Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 10: Problem 10

If \(T: V \rightarrow V\) is an isomorphism of the inner product space \(V\), show that $$ \langle\mathbf{v}, \mathbf{w}\rangle_{1}=\langle T(\mathbf{v}), T(\mathbf{w})\rangle $$ defines a new inner product \(\langle,\rangle_{1}\) on \(V\).

Short Answer

Expert verified
The function \( \langle \mathbf{v}, \mathbf{w} \rangle_{1} = \langle T(\mathbf{v}), T(\mathbf{w}) \rangle \) defines a new inner product on \( V \).

Step by step solution

01

Understand the properties of inner products

An inner product on a vector space must satisfy three properties: (1) Linearity in the first argument, (2) Symmetry, and (3) Positive definiteness. We need to show that \( \langle \mathbf{v}, \mathbf{w} \rangle_{1} = \langle T(\mathbf{v}), T(\mathbf{w}) \rangle \) satisfies these conditions.
02

Check linearity in the first argument

To prove linearity, consider vectors \( \mathbf{v}_1, \mathbf{v}_2 \) in \( V \) and scalars \( a, b \). We need to show that \( \langle a\mathbf{v}_1 + b\mathbf{v}_2, \mathbf{w} \rangle_{1} = a \langle \mathbf{v}_1, \mathbf{w} \rangle_{1} + b \langle \mathbf{v}_2, \mathbf{w} \rangle_{1} \). Since \( T \) is linear, we have \( T(a\mathbf{v}_1 + b\mathbf{v}_2) = aT(\mathbf{v}_1) + bT(\mathbf{v}_2) \). Thus, \[ \langle a\mathbf{v}_1 + b\mathbf{v}_2, \mathbf{w} \rangle_{1} = \langle T(a\mathbf{v}_1 + b\mathbf{v}_2), T(\mathbf{w}) \rangle = \langle aT(\mathbf{v}_1) + bT(\mathbf{v}_2), T(\mathbf{w}) \rangle. \] By linearity of the original inner product, \[ \langle aT(\mathbf{v}_1) + bT(\mathbf{v}_2), T(\mathbf{w}) \rangle = a \langle T(\mathbf{v}_1), T(\mathbf{w}) \rangle + b \langle T(\mathbf{v}_2), T(\mathbf{w}) \rangle. \] This proves linearity.
03

Check symmetry

We must show that \( \langle \mathbf{v}, \mathbf{w} \rangle_{1} = \langle \mathbf{w}, \mathbf{v} \rangle_{1} \). Using symmetry of the original inner product: \( \langle T(\mathbf{v}), T(\mathbf{w}) \rangle = \langle T(\mathbf{w}), T(\mathbf{v}) \rangle \), which implies \[ \langle \mathbf{v}, \mathbf{w} \rangle_{1} = \langle T(\mathbf{v}), T(\mathbf{w}) \rangle = \langle T(\mathbf{w}), T(\mathbf{v}) \rangle = \langle \mathbf{w}, \mathbf{v} \rangle_{1}. \] This proves symmetry.
04

Check positive definiteness

Positive definiteness requires \( \langle \mathbf{v}, \mathbf{v} \rangle_{1} > 0 \) for all nonzero \( \mathbf{v} \). Since \( T \) is an isomorphism (and hence bijective), \( T(\mathbf{v}) eq \mathbf{0} \) if \( \mathbf{v} eq \mathbf{0} \). Therefore, the positive definiteness of the original inner product implies \( \langle T(\mathbf{v}), T(\mathbf{v}) \rangle > 0 \). Thus, \( \langle \mathbf{v}, \mathbf{v} \rangle_{1} = \langle T(\mathbf{v}), T(\mathbf{v}) \rangle > 0 \), proving positive definiteness.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Isomorphism
In the context of inner product spaces, an isomorphism is a special kind of linear map. It is both bijective (one-to-one and onto) and preserves the structure of the vector space.
In simpler terms, if you have a map \( T: V \rightarrow V \), this map transforms vectors from the space \( V \) without losing any dimension or information because it can be reversed.
If \( T \) is an isomorphism, then \( T \) and its inverse maintain the vector operations such as addition and scalar multiplication.
  • An isomorphism keeps the essence of operations and properties intact in the new space it maps to.
  • It's like rebranding a flavor – remains the same taste, just with a new packaging.
  • In mathematics, knowing a function is isomorphic means transformations won't distort essential properties.
Linearity
Linearity is a fundamental aspect of inner products, ensuring that operations within the inner product remain consistent. To say that an operation, like an inner product, is linear, means that it behaves predictably when we perform operations like addition and scalar multiplication.
For a map \( T \) that is linear, like an isomorphism, one important idea is that linearity is preserved: \<\( T(a\mathbf{v} + b\mathbf{w}) = aT(\mathbf{v}) + bT(\mathbf{w}) \).
When we say an inner product is linear in one argument, it means:
  • The result is affected directly in a proportional manner when one of the vectors in the product is scaled or changed by addition.
  • Linearity ensures calculations are time-saving and predictable, crucial for complex mathematical problem-solving.
Symmetry
Symmetry in inner product spaces is a straightforward yet elegant property. In essence, it states that the order of the vectors in the inner product does not change the result.
Mathematically, this property is shown as \( \langle \mathbf{v}, \mathbf{w} \rangle = \langle \mathbf{w}, \mathbf{v} \rangle \).
Symmetry is important because it simplifies the handling of inner products and assures consistency, making it possible to swap vectors without concern.
  • Think of symmetry as the attendance of reciprocity in operations, enforcing fairness and order.
  • In practical use, symmetry reduces the risk of error in calculations involving multiple vector operations.
  • This property also highlights that inner product spaces are inherently a step towards understanding more profound symmetries in geometry and physics.
Positive Definiteness
Positive definiteness is crucial in defining a valid inner product, ensuring that the inner product of a vector with itself yields a positive number as long as the vector is nonzero.
This condition mathematically translates to \( \langle \mathbf{v}, \mathbf{v} \rangle > 0 \) for all nonzero \( \mathbf{v} \).
Why does this matter? Positive definiteness guarantees that our space behaves in ways that align with notions of length and distance, making the geometry in the space meaningful.
  • Think of it as guaranteeing there's a measurable, real-world-like sense of distance between points or vectors.
  • This property rules out anomalies where measurements could behave unexpectedly (like having zero or negative distances).
  • For engineers and physicists alike, positive definiteness is foundational when these principles are applied to model real scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In each case, show that \(T\) is symmetric and find an orthonormal basis of eigenvectors of \(T\). a. \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) \(T(a, b, c)=(2 a+2 c, 3 b, 2 a+5 c) ;\) use the dot product b. \(T: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}\) \(T(a, b, c)=(7 a-b,-a+7 b, 2 c) ;\) use the dot product c. \(T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2}\) \(\quad T\left(a+b x+c x^{2}\right)=3 b+(3 a+4 c) x+4 b x^{2}\) inner product \(\left\langle a+b x+c x^{2}, a^{\prime}+b^{\prime} x+c^{\prime} x^{2}\right\rangle=a a^{\prime}+b b^{\prime}+c c^{\prime}\) d. \(T: \mathbf{P}_{2} \rightarrow \mathbf{P}_{2}\) \(\quad T\left(a+b x+c x^{2}\right)=(c-a)+3 b x+(a-c) x^{2} ;\) inner product as in part (c)

