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Chapter 6: 5E (page 331)

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{array}{*{20}{c}}3\\{-2}\\1\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{-1}\\3\\{-3}\\4\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\8\\7\\0\end{array}} \right]\)

Short Answer

Expert verified

The given set is orthogonal.

Step by step solution

01

Definition of an orthogonal set

If \({u_i} \cdot {u_j}\) for \[i \ne j\], then the set of vectors \[\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\} \in {\mathbb{R}^n}\] is said to be orthogonal.

02

Check for orthogonality of vectors

Let the given vectors be, \({u_1} = \left[ {\begin{align}3\\{ - 2}\\1\\3\end{align}} \right]\), \({u_2} = \left[ {\begin{align}{-1}\\{3}\\{-3}\\4\end{align}} \right]\) and \({u_3} = \left[ {\begin{align}3\\8\\7\\0\end{align}} \right]\).

First, find \({u_1} \cdot {u_2}\):

\(\begin{align}{c}{u_1} \cdot {u_2} = \left( 3 \right)\left( { - 1} \right) + \left( { - 2} \right)\left( 3 \right) + \left( 1 \right)\left( { - 3} \right) + \left( 3 \right)\left( 4 \right)\\ = - 3 - 6 - 3 + 12\\ = 0\end{align}\)

Now, find \({u_2} \cdot {u_3}\)::

\(\begin{align}{c}{u_2} \cdot {u_3} = \left( { - 1} \right)\left( 3 \right) + \left( 3 \right)\left( 8 \right) + \left( { - 3} \right)\left( 7 \right) + \left( 4 \right)\left( 0 \right)\\ = - 3 + 24 - 21 + 0\\ = 0\end{align}\)

And find \({u_1} \cdot {u_3}\)::

\(\begin{align}{c}{u_1} \cdot {u_3} = \left( 3 \right)\left( 3 \right) + \left( { - 2} \right)\left( 8 \right) + \left( 1 \right)\left( 7 \right) + \left( 3 \right)\left( 0 \right)\\ = 9 - 16 + 7 + 0\\ = 0\end{align}\)

Since all the pairs are orthogonal hence, the given set is orthogonal.

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Most popular questions from this chapter

In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{array}{*{20}{c}}5\\{ - 4}\\0\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{ - 4}\\1\\{ - 3}\\8\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\3\\5\\{ - 1}\end{array}} \right]\)

Find an orthogonal basis for the column space of each matrix in Exercises 9-12.

12. \(\left( {\begin{aligned}{{}{}}1&3&5\\{ - 1}&{ - 3}&1\\0&2&3\\1&5&2\\1&5&8\end{aligned}} \right)\)

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\[{\bf{y}}\]onto Span\[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\].

5.\[y = \left[ {\begin{aligned}{ - 1}\\2\\6\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}3\\{ - 1}\\2\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\{ - 2}\end{aligned}} \right]\]

In Exercises 17 and 18, all vectors and subspaces are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

17. a.If \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is an orthogonal basis for\(W\), then multiplying

\({v_3}\)by a scalar \(c\) gives a new orthogonal basis \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},c{{\bf{v}}_3}} \right\}\).

b. The Gram–Schmidt process produces from a linearly independent

set \(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)an orthogonal set \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) with the property that for each \(k\), the vectors \({{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}\) span the same subspace as that spanned by \({{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}\).

c. If \(A = QR\), where \(Q\) has orthonormal columns, then \(R = {Q^T}A\).

In Exercises 11 and 12, find the closest point to\[{\bf{y}}\]in the subspace\[W\]spanned by\[{{\bf{v}}_1}\], and\[{{\bf{v}}_2}\].

11.\[y = \left[ {\begin{aligned}3\\1\\5\\1\end{aligned}} \right]\],\[{{\bf{v}}_1} = \left[ {\begin{aligned}3\\1\\{ - 1}\\1\end{aligned}} \right]\],\[{{\bf{v}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]
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