf{u}}_{\bf{1}}^T \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \frac{{{{\bf{y}}^T} \cdot {{\bf{u}}_{\bf{2}}}}}{{{\bf{u}}_2^T \cdot {{\bf{u}}_{\bf{2}}}}}{{\bf{u}}_2}\\ &= \frac{{\left[ {\begin{aligned}6&3&{ - 2}\end{aligned}} \right]\left[ {\begin{aligned}3\\4\\0\end{aligned}} \right]}}{{\left[ {\begin{aligned}3&4&0\end{aligned}} \right]\left[ {\begin{aligned}3\\4\\0\end{aligned}} \right]}}{{\bf{u}}_1} + \frac{{\left[ {\begin{aligned}6&3&{ - 2}\end{aligned}} \right]\left[ {\begin{aligned}{ - 4}\\3\\0\end{aligned}} \right]}}{{\left[ {\begin{aligned}{ - 4}&3&0\end{aligned}} \right]\left[ {\begin{aligned}{ - 4}\\3\\0\end{aligned}} \right]}}{{\bf{u}}_2}\\ &= \frac{{30}}{{25}}\left[ {\begin{aligned}3\\4\\0\end{aligned}} \right] - \frac{3}{5}\left[ {\begin{aligned}{ - 4}\\3\\0\end{aligned}} \right]\\ &= \left[ {\begin{aligned}6\\3\\0\end{aligned}} \right]\end{aligned}\)

Thus, the orthogonal projection is \(\left[ {\begin{aligned}6\\3\\0\end{aligned}} \right]\).