Let \(\langle,\rangle\) be an inner product on a vector space \(V\). Show that the corresponding distance function is translation invariant. That is, show that \(\mathrm{d}(\mathbf{v}, \mathbf{w})=\mathrm{d}(\mathbf{v}+\mathbf{u}, \mathbf{w}+\mathbf{u})\) for all \(\mathbf{v}, \mathbf{w},\) and \(\mathbf{u}\) in \(V\)

\(V\) denotes a finite dimensional inner product space. Let \(T: \mathbb{R}^{2} \rightarrow \mathbb{R}^{2}\) be an isometry. A vector \(\mathbf{x}\) in \(\mathbb{R}^{2}\) is said to be fixed by \(T\) if \(T(\mathbf{x})=\mathbf{x}\). Let \(E_{1}\) denote the set of all vectors in \(\mathbb{R}^{2}\) fixed by \(T\). Show that: a. \(E_{1}\) is a subspace of \(\mathbb{R}^{2}\). b. \(E_{1}=\mathbb{R}^{2}\) if and only if \(T=1\) is the identity map. c. \(\operatorname{dim} E_{1}=1\) if and only if \(T\) is a reflection (about the line \(E_{1}\) ). d. \(E_{1}=\\{0\\}\) if and only if \(T\) is a rotation \((T \neq 1)\).

Let \(\mathbb{R}^{3}\) have the inner product \(\left\langle(x, y, z),\left(x^{\prime}, y^{\prime}, z^{\prime}\right)\right\rangle=2 x x^{\prime}+y y^{\prime}+3 z z^{\prime} .\) In each case use the Gram-Schmidt algorithm to transform \(B\) into an orthogonal basis. a. \(B=\\{(1,1,0),(1,0,1),(0,1,1)\\}\) b. \(B=\\{(1,1,1),(1,-1,1),(1,1,0)\\}\)

Exercise \(\mathbf{1 0 . 1 . 1 3}\) In each case, find a symmetric matrix \(A\) such that \(\langle\mathbf{v}, \mathbf{w}\rangle=\mathbf{v}^{T} A \mathbf{w}\) $$ \begin{array}{l} \text { a. }\left\langle\left[\begin{array}{l} v_{1} \\ v_{2} \end{array}\right],\left[\begin{array}{c} w_{1} \\ w_{2} \end{array}\right]\right\rangle=v_{1} w_{1}+2 v_{1} w_{2}+2 v_{2} w_{1}+5 v_{2} w_{2} \\ \text { b. }\left\langle\left[\begin{array}{l} v_{1} \\ v_{2} \end{array}\right],\left[\begin{array}{c} w_{1} \\ w_{2} \end{array}\right]\right\rangle=v_{1} w_{1}-v_{1} w_{2}-v_{2} w_{1}+2 v_{2} w_{2} \end{array} $$ $$ \begin{array}{l} \text { c. }\left\langle\left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \end{array}\right],\left[\begin{array}{l} w_{1} \\ w_{2} \\ w_{3} \end{array}\right]\right\rangle=\begin{aligned} & 2 v_{1} w_{1}+v_{2} w_{2}+v_{3} w_{3}-v_{1} w_{2} \\ &-v_{2} w_{1}+v_{2} w_{3}+v_{3} w_{2} \\ \text { d. }\left\langle\left[\begin{array}{l} v_{1} \\ v_{2} \\ v_{3} \end{array}\right],\left[\begin{array}{l} w_{1} \\ w_{2} \\ w_{3} \end{array}\right]\right\rangle=& v_{1} w_{1}+2 v_{2} w_{2}+5 v_{3} w_{3} \\ &-2 v_{1} w_{3}-2 v_{3} w_{1} \end{aligned} \end{array} $$
See all solutions

Recommended explanations on Economics Textbooks

Taxation Economics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free