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Use the inner product axioms and other results of this section to verify the statements in Exercises 15–18.

15.\(\left\langle {{\rm{u,}}\,c{\rm{v}}} \right\rangle = c\left\langle {{\rm{u,}}\,{\rm{v}}} \right\rangle \) for all scalars \(c\).

Short Answer

Expert verified

The statement \(\left\langle {{\rm{u}},c{\rm{v}}} \right\rangle = c\left\langle {{\rm{u}},{\rm{v}}} \right\rangle \) is verified.

Step by step solution

01

Apply axiom 1

According to axiom 1, if\({\bf{u}}\)and\({\rm{v}}\)bepair of vectors in a vector space\(V\), then, the inner product on\(V\),relates a real number\(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle \)and satisfies the axiom,\(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle = \left\langle {{\rm{v}},{\rm{u}}} \right\rangle \)for all \({\bf{u}}\),\({\rm{v}}\),\({\rm{w}}\)and scalars\(c\).

Apply axiom 1 to the left side of the given equation, as follows:

\(\left\langle {{\rm{u}},c{\rm{v}}} \right\rangle = \left\langle {c{\rm{v}},{\rm{u}}} \right\rangle \)

02

Apply axiom 3

According to axiom 3, if\({\bf{u}}\)and\({\rm{v}}\)bepair of vectors in a vector space\(V\), then, the inner product on\(V\),relates a real number\(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle \)and satisfies the axiom,\(\left\langle {c{\rm{u}},{\rm{v}}} \right\rangle = c\left\langle {{\rm{u}},{\rm{v}}} \right\rangle \)for all \({\bf{u}}\),\({\rm{v}}\),\({\rm{w}}\)and scalars\(c\).

Apply axiom 3 to the resulting equation, as follows:

\(\begin{aligned}\left\langle {{\rm{u}},c{\rm{v}}} \right\rangle = \left\langle {c{\rm{v}},{\rm{u}}} \right\rangle \\ = c\left\langle {{\rm{v}},{\rm{u}}} \right\rangle \end{aligned}\)

Again, apply axiom 1 to the resulting equation:

\(\begin{aligned}\left\langle {{\rm{u}},c{\rm{v}}} \right\rangle = c\left\langle {{\rm{v}},{\rm{u}}} \right\rangle \\ = c\left\langle {{\rm{u}},{\rm{v}}} \right\rangle \end{aligned}\)

Thus, the statement \(\left\langle {{\rm{u}},c{\rm{v}}} \right\rangle = c\left\langle {{\rm{u}},{\rm{v}}} \right\rangle \) is verified.

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Most popular questions from this chapter

24. Question: In Exercises 23 and 24, all vectors are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

  1. Not every orthogonal set in \({\mathbb{R}^n}\) is linearly independent.
  2. If a set \(S = \left\{ {{{\mathop{\rm u}\nolimits} _1}, \ldots ,{{\mathop{\rm u}\nolimits} _p}} \right\}\) has the property that \({{\mathop{\rm u}\nolimits} _i} \cdot {{\mathop{\rm u}\nolimits} _j} = 0\) whenever \(i \ne j\), then \(S\) is an orthonormal set.
  3. If the columns of a \(m \times n\) matrix A are orthonormal, then the linear mapping \({\mathop{\rm x}\nolimits} \mapsto A{\mathop{\rm x}\nolimits} \) preserves lengths.
  4. The orthogonal projection of y onto v is the same as the orthogonal projection of y onto \(c{\mathop{\rm v}\nolimits} \) whenever \(c \ne 0\).
  5. An orthogonal matrix is invertible.

Suppose \(A = QR\), where \(R\) is an invertible matrix. Showthat \(A\) and \(Q\) have the same column space.

A simple curve that often makes a good model for the variable costs of a company, a function of the sales level \(x\), has the form \(y = {\beta _1}x + {\beta _2}{x^2} + {\beta _3}{x^3}\). There is no constant term because fixed costs are not included.

a. Give the design matrix and the parameter vector for the linear model that leads to a least-squares fit of the equation above, with data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\).

b. Find the least-squares curve of the form above to fit the data \(\left( {4,1.58} \right),\left( {6,2.08} \right),\left( {8,2.5} \right),\left( {10,2.8} \right),\left( {12,3.1} \right),\left( {14,3.4} \right),\left( {16,3.8} \right)\) and \(\left( {18,4.32} \right)\), with values in thousands. If possible, produce a graph that shows the data points and the graph of the cubic approximation.

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

10. \(\left( {\begin{aligned}{{}{}}{ - 1} & 6 & 6 \\ 3 & { - 8}&3\\1&{ - 2}&6\\1&{ - 4}&{ - 3}\end{aligned}} \right)\)

In exercises 7-10, show that {u1, u2} or {u1,u2,u3} is an orthogonal basis for \({\mathbb{R}^2}\) or \({\mathbb{R}^3}\), respectively. Then express x as a linear combination of the u.

7. \[{u_1} = \left[ {\begin{align}2\\{ - 3}\end{align}} \right]\], \[{u_2} = \left[ {\begin{align}6\\4\end{align}} \right]\], and \[x = \left[ {\begin{align}9\\{ - 7}\end{align}} \right]\]

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