LINEAR COMBINATIONS AND SUBSPACES

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1 CS131 Part II, Linear Algebra and Matrices CS131 Mathematics for Compter Scientists II Note 5 LINEAR COMBINATIONS AND SUBSPACES Linear combinations. In R 2 the vector (5, 3) can be written in the form (5, 3) = 5(1, 0) + 3(0, 1) and also in the form (5, 3) = 1(2, 0) + 3(1, 1). In each case we say that (5, 3) is a linear combination of the two vectors on the right hand side. If, v R 2 and α, β R, then a vector of the form α + βv is a linear combination of and v. Problem. Express the vector (6, 6) as a linear combination of (0, 3) and (2, 1). Soltion. We want to find nmbers α and β with Now from the properties of vectors: (6, 6) = α(0, 3) + β(2, 1). (6, 6) = α(0, 3) + β(2, 1) (6, 6) = (2β, 3α + β) 2β = 6 and 3α + β = 6 β = 3 and α = 1 so we have (6, 6) = 1(0, 3) + 3(2, 1). We can similarly define linear combinations of more than two vectors and of vectors in R n. If 1, 2,..., m are vectors in R n and α 1, α 2,..., α m are real nmbers, then any vector of the form α α α m m is called a linear combination of 1, 2,..., m. Problem. Express the vector (3, 0) as a linear combination of the vectors (1, 1), (1, 0) and (1, 1) in two different ways. 5 1

2 Soltion. We have (3, 0) = α(1, 1) + β(1, 0) + γ(1, 1) (3, 0) = (α + β + γ, α γ) α + β + γ = 3 and α γ = 0 β = 3 2γ, α = γ. Choosing any vale for γ then determines the vales of α and β and gives s a linear combination. There are infinitely many choices of γ and therefore infinitely many ways of writing (3, 0) as a linear combination of the three vectors. We are asked for only two so to make things easy we can take γ = 0 and γ = 1 giving (3, 0) = 0(1, 1)+3(1, 0)+0(1, 1) and (3, 0) = 1(1, 1)+1(1, 0)+1(1, 1). Examples of linear combinations. (1) A linear combination of a single vector v is defined as a mltiple αv (α R) of v. (2) In R 3 if and v are not parallel, then α + βv represents a vertex of a parallelogram having α and βv as sides. Hence a linear combination of and v is a vector in the plane containing, v and 0. Some linear combinations of and v in R 3 : v O α + βv α + βv v α + βv O O (α, β > 0) (0 < α, β < 1) (α < 0 < β) v Problem. Let = (1, 0, 3), v = (0, 2, 0) and w = (0, 3, 1) (i) Find the linear combination 2 + 3v + 4w (ii) express (1, 5, 4) as a linear combination of, v and w (iii) Can (1, 5, 4) be expressed as a linear combination of and v? 5 2

3 Soltion. (i) 2 + 3v + 4w = (2, 0, 6) + (0, 6, 0) + (0, 12, 4) = (2, 18, 10). (ii) We have (1, 5, 4) = α + βv + γw (1, 5, 4) = α(1, 0, 3) + β(0, 2, 0) + γ(0, 3, 1) (1, 5, 4) = (α, 2β + 3γ, 3α + γ) α = 1, 2β + 3γ = 5, 3α + γ = 4 α = 1, γ = 1, β = 1 so (1, 5, 4) = (1, 0, 3) + (0, 2, 0) + (0, 3, 1). (iii) We have (1, 5, 4) = α + βv (1, 5, 4) = α(1, 0, 3) + β(0, 2, 0) = (α, 2β, 3α) α = 1, 2β = 5, 3α = 4 Bt we cannot have both α = 1 and 3α = 4 so these eqations cannot be solved for α and β. Hence (1, 5, 4) cannot be written as a linear combination of and v. Span. If U = { 1, 2,..., m } is a finite set of vectors in R n, then the span of U is the set of all linear combinations of 1, 2,..., m and is denoted by span U. Hence span U = {α α α m m α 1, α 2,..., α m R} Examples. (1) If U = {} contains jst a single vector, then span {} = {α α R} is the set of all mltiples of. (2) In R 2 if U = {(1, 0), (0, 1)} then the span of U is R 2. To see this note that we can write an arbitrary vector (x, y) in R 2 as a linear combination of (1, 0) and (0, 1) as follows: (x, y) = x(1, 0) + y(0, 1). (3) In R 3 the span of the set {(1, 0, 0), (0, 1, 0), (0, 0, 1)} is R 3. (4) In R 3 let = (1, 0, 1) and v = (2, 0, 3). Then α + βv = α(1, 0, 1) + β(2, 0, 3) = (α + 2β, 0, α + 3β) so any linear combination of and v has 0 for its middle component. In fact any vector with middle component 0 is a linear combination of and v. To see this note that for any x, z R (x, 0, z) = α + βv (x, 0, z) = α(1, 0, 1) + β(2, 0, 3) (x, 0, z)) = (α + 2β, 0, α + 3β) α + 2β = x, α + 3β = z β = z x, α = 3x 2z 5 3

4 giving (x, 0, z) = (3x 2z) + (z x)v so any vector whose middle component is 0 is in span {, v}. Hence span {, v} = {(x, 0, z) x, z R}. Geometrically this is the set of all points whose y-coordinate is 0 i.e. the xz-plane. (5) In general, if and v are not parallel, then the span of {, v} is the plane determined by the three points, v and 0. Sbspace. A sbspace of R n is a nonempty sbset S of R n with the following properties: (1), v S + v S; (2) S, λ R λ S. Ths a sbspace is a set which is closed nder addition and scalar mltiplication. It follows by indction on m that if S is a sbspace and 1, 2,..., m S then any linear combination of 1, 2,..., m also belongs to S. We always have two simple examples of sbspaces. The set {0} consisting of jst the zero vector is a sbspace of R n. The set R n itself is a sbspace of R n. Problem. In each of the following cases determine if the set S is a sbspace of R n. (1) S = {(x, y, 0) x, y R} R 3 ; (2) S = {(1, 1)} R 2 ; (3) S = {(x, y) x 2 + y 2 1} R 2. Soltion. (1) To prove this is a sbspace we need to show that the sm of two vectors in S also belongs to S and also that any mltiple of a vector in S also belongs to S. Let, v S then we have = (a, b, 0) and v = (c, d, 0) for some a, b, c, d R. Now + v = (a, b, 0) + (c, d, 0) = (a + c, b + d, 0) which belongs to S. Also for any λ R λ = λ(a, b, 0) = (λ a, λ b, 0) S. This proves that S is a sbspace. (2) The set S = {(1, 1)} is not a sbspace of R 2 since for example we have 5 4

5 (1, 1) S bt the mltiple 2(1, 1) = (2, 2) does not belong to S. (3) The set {(x, y) x 2 + y 2 1} is not a sbspace since, for example, we have = (1, 0) S and v = (0, 1) S bt + v = (1, 1) does not belong to S since = 2 1. Properties of Sbspaces (1) Every sbspace of R n contains the zero vector. (2) If U is a nonempty finite sbset of R n then the span of U is a sbspace of R n and is called the sbspace spanned or generated by U. The first property of a sbspace listed above is often sefl for showing that a certain set cannot be a sbspace. For example the set {(1, 1)} is not a sbspace of R 2 since it does not contain the zero vector. ABSTRACT Content Linear combinations, definitions, span, sbspace, properties of sbspaces. In this Note linear combinations and sbspaces are defined as well as the important concept of span. These entities appear freqently in areas sch as linear algebra and they form an important part of the theory of Least Sqares approximation. History The German mathematician Hermann Grassman ( ) was the first to introdce the ideas of linear combinations, linear independence, sbspace, basis and dimension. Grassman s ideas derived from his attempts to translate geometric concepts abot n-dimensional space into the langage of algebra withot dealing with coordinates. Althogh his ideas were initially difficlt to nderstand, they have become important in fields sch as vector analysis. 5 5