BASIS AND DIMENSION Budi Murtiyasa Muhammadiyah University of Surakarta
Plan Linearly dependence and Linearly Independece Basis and Dimension Sum Space and Intersection Space Row/Column Space
Linearly Dependence and Linearly Independence
linearly dependence and Linearly Independence A vector space V over field F. Vectors u, u, u,, u n V called linearly dependence (bergantung linear) or simply dependent if there are exist scalars a, a, a,, a n F which not all zero such that : a u + a u + + a n u n =
From, a u + a u + + a n u n = if it is only satisfy for all a i = (a = a = = a n = ), then vectors u, u, u,, u n V called linearly independence (bebas linear) or simply independent.
vectors u, v, w R, where : u =, v =, and w = 6. evaluate, the vectors dependent or independent?. solution: x u + y v + z w = -x + y + 5z = x y 6z = x + y z = 5 It can obtained : x = -, y =, and z = - thus : - u + v w = There are exist a scalars not zero, so vectors u, v, and w are dependent.
solution: (using a matrices) w v u = 6 5 u w u v u 5 = 4 4 4 4 ) ( ) 5 ( u v u w u v u = 4 4 From the row echelon form, the last row can be read as : (w + 5u) (v + u) = or : u v + w = There are exist a scalars not zero, so vectors u, v, and w are dependent. Observe that the echelon form have a zero row.
Vectors u, v, w R, where : u =, v =, and w =. evaluate, the vectors dependent or independent?. solution: x u + y v + z w = x + y z = -x + y + z = x y z = Its only obtained: x =, y =, and z = thus: u + v + w = There are only a scalar zero (), So vectors u, v, and w are independent.
Solution: (using a matrices) w v u = u w u v u = 6 ) ( 6 ) ( u v u w u v u = 6 From the row echelon form, there aren t zero row. Vectors u, v, and w are independent. Observe that the echelon form haven t a zero row.
Theorem The nonzero rows of the echelon form of matrices are independent.
Theorem A vectors u, u, u,, u n V called dependent if one of the vectors can be expressed as linear combination of the other vectors.
Notes : If u =, then u must be dependent. If u, then u must be independent. A set of vectors that containing zero vector (), is dependent. A set of vector that containing two vectors which is same or multiplicative, it is dependent. Suppose U V. If U dependent, then V dependent. Suppose W V. If V independent, then W independent. Geometrically, two vectors which is dependent lies on the same line ( same plane).
two vectors dependence y v kv x
Basis and Dimension
Basis and Dimension a vector space V have finite dimensional n (denote dim V = n) if there are independent vectors e, e,, e n V. Set of { e, e,, e n } called basis of V; and number of independent vectors is maximum n. // budi murtiyasa ums surakarta 5
Suppose V = {u, u, u }, where - u = u = - and u = - observe that set {u, u, u } is dependent; hence, set {u, u, u } is not basis of V. But, set {u, u } is independent. Thus, set {u, u } is basis of V. Hence, dim V =. Besides, set {u, u } is also independent, so set {u, u } is basis of V, and dim V =. From the above discussion, it was shown that basis of a vector space is not unique. // budi murtiyasa ums surakarta 6
Because of The nonzero rows of the echelon form of matrices are independent, so algorithm to determine a basis can be done as follow.. Develop a matrix M which rows are the given vectors. Reduce M to echelon form. The non zero rows of the echelon form matrices are the basis.
