Unit13

# Unit13 - Unit 13 Euclidean n-Space and Linear Equations...

This preview shows pages 1–3. Sign up to view the full content.

Unit 13 Euclidean n -Space and Linear Equations Euclidean n -Space: Rfractur n Much of this half of the course will involve objects called vectors. In order to introduce this subject we begin with some definitions and notation: Definition 13.1. Rfractur denotes the set of all real numbers. Rfractur n , called n-space , is defined to be the set of all n -tuples: vectoru = ( u 1 , u 2 , u 3 , ..., u n - 1 , u n ) where u i is some real number, for all values of i , i.e., u i ∈ Rfractur for i = 1 , ..., n . An element of Rfractur n , vectoru , is called an n-vector or just a vector . The number u i is called the i th component of the vector vectoru . We say that two vectors vectoru ∈ Rfractur n and vectorv ∈ Rfractur n are equal , written vectoru = vectorv , if each pair of corresponding components are equal. That is, vectoru = vectorv if and only if u i = v i for every i , i.e. for i = 1 , 2 , ..., n . For instance, for vectoru,vectorv and vectorw ∈ Rfractur 3 , if we have vectoru = (1 , 1 , 2) ,vectorv = (1 , 2 , 1) and vectorw = (1 , 1 , 2), we see that the first components of vectoru and vectorv are identical, but they differ in their second and third components, so vectoru and vectorv are not equal vectors. However, comparing vectoru and vectorw , we see that u 1 = w 1 = 1, u 2 = w 2 = 1 and u 3 = w 3 = 2, so all of the corresponding components of these vectors are equal and we have vectoru = vectorw . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Definition 13.2. A number, i.e., an element of Rfractur , is referred to as a scalar . So 1, 0, π , e and - 5 are all examples of scalars. We have defined two kinds of objects, vectors and scalars. The question arises as to how we may combine these two things (e.g., can we add them together?, multiply them together?, etc.). We now address this question. Definition 13.3. For any vectoru = ( u 1 , u 2 , ..., u n ) ∈ Rfractur n and any c ∈ Rfractur (i.e., vectoru is any vector and c is any scalar), the operation of scalar multiplication of a vector is defined as: cvectoru = ( cu 1 , cu 2 , cu 3 , ..., cu n - 1 , cu n ) That is, to multiply the vector vectoru by the scalar c , we multiply each com- ponent of the vector by that scalar. Example 1 . For vectoru = (1 , 2 , 3), vectorv = (1 , - 3 , 4 , 0) and vectorw = (1 , 1 , 2 , 2 , 3), find 3 vectoru , - 2 vectorv and π vectorw . Solution: 3 vectoru = 3(1 , 2 , 3) = (3 , 6 , 9) - 2 vectorv = - 2(1 , - 3 , 4 , 0) = ( - 2 , 6 , - 8 , 0) and π vectorw = π (1 , 1 , 2 , 2 , 3) = ( π, π, 2 π, 2 π, 3 π ) Definition 13.4. The zero vector in Rfractur n , denoted vector 0, is the n -vector whose components are all zero. For instance, in Rfractur 2 , we have vector 0 = (0 , 0), whereas in Rfractur 3 , vector 0 = (0 , 0 , 0) and for vector 0 ∈ Rfractur 4 , we have vector 0 = (0 , 0 , 0 , 0). Definition 13.5. The operation of vector addition is defined as follows: For any vectoru = ( u 1 , u 2 , u 3 , ..., u n - 1 , u n ) and vectorv = ( v 1 , v 2 , v 3 , ..., v n - 1 , v n ), i.e., any vectoru , vectorv ∈ Rfractur n , we define the sum of the two vectors to be: vectoru + vectorv = ( u 1 + v 1 , u 2 + v 2 , u 3 + v 3 , ..., u n - 1 + v n - 1 , u n + v n ) That is, for vectors in the same space, we can perform the vector addition vectoru + vectorv by adding corresponding components of the vectors vectoru and vectorv .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern