What is a Linear Space/Vector Space?

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Agnes Lee
5 years ago
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1 What is a Linear Space/Vector Space? The terms linear space and vector space mean the same thing and can be used interchangeably. I have used the term linear space in the discussion below because I prefer it, but that is a personal preference. To start with, a linear space consists of a non-empty set, and 2 operations, addition, and multiplication by a scalar. The elements of the set and these operations must satisfy 10 properties which will be discussed below. The elements of the set are traditionally referred to as "vectors" but are not necessarily the type of vectors with which you are probably most familiar (i.e. standard 2- or 3-dimensional vectors such as 3i + 4j or <1, 2, 5>). Some examples of common elements of linear spaces include traditional n- dimensional vectors, functions of a real variable, convergent infinite series, matrices, and sequences of real numbers. Most often, the operations of addition (of elements in the linear space) and multiplication (of the elements in the linear space) by a scalar are the common or standard operations associated with the particular type of element. (For example, if the elements in the linear space are symmetric 4 x 4 matrices, then addition would be defined as standard addition for matrices and scalar multiplication as the common multiplication of a matrix by a scalar.) It should be noted however, that there is no requirement that the two operations resemble traditional addition or multiplication, as long as they satisfy the properties discussed below. If the scalars we are using are real numbers, then the linear space is called a real linear space. If the scalars are complex numbers then the space is called a complex linear space. Thus whether a linear space is a real linear space or a complex linear space is determined entirely by the types of scalars being used, and does not depend upon what the elements of the space are.

2 A linear space consists of a non-empty set, and 2 operations, addition, and multiplication by a scalar. The elements of the set and these operations must satisfy 10 properties. The 10 properties can be conveniently grouped into 3 categories as follow: 2 Closure Properties: Closed under addition. Closed under multiplication by scalars 4 Properties of Addition: Addition is commutative Addition is associative Existence of an additive identity Existence of additive inverses 4 Properties of Multiplication by a Scalar: Multiplication by scalars is associative Distributive law for addition of elements of the set Distributive law for addition of scalars Existence of a multiplicative identity

3 What do these properties mean? Let's call our set of elements V and we'll let u, v, and w represent elements in V (i.e. u, v, and w are vectors). We'll let c and k represent scalars. Closure Properties Closed under addition means that if v and u belong to V, so does v + u. For example, the set of integers is closed under addition because anytime one adds two integers together, the result is still an integer. The set of odd numbers is NOT closed under addition because it is possible to add two odd numbers together and get a number which is not odd. Closed under multiplication by scalars means that if v belongs to V, so does kv. For example, consider the set of Pythagorean triples (i.e. the set consisting of triples (A,B,C) such that A 2 + B 2 = C 2 ). If you take any element of that set and multiply it by k, the new triple will also be a Pythagorean triple and will be an element of the set. That is, if (A,B,C) is an element of the set, so is k(a,b,c) = (ka,kb,kc) because (ka) 2 + (kb) 2 = (kc) 2. The set of integers is NOT closed under scalar multiplication because it is possible to multiply an integer by a scalar and get a result that is not an integer (e.g. if one multiplies the integer 5 by the scalar 1/2, the result is 2.5 which is not an integer). 4 Properties of Addition: Addition is commutative simply means that u + v = v + u. For example the traditional addition of functions is commutative because f(x) + g(x) = g(x) + f(x) for all values of x and so f + g = g + f. However, someone could define the addition of functions differently. If we use the symbol +O to represent this new addition and define it as f(x) +O g(x) = f(x) + g(x), then it is no longer commutative because f(x) +O g(x) = f(x) + g(x) g(x) + f(x) = g(x) +O f(x).

4 Addition is associative means that if one is adding more than 2 elements, it doesn't matter which pair(s) of elements are added first: (u + v) + w = u + (v + w). Existence of an additive identity (i.e. a zero element) means that the set V contains an element "0" such that "0" + v = v for each v in V. [ Or, more concisely, "0" V "0" + v = v, v V.] Since "0" is an element of V it has to be an appropriate type of element. That is to say, if V is a set of 2 by 2 matrices, then "0" would be the 2 x 2 zero matrix. If V were to be a set of infinite sequences, then "0" would be the zero sequence, i.e. 0, 0, 0, 0,... etc. (Actually, in the examples above, whether or not the zero element would be the zero matrix or the zero sequence would also depend upon how addition is defined.) Existence of additive inverses (i.e. negatives) means that the set V contains an element -v such that -v + v = "0" for each v in V. [Or, v V, -v V -v + v = "0".] It should be noted that although in many/most cases -v = (-1)v, this is not necessarily true, depending upon how multiplication by a scalar is defined. 4 Properties of Multiplication by a Scalar: Multiplication by scalars is associative means that (ck)v = c(kv). Distributive law for addition of elements of the set means c(u + v) = cu + cv. Notice that we are adding u and v, elements of V. Distributive law for addition of scalars means (c + k)v = cv + kv. Here we are adding the two scalars c and k. Existence of a multiplicative identity means that there is a scalar "1" such that "1"v = v. Since most of the time the operation of scalar multiplication is standard multiplication by a scalar, "1" is usually just the number 1.

