/*! 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 Use mathematical induction to pr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 12

Use mathematical induction to prove that if \(L\) is a linear transformation from \(V\) to \(W\), then \\[ \begin{array}{l} L\left(\alpha_{1} \mathbf{v}_{1}+\alpha_{2} \mathbf{v}_{2}+\cdots+\alpha_{n} \mathbf{v}_{n}\right) \\ \quad=\alpha_{1} L\left(\mathbf{v}_{1}\right)+\alpha_{2} L\left(\mathbf{v}_{2}\right)+\cdots+\alpha_{n} L\left(\mathbf{v}_{n}\right) \end{array} \\]

Short Answer

Expert verified
Using mathematical induction, we can prove the statement as follows: 1. Establish the base case for n=1: Since L is a linear transformation, \(L(\alpha_{1} \mathbf{v}_{1}) = \alpha_{1} L(\mathbf{v}_{1})\) holds true. 2. Assume the statement is true for n=k: \(L(\sum_{i=1}^{k}\alpha_{i}\mathbf{v}_{i}) = \sum_{i=1}^{k}\alpha_{i} L(\mathbf{v}_{i})\). 3. Prove the statement for n=k+1: \(L(\sum_{i=1}^{k+1}\alpha_{i}\mathbf{v}_{i}) = L(\sum_{i=1}^{k}\alpha_{i}\mathbf{v}_{i}+\alpha_{k+1}\mathbf{v}_{k+1}) = L(\sum_{i=1}^{k}\alpha_{i}\mathbf{v}_{i}) + L(\alpha_{k+1}\mathbf{v}_{k+1}) = \sum_{i=1}^{k+1}\alpha_{i} L(\mathbf{v}_{i})\). 4. Hence, the statement holds true for all n: \(L(\sum_{i=1}^{n}\alpha_{i}\mathbf{v}_{i}) = \sum_{i=1}^{n}\alpha_{i} L(\mathbf{v}_{i})\).

Step by step solution

01

Base Case (n=1)

When n=1, we have: \[ L\left(\alpha_{1} \mathbf{v}_{1}\right) = \alpha_{1} L\left(\mathbf{v}_{1}\right) \] This is true since L is a linear transformation. By definition, a linear transformation satisfies the two properties: 1. Additivity: \(L(\mathbf{v}+\mathbf{w})=L(\mathbf{v})+L(\mathbf{w})\) 2. Scalar multiplication: \(L(c \mathbf{v})=c L(\mathbf{v})\) Since our base case is a scalar multiplication, it holds true for 饾憶=1.
02

Inductive Hypothesis

Assume the statement is true for n=k: \[ L\left(\alpha_{1} \mathbf{v}_{1}+\alpha_{2} \mathbf{v}_{2}+\cdots+\alpha_{k}\mathbf{v}_{k}\right) =\alpha_{1} L\left(\mathbf{v}_{1}\right)+\alpha_{2} L\left(\mathbf{v}_{2}\right)+\cdots+\alpha_{k} L\left(\mathbf{v}_{k}\right) \]
03

Inductive Step (n=k+1)

Now, we need to prove the statement for n=k+1: \[ L\left(\alpha_{1} \mathbf{v}_{1}+\alpha_{2} \mathbf{v}_{2}+\cdots+\alpha_{k}\mathbf{v}_{k} + \alpha_{k+1} \mathbf{v}_{k+1} \right) =\alpha_{1} L\left(\mathbf{v}_{1}\right)+\alpha_{2} L\left(\mathbf{v}_{2}\right)+\cdots+\alpha_{k} L\left(\mathbf{v}_{k}\right) + \alpha_{k+1} L\left(\mathbf{v}_{k+1}\right) \] Using the additivity property of the linear transformation and the inductive hypothesis, we get: \[ \begin{array}{l} L\left(\alpha_{1} \mathbf{v}_{1}+\alpha_{2} \mathbf{v}_{2}+\cdots+\alpha_{k} \mathbf{v}_{k}+\alpha_{k+1} \mathbf{v}_{k+1}\right) \\ \quad=L\left(\left(\alpha_{1} \mathbf{v}_{1}+\alpha_{2} \mathbf{v}_{2}+\cdots+\alpha_{k} \mathbf{v}_{k}\right)+\alpha_{k+1} \mathbf{v}_{k+1}\right)\\ \quad=L\left(\alpha_{1} \mathbf{v}_{1}+\alpha_{2} \mathbf{v}_{2}+\cdots+\alpha_{k} \mathbf{v}_{k}\right) + L\left(\alpha_{k+1} \mathbf{v}_{k+1}\right)\\ \quad=\alpha_{1} L\left(\mathbf{v}_{1}\right)+\alpha_{2} L\left(\mathbf{v}_{2}\right)+\cdots+\alpha_{k} L\left(\mathbf{v}_{k}\right) +\alpha_{k+1} L\left(\mathbf{v}_{k+1}\right) \end{array} \] Thus, the statement holds true for n=k+1.
04

