/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Let \(V\) be a finite-dimensiona... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 12

Let \(V\) be a finite-dimensional vector space with the ordered basis \(\beta\). Prove that \(\psi(\beta)=\beta^{* *}\), where \(\psi\) is defined in Theorem \(2.26\).

Short Answer

Expert verified
The proof is carried out in four steps. Step 1: The linear transformation \(\psi: V \to V^{**}\), is given by \(\psi(v)(f) = f(v)\). Step 2: Given an ordered basis \(\beta = \{v_1,v_2,\dots,v_n\}\) of a finite-dimensional vector space \(V\), the dual basis \(\beta^* = \{f_1,f_2,\dots,f_n\}\) of the dual space \(V^*\) is defined with the property \(f_i(v_j) = \delta_{ij}\). Step 3: The double-dual basis \(\beta^{**} = \{f_1^*,f_2^*,\dots,f_n^*\}\) of the double-dual space \(V^{**}\) is defined with the property \(f_i^*(f_j) = \delta^{*}_{ij}\). Step 4: Upon applying the function \(\psi\) to each basis element \(\beta\), we obtain \(\psi(v_i) = f_i^{**}\) for all \(i\). It follows that \((\psi(v_i))(f_j) = f_i^{**}(f_j)\), which simplifies to \(\delta_{ij}\) as per the definition of the dual basis, verifying \(\psi(\beta) = \beta^{**}\).

Step by step solution

01

Recall Theorem 2.26 and the definition of \(\psi\)

In Theorem 2.26, we have a linear transformation \(\psi: V \to V^{**}\) defined as \[\psi: V \to V^{**},\quad \psi(v)(f) = f(v),\] where \(V^{**}\) denotes the double-dual space of the original vector space \(V\). The function \(\psi\) maps each vector \(v\) in V to a corresponding functional \(\psi(v)\) in \(V^{**}\). The goal is to show that for a basis \(\beta\) of V, the ordered basis \(\psi(\beta) = \beta^{**}\).
02

Find the dual 螔asis

For any ordered basis \(\beta = \{v_1,v_2,\dots,v_n\}\) of a finite-dimensional vector space \(V\), we can find the dual basis \(\beta^* = \{f_1,f_2,\dots,f_n\}\) of the dual space \(V^*\). The dual basis satisfies \(f_i(v_j) = \delta_{ij}\), where \(\delta_{ij}\) is the Kronecker delta: \[\delta_{ij} = \begin{cases} 1 & \text{if } i=j,\\ 0 & \text{if } i \neq j. \end{cases}\]
03

Find the double-dual 螔asis

Next, we need to find the double-dual basis \(\beta^{**} = \{f_1^*,f_2^*,\dots,f_n^*\}\) of the double-dual space \(V^{**}\). The double-dual basis can be constructed from the dual basis \(\beta^*\). Each double-dual basis element \(f_i^*\) satisfies the following property: \[f_i^*(f_j) = \delta^{*}_{ij},\] where the Kronecker delta \(\delta^{*}_{ij}\) is defined similarly to \(\delta_{ij}\).
04

Apply the function \(\psi\) to the ordered basis of \(V\)

Now we will apply the function \(\psi\) to each element of the ordered basis \(\beta\): \[\psi(v_i) = f_i^{**}, \quad i = 1,\dots,n.\] We will show that this application of \(\psi\) gives us the double-dual basis \(\beta^{**}\) of \(V^{**}\). To prove that, we need to show that for all \(i, j = 1,\dots,n\), we have: \[(\psi(v_i))(f_j) = f_i^{**}(f_j).\] By the definition of \(\psi\) (from Step 1), we have \[(\psi(v_i))(f_j) = f_j(v_i).\] According to the property of the dual basis (from Step 2), this expression simplifies to: \[f_j(v_i) = \delta_{ij}.\] Finally, recall that the double-dual basis \(f_i^*\) from Step 3 satisfies \[f_i^*(f_j) = \delta^{*}_{ij}.\] Since \(\delta^{*}_{ij} = \delta_{ij}\), we have proved that the application of \(\psi\) to the ordered basis \(\beta\) gives us the double-dual basis \(\beta^{**}\): \[\psi(\beta) = \beta^{**}.\]

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Suppose \(A\) and \(B\) are orthogonal matrices. Show that \(A^{T}, A^{-1}, A B\) are also orthogonal.

Let \(A\) be an \(n \times n\) matrix. (a) Suppose that \(A^{2}=O .\) Prove that \(A\) is not invertible. (b) Suppose that \(A B=O\) for some nonzero \(n \times n\) matrix \(B\). Could \(A\) be invertible? Explain.

Let \(V\) be a finite-dimensional vector space, and let \(T: V \rightarrow V\) be linear. (a) If \(\operatorname{rank}(T)=\operatorname{rank}\left(T^{2}\right)\), prove that \(R(T) \cap N(T)=\\{0\\}\). Deduce that $\mathrm{V}=\mathrm{R}(\mathrm{T}) \oplus \mathrm{N}(\mathrm{T})$ (see the exercises of Section 1.3). (b) Prove that $\mathrm{V}=\mathrm{R}\left(\mathrm{T}^{k}\right) \oplus \mathrm{N}\left(\mathrm{T}^{k}\right)\( for some positive integer \)k$.

Assume the notation in Theorem \(2.13 .\) (a) Suppose that \(z\) is a (column) vector in \(\mathrm{F}^{p}\). Use Theorem 2.13(b) to prove that \(B z\) is a linear combination of the columns of \(B\). In particular, if \(z=\left(a_{1}, a_{2}, \ldots, a_{p}\right)^{t}\), then show that $$ B z=\sum_{j=1}^{p} a_{j} v_{j} . $$ (b) Extend (a) to prove that column \(j\) of \(A B\) is a linear combination of the columns of \(A\) with the coefficients in the linear combination being the entries of column \(j\) of \(B\). (c) For any row vector \(w \in \mathrm{F}^{m}\), prove that \(w A\) is a linear combination of the rows of \(A\) with the coefficients in the linear combination being the coordinates of \(w\). Hint: Use properties of the transpose operation applied to (a). (d) Prove the analogous result to (b) about rows: Row \(i\) of \(A B\) is a linear combination of the rows of \(B\) with the coefficients in the linear combination being the entries of row \(i\) of \(A\).

Suppose \(U=\left[U_{i k}\right]\) and \(V=\left[V_{k j}\right]\) are block matrices for which \(U V\) is defined and the number of columns of each block \(U_{i k}\) is equal to the number of rows of each block \(V_{k j}\). Show that \(U V=\left[W_{i j}\right]\) where \(W_{i j}=\sum_{k} U_{i k} V_{k j}\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

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.