/*! 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 4 If \(\Sigma\) is linear, then \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 4

If \(\Sigma\) is linear, then \(\Lambda_{\Sigma, 0}\) is linear.

Short Answer

Expert verified
To show that the transformation \(\Lambda_{\Sigma, 0}: V\times W \rightarrow V\times W\), defined by \(\Lambda_{\Sigma, 0}(v, w) = (v, \Sigma(v) + w)\), is linear when \(\Sigma\) is linear, we need to check linearity with respect to both addition and scalar multiplication. For addition, let \((v_1, w_1)\) and \((v_2, w_2) \in V\times W\). We get: \(\Lambda_{\Sigma, 0}((v_1, w_1) + (v_2, w_2)) = \Lambda_{\Sigma, 0}(v_1, w_1) + \Lambda_{\Sigma, 0}(v_2, w_2)\) For scalar multiplication, let \((v,w) \in V\times W\) and \(c\) be a scalar. We get: \(\Lambda_{\Sigma, 0}(c(v, w)) = c\Lambda_{\Sigma, 0}(v, w)\) Since \(\Lambda_{\Sigma, 0}\) satisfies both conditions, it is a linear transformation when \(\Sigma\) is linear.

Step by step solution

01

1. Definitions:

Let's define the given transformation 危 and the corresponding transformation 螞_{危, 0}. We are given that 危: V 鈫 W is linear, where V and W are vector spaces. We need to show that the transformation 螞_{危, 0}: V脳W 鈫 V脳W, defined by 螞_{危, 0}(v, w) = (v, 危(v) + w), is linear.
02

2. Linearity with respect to addition:

To prove the linearity of 螞_{危, 0}, we first need to show that it is linear with respect to addition. Let (v鈧, w鈧) and (v鈧, w鈧) 鈭 V脳W. Then, we have: \( \Lambda_{危, 0}((v鈧, w鈧) + (v鈧, w鈧)) = \Lambda_{\Sigma, 0}(v鈧+v鈧, w鈧+w鈧) = (v鈧+v鈧, 危(v鈧+v鈧) + (w鈧+w鈧)) \) Since 危 is linear, we know that 危(v鈧+v鈧) = 危(v鈧) + 危(v鈧). Hence: \( (v鈧+v鈧, 危(v鈧+v鈧) + (w鈧+w鈧)) = (v鈧+v鈧, (危(v鈧) + 危(v鈧)) + (w鈧+w鈧)) \newline = (v鈧+v鈧, (\Sigma(v鈧) + w鈧) + (\Sigma(v鈧) + w鈧)) = (v鈧, 危(v鈧) + w鈧) + (v鈧, 危(v鈧) + w鈧) \) This shows that 螞_{危, 0} is linear with respect to addition: \( \Lambda_{危, 0}((v鈧, w鈧) + (v鈧, w鈧)) = \Lambda_{\Sigma, 0}(v鈧, w鈧) + \Lambda_{\Sigma, 0}(v鈧, w鈧) \)
03

3. Linearity with respect to scalar multiplication:

Next, we need to show that 螞_{危, 0} is linear with respect to scalar multiplication. Let (v, w) 鈭 V脳W and c 鈭 鈩 (or the respective scalar field). Then, we have: \( \Lambda_{危, 0}(c(v, w)) = \Lambda_{\Sigma, 0}(cv, cw) = (cv, 危(cv) + cw) \) Because 危 is linear, we know that 危(cv) = c危(v). Hence, \( (cv, 危(cv) + cw) = (cv, c危(v) + cw) = c(v, 危(v) + w) = c\Lambda_{\Sigma, 0}(v, w) \) This shows that 螞_{危, 0} is linear with respect to scalar multiplication: \( \Lambda_{\Sigma, 0}(c(v, w)) = c\Lambda_{\Sigma, 0}(v, w) \)
04

Conclusion:

Since we have shown that 螞_{危, 0} is linear with respect to both addition and scalar multiplication, we can conclude that 螞_{危, 0} is a linear transformation when 危 is linear.

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!

Key Concepts

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

Linear Transformations
When you hear linear transformations, think of it as a way to move things around in a space without bending or twisting them. More formally, a linear transformation is a rule that transfers vectors from one vector space to another, while preserving the operations of vector addition and scalar multiplication.

