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

Expert verified

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

01

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 \({{\bf{u}}_1}\).

02

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 \(\left[ {\begin{aligned}2\\4\\2\\2\end{aligned}} \right] + \left[ {\begin{aligned}2\\1\\{ - 5}\\1\end{aligned}} \right]\).

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Most popular questions from this chapter

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

2. \(\left( {\begin{aligned}{{}{}}0\\4\\2\end{aligned}} \right),\left( {\begin{aligned}{{}{}}5\\6\\{ - 7}\end{aligned}} \right)\)

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

6. \(\left( {\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} }}} \right){\mathop{\rm x}\nolimits} \)

To measure the take-off performance of an airplane, the horizontal position of the plane was measured every second, from \(t = 0\) to \(t = 12\). The positions (in feet) were: 0, 8.8, 29.9, 62.0, 104.7, 159.1, 222.0, 294.5, 380.4, 471.1, 571.7, 686.8, 809.2.

a. Find the least-squares cubic curve \(y = {\beta _0} + {\beta _1}t + {\beta _2}{t^2} + {\beta _3}{t^3}\) for these data.

b. Use the result of part (a) to estimate the velocity of the plane when \(t = 4.5\) seconds.

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

3. \(\left( {\begin{aligned}{{}{}}2\\{ - 5}\\1\end{aligned}} \right),\left( {\begin{aligned}{{}{}}4\\{ - 1}\\2\end{aligned}} \right)\)

In Exercises 3–6, verify that\[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\]is an orthogonal set, and then find the orthogonal projection of\[y\]onto Span\[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\].

6.\[{\rm{y}} = \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\]

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