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Chapter 6: Q1E (page 331)

Let \({\mathbb{R}^{\bf{2}}}\) have the inner product of Example 1, and let \({\bf{x}} = \left( {{\bf{1}},{\bf{1}}} \right)\) and \({\bf{y}} = \left( {{\bf{5}}, - {\bf{1}}} \right)\).

a. Find\(\left\| {\bf{x}} \right\|\),\(\left\| {\bf{y}} \right\|\), and\({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^{\bf{2}}}\).

b. Describe all vectors\(\left( {{z_{\bf{1}}},{z_{\bf{2}}}} \right)\), that are orthogonal to y.

Short Answer

Expert verified

(a) \(\left\| {\bf{x}} \right\| = 3\),\(\left\| {\bf{y}} \right\| = \sqrt {105} \)and \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^2} = 225\)

(b) The set of vectors \(\left( {{z_1},{z_2}} \right)\) are in the form of \(\frac{{{z_1}}}{1} = \frac{{{z_2}}}{4}\).

Step by step solution

01

Find answer for part (a)

Find the value of \(\left\| {\bf{x}} \right\|\).

\(\begin{aligned}\left\| {\bf{x}} \right\| &= \sqrt {\left\langle {{\bf{x}},{\bf{x}}} \right\rangle } \\ &= \sqrt {\left\langle {\left( {1,1} \right),\left( {1,1} \right)} \right\rangle } \\ &= \sqrt {4\left( 1 \right)\left( 1 \right) + 5\left( 1 \right)\left( 1 \right)} \\ &= \sqrt 9 \\ &= 3\end{aligned}\)

Thus, the value of \(\left\| {\bf{x}} \right\|\) is 3.

Find the value of \(\left\| {\bf{y}} \right\|\).

\(\begin{aligned}\left\| {\bf{y}} \right\| &= \sqrt {\left\langle {{\bf{y}},{\bf{y}}} \right\rangle } \\ &= \sqrt {\left\langle {\left( {5, - 1} \right),\left( {5, - 1} \right)} \right\rangle } \\ &= \sqrt {4\left( 5 \right)\left( 5 \right) + 5\left( { - 1} \right)\left( { - 1} \right)} \\ &= \sqrt {105} \end{aligned}\)

Thus, the value of \(\left\| {\bf{y}} \right\|\) is \(\sqrt {105} \).

Find the value of \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^2}\).

\(\begin{aligned}{\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^2} &= {\left| {\left\langle {\left( {1,1} \right),\left( {5, - 1} \right)} \right\rangle } \right|^2}\\ &= {\left| {4\left( 1 \right)\left( 5 \right) + 5\left( 1 \right)\left( { - 1} \right)} \right|^2}\\ &= {\left| {15} \right|^2}\\ &= 225\end{aligned}\)

Thus, the value of \({\left| {\left\langle {{\bf{x}},{\bf{y}}} \right\rangle } \right|^2}\) is 225.

02

Find answer for part (b)

The vectors \(\left( {{z_1},{z_2}} \right)\) are orthogonal to \(y\), if and only if \(\left\langle {z,{\bf{y}}} \right\rangle = 0\). Therefore,

\(\begin{aligned}\left\langle {z,{\bf{y}}} \right\rangle &= 0\\\left\langle {\left( {{z_1},{z_2}} \right),\left( {5, - 1} \right)} \right\rangle &= 0\\4\left( {{z_1}} \right)\left( 5 \right) + 5\left( { - 1} \right)\left( {{z_2}} \right) &= 0\\20{z_1} &= 5{z_2}\\\frac{{{z_1}}}{1} &= \frac{{{z_2}}}{4}\end{aligned}\)

Thus, the set of vectors \(\left( {{z_1},{z_2}} \right)\) are in the form of \(\frac{{{z_1}}}{1} = \frac{{{z_2}}}{4}\).

So, all the multiples of the vector \(\left( {\begin{aligned}1\\4\end{aligned}} \right)\) are orthogonal to the vector y.

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Most popular questions from this chapter

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

Let \({\bf{u}} = \left( {\begin{aligned}5\\{ - 6}\\7\end{aligned}} \right)\), and let \(W\) be the set of all \({\bf{x}}\) in \({\mathbb{R}^3}\) such that \({\bf{u}} \cdot {\bf{x}} = 0\). What theorem in Chapter 4 can be used to show that \(W\) is a subspace of \({\mathbb{R}^3}\)? Describe \(W\) in geometric language.

Question 16: Let \({\mathop{\rm y}\nolimits} = \left( {\begin{array}{*{20}{c}}{ - 3}\\9\end{array}} \right)\), and \({\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right)\). Compute the distance from y to the line through u and the origin.

In Exercises 9-12 find (a) the orthogonal projection of b onto \({\bf{Col}}A\) and (b) a least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

12. \(A = \left[ {\begin{array}{{}{}}{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{ - {\bf{1}}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}\\{ - {\bf{1}}}&{\bf{1}}&{ - {\bf{1}}}\end{array}} \right]\), \({\bf{b}} = \left( {\begin{array}{{}{}}{\bf{2}}\\{\bf{5}}\\{\bf{6}}\\{\bf{6}}\end{array}} \right)\)

Suppose the x-coordinates of the data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\) are in mean deviation form, so that \(\sum {{x_i}} = 0\). Show that if \(X\) is the design matrix for the least-squares line in this case, then \({X^T}X\) is a diagonal matrix.
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