5 Here are a few examples of linear spaces. Unless otherwise specified, addition and multiplication by a scalar are defined in the usual way for the particular elements of the indicated set. The set of ordered triples of real numbers. This is often denoted by 3 and referred to as "real three-dimensional space" or "three-dimensional Euclidean space". Elements of this set are sometimes written as (1, 2, 3) or <1, 2, 3> or i +2j + 3k. The set of all polynomials of degree n, where n is fixed The set of all bounded real sequences The set of all 4 x 4 matrices with real elements The set of all functions f which are integrable on [0,1] with f(x) dx = 0 The set of all vectors (x,y,z) in 3 with y = 5x The set of all absolutely convergent real series Here are some examples which are not linear spaces. As above, unless otherwise specified, addition and multiplication by a scalar are defined in the usual way for the particular elements of the indicated set. Can you find which property/properties is/are not satisfied in each of these examples? The set of all functions defined on [1,3] such that f(1) = 1 + f(3) 1 0 The set of all vectors (x,y,z) in 3 with x = 0 or y = 0 The set of all 2 x 2 matrices with a determinant equal to zero The set of all functions f which are integrable on [0,1] with f(x) dx 0 1 0

6 EXAMPLE: Let V = + the set of positive real numbers Define sum + as x + y = xy, the usual product (x,y V) Define scalar multiplication as c x = x c [note: for scalars, a + b and ab are defined as usual] a) Prove that V is a real linear space with 1 as the zero element b) What is the multiplicative identity in this vector space? c) What corresponds to x (the additive inverse) in this space? It is closed under addition because x + y = xy and xy + It is closed under scalar multiplication because c x = x c and x c + x + y = y + x? x + y = xy (from the definition of + ) xy = yx (standard multiplication is commutative) yx = y + x (from the definition of + ) So + is commutative (x + y) + z = x + (y + z)? (x + y) + z = (xy) z (from the definition of + ) (xy)z = x(yz) (standard multiplication is associative) x(yz) = x + (y + z) (from the definition of + ) So + is associative Is there an additive identity? x + 1 = x1 = x So 1 is the additive identity [i.e. 0 = 1]

7 Does each element of V have an additive inverse in V? x + [(-1) x] = x + [1/x] (from the definition of ) x + [1/x] = x(1/x) = 1 (from the definition of + and standard multiplication) So there is an additive inverse for each x + and it is the reciprocal of x. a (b x) = (ab) x? a (b x) = a (x b ) (from the definition of ) a (x b ) = (x b ) a (from the definition of ) (x b ) a = x ab (standard algebra) x ab = (ab) x (from the definition of ) So a (b x) = (ab) x (Multiplication by a scalar is associative.) a ( x + y) = (a x) + (a y)? a ( x + y) = a (xy) (from the definition of + ) a (xy) = (xy) a (from the definition of ) (xy) a = x a y a (standard algebra) x a y a =(x a ) + (y a ) (from the definition of + ) (x a ) + (y a ) = (a x) + (a y) (from the definition of ) So, a ( x + y) = (a x) + (a y) (There is a distributive law for addition of elements in the set.)

8 (a + b) x = (a x) + (b x)? (a + b) x = x (a+b) (from the definition of ) x (a+b) = x a x b (standard algebra) x a x b = x a + x b (from the definition of + ) x a + x b = (a x) + (b x) (from the definition of ) So, (a + b) x = (a x) + (b x) (There is a distributive law for addition of scalars) Is there a multiplicative identity? 1 x = x 1 = x, so there is a multiplicative identity and it is 1. We have shown that the set + and the operations of + and satisfy all 10 required properties and thus constitute a linear space. In our demonstration of the properties we assumed that the scalars are real and so it is a real linear space. Additionally, in showing that there is an additive identity (i.e. zero element) we found that it equals 1. b) The multiplicative identity is 1, since 1 x = x c) We will let the additive inverse of x be Z and so we need to find Z such that x + Z = 0. x + Z = 0 means that xz = "0" = 1, thus Z = 1/x. So in this linear space, the additive inverse is the reciprocal.

EXERCISE SET 5.1. = (kx + kx + k, ky + ky + k ) = (kx + kx + 1, ky + ky + 1) = ((k + )x + 1, (k + )y + 1)

EXERCISE SET 5.1. = (kx + kx + k, ky + ky + k ) = (kx + kx + 1, ky + ky + 1) = ((k + )x + 1, (k + )y + 1) EXERCISE SET 5. 6. The pair (, 2) is in the set but the pair ( )(, 2) = (, 2) is not because the first component is negative; hence Axiom 6 fails. Axiom 5 also fails. 8. Axioms, 2, 3, 6, 9, and are easily