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Chapter 6: Q13E (page 331)

Let A be any invertible \(n \times n\) matrix. Show that for u, v in \({\mathbb{R}^n}\), the formula \(\left\langle {{\bf{u}},{\bf{v}}} \right\rangle = \left( {A{\bf{u}}} \right) \cdot \left( {A{\bf{v}}} \right) = {\left( {A{\bf{u}}} \right)^T}\left( {A{\bf{v}}} \right)\) defines an inner product on \({\mathbb{R}^n}\).

Short Answer

Expert verified

The formula \(\left\langle {{\bf{u}},{\bf{v}}} \right\rangle = \left( {A{\bf{u}}} \right) \cdot \left( {A{\bf{v}}} \right) = {\left( {A{\bf{u}}} \right)^T}\left( {A{\bf{v}}} \right)\) defines an inner product on \({\mathbb{R}^n}\).

Step by step solution

01

Check for the first Axiom

Consider the following expression:

\(\begin{aligned}\left\langle {{\bf{u}},{\bf{v}}} \right\rangle &= \left( {A{\bf{u}}} \right) \cdot \left( {A{\bf{v}}} \right)\\ &= \left( {A{\bf{v}}} \right) \cdot \left( {A{\bf{u}}} \right)\\ &= \left\langle {{\bf{v}},{\bf{u}}} \right\rangle \end{aligned}\)

Thus, Axiom (1) is proved.

02

Check for the second Axiom

Consider the following expression:

\(\begin{aligned}\left\langle {{\bf{u}} + {\bf{v}},{\bf{w}}} \right\rangle &= \left[ {A\left( {{\bf{u}} + {\bf{v}}} \right)} \right] \cdot \left( {A{\bf{w}}} \right)\\ &= \left( {A{\bf{u}} + A{\bf{v}}} \right) \cdot \left( {A{\bf{w}}} \right)\\ &= \left( {A{\bf{u}}} \right) \cdot \left( {A{\bf{w}}} \right) + \left( {A{\bf{v}}} \right) \cdot \left( {A{\bf{w}}} \right)\\ &= \left\langle {{\bf{u}},{\bf{w}}} \right\rangle + \left\langle {{\bf{v}},{\bf{w}}} \right\rangle \end{aligned}\)

Thus, Axiom (2) is proved.

03

Check for the third Axiom

Consider the following expression:

\(\begin{aligned}\left\langle {c{\bf{u}},{\bf{v}}} \right\rangle &= \left( {Ac{\bf{u}}} \right) \cdot \left( {A{\bf{v}}} \right)\\ &= c\left( {A{\bf{u}}} \right) \cdot \left( {A{\bf{v}}} \right)\\ &= c\left\langle {{\bf{u}},{\bf{v}}} \right\rangle \end{aligned}\)

Thus, Axiom (3) is proved.

04

Check for the fourth Axiom

Consider the following expression:

\(\begin{aligned}\left\langle {{\bf{u}},{\bf{u}}} \right\rangle &= \left( {A{\bf{u}}} \right) \cdot \left( {A{\bf{u}}} \right)\\ &= {\left\| {A{\bf{u}}} \right\|^2}\end{aligned}\)

Here, \({\left\| {A{\bf{u}}} \right\|^2} \ge 0\), and this quantity is zero if the vector \(A{\bf{u}}\) is a zero vector. It also implies that \({\bf{u}} = 0\), since A is an invertible matrix.

Thus, Axiom (4) is proved.

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Most popular questions from this chapter

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

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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 \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\).

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