Suppose a vector space V = {u, v, w, s}, where: u =, v =, w =, and s =. find basis and dimension of V! solution: (using matrices) = ~ ~ Basis of V = {(-,, ) T, (, -, ) T }. Dim V =. 5 8 s w v u 8 5 6 // 8 budi murtiyasa ums surakarta
Find basis and dimension of V = {u, v, w}; if // budi murtiyasa ums surakarta 9
For R n ; with e =, e =,, e n = are basis of R n, and dim R n = n. Basis {e, e,, e n } called natural basis or standard basis. So, natural basis of R are e = and e =. Dim R =. : : : // budi murtiyasa ums surakarta
Theorem Set {u, u,, u n } which is independent from a n-dimensional vector space V is generating system for the vector space V. // budi murtiyasa ums surakarta
Notes : Every generating system which is independent is a basis of a vector space. every set {u, u,, u n } which is independent is a basis of n-dimensional vector space. // budi murtiyasa ums surakarta
example: u, v, w R, with u =, v =, w =. it can be verified that {u, v, w} is dependent; it is also generating system for R. but, set {u, v, w} is not basis for R. meanwhile, set {u, w} is independent, it is also generating system for R. thus, set {u, w} is basis of R. // budi murtiyasa ums surakarta
Theorem If there is n-dimensional vector space V, then every set that contains n+ (or more) of vectors are dependent. // budi murtiyasa ums surakarta 4
example: u, v, w R, with u =, v =, w =. it can be verified that set {u, v, w} is dependent; why?. - V = - - - - Dependent or Independent? // budi murtiyasa ums surakarta 5
Application on Homogenous System Remember for homogenous system AX = which n variables. Dimension the solution space of AX = is n r, where r is rank of coefficient matrix A.
7 Find Basis and dimension for solution space of the system: -x + x x + x 4 = x x + x x 4 = x x + x 8x 4 = Solution: A = 8 ~ 6 ~ r(a) = n = 4 Number free variables= n r = 4 = Free variable are: x and x 4 New equation: - x + x x + x 4 = x x 4 = Suppose x = α, and x 4 = β With α, β are Real number x x 4 = x β = x = - β -x + x x + x 4 = -x + α (-β) + β = x = α + 7β General solution: Basis for the solution space are : 7 7 7 4 x x x x 7 and The dimension = n r = 4 =.
Sum Space
Sum Space If U and W are subspace of V, sum space of U and W are U + W = {u+w u U and w W}. // budi murtiyasa ums surakarta 9
Find basis and dimension of U + W // budi murtiyasa ums surakarta
Solution : U + W = To find basis of U + W, it can be done as follow : // budi murtiyasa ums surakarta
4 5 4 5 9 6 7 4 5 8 8 4 5 4 5 Basis U+W ={(,-,,)T,(,-,5,4)T,(,,,)T,(,,,)T} Dim U+W = 4. ~ ~ ~ ~ // budi murtiyasa ums surakarta H
Theorem If U and W are subspace of V, then U+W are subspace of V. // budi murtiyasa ums surakarta
-- // 4 budi murtiyasa ums surakarta Suppose U and W subspace of V = R 4, where : U= { a + b c = ; a, b, c, d ϵ R} W= { a d = ; b + c = ; a,b,c,d ϵ R} Find basis and dimension of : (i) U (ii) W (iii) U W (iv) U + W d c b a d c b a
Observe that : dim(u + W) = dim(u) + dim(w) dim (U W)
Direct Sum Suppose U and W are subspace of V. The vector space V is direct sum of U and W, denote U W, if every v V can be expressed by one way and the only as v = u + w; where u U and w W. // budi murtiyasa ums surakarta 6
Suppose U and W are subspace of V, where a U = b and W = c a V = b is direct sum of U and W, c 4 4 for example : 8 = 8 + 7 7 // budi murtiyasa ums surakarta 7
Suppose U and W are subspace of V, where a a U = b and W = b, then V = b c c is not direct sum of U and W, because 4 4 For example : 8 = 5 + or 7 7 4 4 8 = + 6 etc. 7 7 // budi murtiyasa ums surakarta 8
Theorem a vector space V is called direct sum of subspace U and W if and only if () U + W = V, and () U W = { }. // budi murtiyasa ums surakarta 9
Row/Column Space
Row space and Coloumn Space For a mxn matrix A, then dimension of row space = dimension of coloumn space. // budi murtiyasa ums surakarta 4
Dimension of row space: ~ H 6 6 ~ ~ There are two rows that non zero, its mean the dimension =. Dimension of coloumn space: ~ K 6 6 ~ 6 ~ There are two coloumns that non zero, its mean, the dimension =. // 4 budi murtiyasa ums surakarta