Question

Question: In Exercises 1 and 2, you may assume that\(\left\{ {{{\bf{u}}_{\bf{1}}},...,{{\bf{u}}_{\bf{4}}}} \right\}\)is an orthogonal basis for\({\mathbb{R}^{\bf{4}}}\).

2.\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{2}}}\\{\bf{1}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\\{ - {\bf{1}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({\bf{x}} = \left[ {\begin{aligned}{\bf{4}}\\{\bf{5}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right]\)

Write v as the sum of two vectors, one in\({\bf{Span}}\left\{ {{{\bf{u}}_1}} \right\}\)and the other in\({\bf{Span}}\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_{\bf{4}}}} \right\}\).

Short answer

The vector v is given as \(\left[ {\begin{aligned}2\\4\\2\\2\end{aligned}} \right] + \left[ {\begin{aligned}2\\1\\{ - 5}\\1\end{aligned}} \right]\).

Step-by-step solution

Find the orthogonal projection of v on w

The orthogonal projection of v on w can be calculated as follows:

\(\begin{aligned}{\rm{\hat v}} = \frac{{{\bf{v}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1}\\ = \frac{{4 \times 1 + 5 \times 2 - 3 \times 1 + 3 \times 1}}{{{1^2} + {2^2} + {1^1} + {1^2}}}\left[ {\begin{aligned}1\\2\\1\\1\end{aligned}} \right]\\ = \frac{{14}}{7}\left[ {\begin{aligned}1\\2\\1\\1\end{aligned}} \right]\\ = 2\left[ {\begin{aligned}1\\2\\1\\1\end{aligned}} \right]\\ = \left[ {\begin{aligned}2\\4\\2\\2\end{aligned}} \right]\end{aligned}\)

The vector \(\left[ {\begin{aligned}2\\4\\2\\2\end{aligned}} \right]\) is in the span of \({{\bf{u}}_1}\).

Find the orthogonal projection of v on H

The orthogonal vector to the projection on w in H can be calculated as follows:

\(\begin{aligned}{\bf{v}} - {\bf{\hat v}} = \left[ {\begin{aligned}4\\5\\{ - 3}\\3\end{aligned}} \right] - \left[ {\begin{aligned}2\\4\\2\\2\end{aligned}} \right]\\ = \left[ {\begin{aligned}2\\1\\{ - 5}\\1\end{aligned}} \right]\end{aligned}\)

The vector \(\left[ {\begin{aligned}2\\1\\{ - 5}\\1\end{aligned}} \right]\) is in the span of \(\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_4}} \right\}\).

Thus, the vector v can be expressed as \(\left[ {\begin{aligned}2\\4\\2\\2\end{aligned}} \right] + \left[ {\begin{aligned}2\\1\\{ - 5}\\1\end{aligned}} \right]\).