Imagine you have a photo and you want to resize it or rotate it. If you do these changes without distorting the image, you're applying a linear transformation to the picture: scaling and rotating preserve the 'linearity' of the image. Similarly, in mathematics, a linear transformation between vector spaces V and W, for instance, takes any vectors v and u in V and a scalar c and makes sure two main rules are followed:
  • Addition: The transformation of the sum of two vectors is the sum of their transformations, so T(v + u) = T(v) + T(u).
  • Scalar multiplication: The transformation of a vector multiplied by a scalar is the same as the scalar multiplied by the transformation of the vector, so T(c*v) = c*T(v).
These rules make complex problems manageable and predictable without surprises 鈥 which is why we like linear transformations!
Vector Spaces
Imagine having a blank canvas where you can place points, draw lines, and move them around; this canvas is like a vector space. It's a collection of objects, called vectors, along with rules for adding them together and multiplying them by numbers, known as scalars.

Vector spaces are everywhere in mathematics and physics, as they provide the framework for a wide range of phenomena. In your room, the position of objects can be described within a three-dimensional vector space. In a vector space, you can do two main things:
  • Add vectors: If you take two vectors, you can add them to get a third vector. This is just like moving two displacements one after the other when navigating a city.
  • Scale vectors: You can stretch or shrink a vector by multiplying it with a scalar, which is just a regular number. This is akin to adjusting the volume of your music 鈥 the tune (direction) stays the same, but the loudness (magnitude) changes.
Understanding vector spaces is key to mastering concepts in physics, engineering, and even finance, as they form the backdrop for any linear transformation.
Scalar Multiplication
Scalar multiplication is like the zoom function on a camera; it changes the size of the image while keeping its content intact. In mathematics, multiplying a vector by a scalar will either stretch or compress the vector, but won't change its direction (unless, of course, the scalar is negative, in which case the direction is reversed).

Scalar multiplication is a fundamental operation in vector spaces. It allows us to change the magnitude of vectors without affecting their essential properties. Here's what happens when you multiply a vector v by a scalar c:
  • The length of the vector is multiplied by the absolute value of c.
  • If c is positive, the direction of the vector remains the same; if c is negative, the vector's direction is reversed.
  • If c is zero, the resulting vector is the zero vector (which has zero length and no particular direction).
Understanding scalar multiplication is crucial for grasping more complex concepts like linear transformations, as we often use scalars to combine transformations in useful ways.

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

If \(\Sigma\) is linear and \(\Lambda\) is an integral behavior, then $$ \Lambda_{\Sigma, 0}=\Lambda $$ if and only if $$ \widetilde{K}(t, \tau)=C(t) \Phi(t, \tau) B(\tau) $$ for almost all \((t, \tau)\). Thus, the integral behavior \(\Lambda\) is realizable by some linear system if and only if there exist time-varying matrices $$ (A(t), B(t), C(t)) $$ such that the above conditions hold.

The discrete-time system \(\Sigma\) is linear (over the field \(\mathbb{K}\) ) if: \- It is complete; \- \(X\) and \(\mathcal{U}\) are vector spaces; and \- \(\mathcal{P}(t, \cdot \cdot \cdot)\) is linear for each \(t \in \mathbb{Z}\). The system with outputs \(\Sigma\) is linear if in addition: \- \(y\) is a vector space; and \- \(h(t, \cdot)\) is linear for each \(t \in \mathbb{Z}\). The system is finite dimensional if both \(\mathcal{U}\) and \(X\), as well as \(y\) for a system with outputs, are finite dimensional; the dimension of \(\Sigma\) is in that case the dimension of \(\mathcal{X}\).

Exercise 2.7.15 simplifying assumptions, and choosing appropriate units and time scales, we have an equation $$ \ddot{\theta}=\sin \theta-u \cos \theta $$

Consider the ("bilinear") system \(\dot{x}=x u\), where \(\mathcal{X}=\mathcal{U}=\mathbb{R}\). (a) Give the linearized system at \(x=2, u=0\). (b) Idem along the trajectory \(\xi(t)=e^{t}, t \geq 0, \omega \equiv 1\).

If \(\left(\Sigma, x^{0}\right)\) is an initialized continuous-time system, then its behavior \(\Lambda\) is a continuous-time behavior.
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Calculus

Read Explanation

Geometry

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.