/*! 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 3 Show that (a) \(k(u-v)=k u-k v,\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 3

Show that (a) \(k(u-v)=k u-k v,\) (b) \(u+u=2 u\)

Short Answer

Expert verified
We proved (a) \(k(u-v)=ku-kv\) by showing that: 1. \(k(u-v)=k(u+(-v))\) 2. Applying distributive property: \(k(u+(-v))=ku+k(-v)\) 3. Associative property: \(ku+k(-v)=ku+(-kv)\) 4. Vector subtraction: \(ku+(-kv)=ku-kv\) For (b) \(u+u=2u\), we proved by: 1. Adding equal vectors: \(u+u=(u_1, u_2, u_3)+ (u_1, u_2, u_3)\) 2. Adding components: \((u_1+u_1 , u_2+u_2, u_3+u_3)\) 3. Simplifying: \((2u_1, 2u_2, 2u_3)\) 4. Scalar multiplication: \((2u_1, 2u_2, 2u_3) = 2(u_1, u_2, u_3)\)

Step by step solution

01

(a) Proving \(k(u-v)=ku-kv\)

First, we need to show that the scalar multiple of the difference of two vectors u and v is equal to the difference of the scalar multiples of these vectors. Let's start by looking at the left side of the expression \(k(u-v)\). According to the definition of vector subtraction, the difference of two vectors u and v can be written as \(u-v=u+(-v)\). Now, let's apply the scalar multiplication on this expression: \(k(u-v)=k(u+(-v))\). By the distributive property of scalar multiplication, we have: \(k(u+(-v))=ku+k(-v)\). Since scalar multiplication is associative, we can write: \(ku+k(-v)=ku+(-kv)\). Finally, using the definition of vector subtraction again, we can rewrite this expression as: \(ku+(-kv)=ku-kv\). Thus, we have proved that \(k(u-v)=ku-kv\).
02

(b) Proving \(u+u=2u\)

Now, let's prove that the sum of a vector u with itself is equal to 2 times the vector u. First, let's note that \(u+u\) is simply the addition of two equal vectors. According to the definition of vector addition, we add the components of the two vectors: \(u+u=(u_1, u_2, u_3)+ (u_1, u_2, u_3)\). Now, we can add the components: \((u_1+u_1 , u_2+u_2, u_3+u_3)\). Which can be simplified to: \((2u_1, 2u_2, 2u_3)\). Now, we can use the definition of scalar multiplication to express this vector as a scalar multiple of u: \((2u_1, 2u_2, 2u_3) = 2(u_1, u_2, u_3)\). Thus, we have proved that \(u+u=2u\).

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.

