Chapter 14: Chapter 14

Problem 320

Let $\left(\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots \mathrm{x}_{\mathrm{n}}\right),\left(\mathrm{y}_{1}, \mathrm{y}_{2}, \ldots, \mathrm{y}_{\mathrm{n}}\right),\left(\mathrm{z}_{1}, \mathrm{z}_{2}, \ldots \ldots \mathrm{z}_{\mathrm{n}}\right)$ be three vectors in $\mathrm{R}^{\mathrm{n}}$. Verify the following properties of the dot product using these vectors : a) $\mathrm{X} \cdot \mathrm{X} \geq 0 ; \mathrm{X} \cdot \mathrm{X}=0$ if and only if $\mathrm{X}=0$ b) $X \cdot Y=Y \cdot X$ c) $(\mathrm{X}+\mathrm{Y}) \cdot \mathrm{Z}=\mathrm{X} \cdot \mathrm{Y}+\mathrm{X} \cdot \mathrm{Z}$ d) $(\mathrm{CX}) \cdot \mathrm{Y}=\mathrm{X} \cdot(\mathrm{cY})=\mathrm{c}(\mathrm{X} \cdot \mathrm{Y})$

Short Answer

Expert verified

In summary, we have verified the properties of the dot product as follows: a) X.X ≥ 0; X.X = 0 if and only if X = 0: The dot product of a vector with itself is always non-negative, and is 0 if and only if the vector is the zero vector. b) X ∙ Y = Y ∙ X: The dot product is commutative, meaning it does not matter which vector we perform the operation on first. c) (X+Y) ∙ Z = X ∙ Z + Y ∙ Z: The dot product is distributive over vector addition, so we can perform the dot product separately on each vector and then add the results. d) (cX) ∙ Y = X ∙ (cY) = c(X ∙ Y): The dot product remains the same if we multiply one of the vectors by a scalar, and it equals the scalar multipled by the dot product of the original vectors.

See the step by step solution

Step by step solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Verify Property a) X.X ≥ 0; X.X = 0 if and only if X = 0

The dot product of a vector with itself is calculated as follows: \[X \cdot X = x_1x_1 + x_2x_2 + ... + x_nx_n\] Notice that each term in the summation is a square of its corresponding component, which means that each term is non-negative. This means the dot product of a vector with itself must be greater than or equal to 0. Furthermore, if $X \cdot X = 0$, this implies that: \[x_1x_1 + x_2x_2 + ... + x_nx_n = 0\] However, since squares of real numbers are non-negative, the only way this equation can be true is if each term is equal to 0: \[x_1 = x_2 = ... = x_n = 0 \implies X = 0\] Thus, Property a) is verified.

Step 2: Verify Property b) X ∙ Y = Y ∙ X

Let's compute both sides of the equation: \[X \cdot Y = x_1y_1 + x_2y_2 + ... + x_ny_n\] \[Y \cdot X = y_1x_1 + y_2x_2 + ... + y_nx_n\] Since multiplication is commutative, the terms for each respective side of the equation are the same. Therefore, Property b) is verified.

Step 3: Verify Property c) (X+Y) ∙ Z = X ∙ Z + Y ∙ Z

Let's compute both sides of the equation: \[(X+Y) \cdot Z = (x_1+y_1)z_1 + (x_2+y_2)z_2 + ... +(x_n+y_n)z_n\] \[= x_1z_1 + x_2z_2 + ... + x_nz_n + y_1z_1 + y_2z_2 + ... + y_nz_n\] \[X \cdot Z + Y \cdot Z = (x_1z_1 + x_2z_2 + ... + x_nz_n) + (y_1z_1 + y_2z_2 + ... + y_nz_n)\] As we can see, both sides of the equation are equal, thus Property c) is verified.

Step 4: Verify Property d) (cX) ∙ Y = X ∙ (cY) = c(X ∙ Y)

Let's compute both sides of the equation: \[(cX) \cdot Y = (cx_1)y_1 + (cx_2)y_2 + ... + (cx_n)y_n\] \[= c(x_1y_1) + c(x_2y_2) + ... + c(x_ny_n)\] Next, compute $X \cdot (cY)$: \[X \cdot (cY) = x_1(cy_1) + x_2(cy_2) + ... + x_n(cy_n)\] \[= c(x_1y_1) + c(x_2y_2) + ... + c(x_ny_n)\] Now, compute c(X ∙ Y): \[c(X \cdot Y) = c(x_1y_1 + x_2y_2 + ... + x_ny_n)\] \[= c(x_1y_1) + c(x_2y_2) + ... + c(x_ny_n)\] We see that all three expressions are equal, therefore Property d) is verified.

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Recommended explanations on Math Textbooks

Select your language

Chapter 14: Chapter 14

Short Answer

Step by step solution

Unlock all solutions

Step 1: Verify Property a) X.X ≥ 0; X.X = 0 if and only if X = 0

Step 2: Verify Property b) X ∙ Y = Y ∙ X

Step 3: Verify Property c) (X+Y) ∙ Z = X ∙ Z + Y ∙ Z

Step 4: Verify Property d) (cX) ∙ Y = X ∙ (cY) = c(X ∙ Y)

Access millions of textbook solutions in one place

Most popular questions from this chapter

More chapters from the book ‘Linear Algebra’

Chapter 1

Chapter 2

Chapter 3

Chapter 4

Chapter 5

Chapter 6

Chapter 7

Chapter 8

Chapter 9

Chapter 10

Chapter 11

Chapter 12

Chapter 13

Chapter 15

Chapter 16

Chapter 17

Chapter 18

Chapter 19

Chapter 20

Chapter 21

Chapter 22

Chapter 23

Chapter 24

Join over 22 million students in learning with our Vaia App

Recommended explanations on Math Textbooks

Pure Maths

Probability and Statistics

Calculus

Mechanics Maths

Statistics

Decision Maths