Conclusion

By the principle of mathematical induction, the statement is proven to hold for all n: \[ L\left(\alpha_{1} \mathbf{v}_{1}+\alpha_{2} \mathbf{v}_{2}+\cdots+\alpha_{n}\mathbf{v}_{n}\right) =\alpha_{1} L\left(\mathbf{v}_{1}\right)+\alpha_{2} L\left(\mathbf{v}_{2}\right)+\cdots+\alpha_{n} L\left(\mathbf{v}_{n}\right) \]

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

Let a be a fixed nonzero vector in \(\mathbb{R}^{2} .\) A mapping of the form \\[ L(\mathbf{x})=\mathbf{x}+\mathbf{a} \\] is called a translation. Show that a translation is not a linear operator. Illustrate geometrically the effect of a translation.

Determine whether the following are linear transformations from \(C[0,1]\) into \(\mathbb{R}^{1}:\) (a) \(L(f)=f(0)\) (b) \(L(f)=|f(0)|\) (c) \(L(f)=[f(0)+f(1)] / 2\) (d) \(L(f)=\left\\{\int_{0}^{1}[f(x)]^{2} d x\right\\}^{1 / 2}\)

Determine the matrix representation of each of the following composite transformations. (a) A yaw of \(90^{\circ}\), followed by a pitch of \(90^{\circ}\) (b) A pitch of \(90^{\circ},\) followed by a yaw of \(90^{\circ}\) (c) A pitch of \(45^{\circ}\), followed by a roll of \(-90^{\circ}\) (d) A roll of \(-90^{\circ},\) followed by a pitch of \(45^{\circ}\) (e) A yaw of \(45^{\circ}\), followed by a pitch of \(-90^{\circ}\) and then a roll of \(-45^{\circ}\) (f) A roll of \(-45^{\circ},\) followed by a pitch of \(-90^{\circ}\) and then a yaw of \(45^{\circ}\)

Let \(L\) be the linear transformation on \(\mathbb{R}^{3}\) defined by \\[ L(\mathbf{x})=\left(\begin{array}{c} 2 x_{1}-x_{2}-x_{3} \\ 2 x_{2}-x_{1}-x_{3} \\ 2 x_{3}-x_{1}-x_{2} \end{array}\right) \\] and let \(A\) be the standard matrix representation of \(L\) (see Exercise 4 of Section 4.2 ). If \(\mathbf{u}_{1}=(1,1,0)^{T}\) \(\mathbf{u}_{2}=(1,0,1)^{T},\) and \(\mathbf{u}_{3}=(0,1,1)^{T},\) then \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) is an ordered basis for \(\mathbb{R}^{3}\) and \(U=\left(\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right)\) is the transition matrix corresponding to a change of basis from \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) to the standard basis \(\left\\{\mathbf{e}_{1}, \mathbf{e}_{2}, \mathbf{e}_{3}\right\\} .\) Determine the matrix \(B\) representing \(L\) with respect to the basis \(\left\\{\mathbf{u}_{1}, \mathbf{u}_{2}, \mathbf{u}_{3}\right\\}\) by calculating \(U^{-1} A U\)

Let \(V\) be the subspace of \(C[a, b]\) spanned by \(1, e^{x}, e^{-x},\) and let \(D\) be the differentiation operator on \(V\) \(\begin{array}{lllll}\text { (a) Find the transition matrix } & S & \text { represent- }\end{array}\) ing the change of coordinates from the ordered basis \(\left[1, e^{x}, e^{-x}\right]\) to the ordered basis \([1, \cosh x, \sinh x] .\left[\cosh x=\frac{1}{2}\left(e^{x}+e^{-x}\right)\right.\) \(\sinh x=\frac{1}{2}\left(e^{x}-e^{-x}\right) \cdot\) (b) Find the matrix \(A\) representing \(D\) with respect to the ordered basis \([1, \cosh x, \sinh x]\) (c) Find the matrix \(B\) representing \(D\) with respect to \(\left[1, e^{x}, e^{-x}\right]\) (d) Verify that \(B=S^{-1} A S\).
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Pure Maths

Read Explanation

Decision 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.