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

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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
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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 .
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