Scalar Multiplication
Scalar multiplication is a fundamental concept in vector algebra. It involves multiplying a vector by a scalar, which is merely a single real number. The effect is that each component of the vector is multiplied by the same scalar.
This operation scales the vector by stretching or shrinking it without altering its direction, unless the scalar is negative. In that case, it also reverses the direction of the vector.
  • For example, if we have a vector \( \mathbf{u} = (u_1, u_2, u_3) \) and a scalar \( k \), the scalar multiplication \( k \mathbf{u} \) is calculated as \( (k \, u_1, k \, u_2, k \, u_3) \).
  • In simple terms, if \( k = 3 \) and \( \mathbf{u} = (1, 2, 3) \), the resulting vector \( 3 \mathbf{u} \) is \( (3, 6, 9) \).
This scaling is at the heart of vector transformations and forms the base of complex vector processes in physics and engineering.
Vector Addition
Vector addition is a crucial aspect of vector algebra. It allows us to combine two vectors to produce a third vector. The process involves adding corresponding components of the vectors. This operation effectively translates one vector to the end of another, acting geometrically like adding steps in a path.
  • Mathematically, if you have vectors \( \mathbf{u} = (u_1, u_2, u_3) \) and \( \mathbf{v} = (v_1, v_2, v_3) \), their addition \( \mathbf{u} + \mathbf{v} \) results in \( (u_1 + v_1, u_2 + v_2, u_3 + v_3) \).
  • For instance, if \( \mathbf{u} = (1, 2, 3) \) and \( \mathbf{v} = (4, 5, 6) \), then \( \mathbf{u} + \mathbf{v} = (5, 7, 9) \).
By understanding vector addition, you can navigate through scenarios such as motion and forces in physics, vector displacement in aircraft navigation, and many more practical applications.
Distributive Property
The distributive property is a key principle in vector algebra that allows us to simplify expressions involving scalar multiplication and vector addition. It states that multiplying a scalar by a sum of vectors equals summing the scalar multiplication of each vector individually.
  • This property can be represented as \( k(\mathbf{u} + \mathbf{v}) = k\mathbf{u} + k\mathbf{v} \), where \( k \) is a scalar, and \( \mathbf{u} \) and \( \mathbf{v} \) are vectors.
  • For example, with \( k = 2 \), \( \mathbf{u} = (1,2) \), and \( \mathbf{v} = (3,4) \), applying the distributive property gives \( 2((1,2) + (3,4)) = 2(1,2) + 2(3,4) = (2,4) + (6,8) = (8,12) \).
Understanding this property is crucial because it helps in simplifying vector equations, making complex vector calculations more manageable.
Associative Property
The associative property is another essential rule in vector algebra that deals with both vector addition and scalar multiplication. It allows you to group vectors or scalars without altering the result of operations.
  • For vector addition, the associative property says that \( (\mathbf{u} + \mathbf{v}) + \mathbf{w} = \mathbf{u} + (\mathbf{v} + \mathbf{w}) \).
  • For instance, with vectors \( \mathbf{u} = (1, 0), \mathbf{v} = (0, 1), \mathbf{w} = (1, 1) \), you will find \( ((1, 0) + (0, 1)) + (1, 1) = (2, 2) \), just as \( (1, 0) + ((0, 1) + (1, 1)) = (2, 2) \).
  • In scalar multiplication, it indicates that any association of scalar multiplication does not affect the result, so \( a(b\mathbf{u}) = (ab)\mathbf{u} \), where \( a \) and \( b \) are scalars and \( \mathbf{u} \) is a vector.
Recognizing these properties helps in arranging and solving complex vector equations in a more seamless manner.

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 \(U\) and \(W\) are subspaces of \(V\) such that \(\operatorname{dim} U=4, \operatorname{dim} W=5\), and \(\operatorname{dim} V=7\). Find the possible dimensions of \(U \cap W\).

Write the polynomial \(f(t)=a t^{2}+b t+c\) as a linear combination of the polynomials \(p_{1}=(t-1)^{2}\), \(p_{2}=t-1, p_{3}=1\). [Thus, \(p_{\downarrow}, p_{2}, p_{3}\) span the space \(\mathbf{P}_{2}(t)\) of polynomials of degree \(\leq 2 .\) ]

Determine whether or not \(u\) and \(v\) are linearly dependent, where (a) \(\quad u=(1,2), v=(3,-5)\) (c) \(u=(1,2,-3), v=(4,5,-6)\) (b) \(u=(1,-3), v=(-2,6)\) (d) \(u=(2,4,-8), v=(3,6,-12)\)

Find a basis and dimension of the subspace \(W\) of \(\mathbf{R}^{3}\) where (a) \(W=\\{(a, b, c): a+b+c=0\\}\), (b) \(W=\\{(a, b, c):(a=b=c)\\}\) (a) Note that \(W \neq \mathbf{R}^{3}\), because, for example, \((1,2,3) \notin W\). Thus, \(\operatorname{dim} W<3\). Note that \(u_{1}=(1,0,-1)\) and \(u_{2}=(0,1,-1)\) are two independent vectors in \(W\). Thus, \(\operatorname{dim} W=2\), and so \(u_{1}\) and \(u_{2}\) form a basis of \(W\) (b) The vector \(u=(1,1,1) \in W\). Any vector \(w \in W\) has the form \(w=(k, k, k)\). Hence, \(w=k u\). Thus, \(u\) spans \(W\) and \(\operatorname{dim} W=1\).

Prove Theorem 4.14: Let \(V\) be a vector space of finite dimension \(n\). Then (i) Any \(n+1\) or more vectors must be linearly dependent. (ii) Any linearly independent set \(S=\left\\{u_{1}, u_{2}, \ldots u_{n}\right\\}\) with \(n\) elements is a basis of \(V\) (iii) Any spanning set \(T=\left\\{v_{1}, v_{2}, \ldots, v_{n}\right\\}\) of \(V\) with \(n\) elements is a basis of \(V\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Calculus

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.