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CHAPTER I VECTOR SPACES Department of Foundation Year, Institute of Technology of Cambodia 2014–2015 VECTOR SPACE ITC 1 / 32

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  • CHAPTER IVECTOR SPACES

    Department of Foundation Year,Institute of Technology of Cambodia

    20142015

    VECTOR SPACE ITC 1 / 32

  • Contents

    1 Definitions

    2 Spanning Sets and Linear Independence

    3 Bases and Dimension

    4 Change of Basis

    5 Inner Product Spaces

    6 Gram-Schmidt Orthogonalisation

    VECTOR SPACE ITC 1 / 32

  • Contents

    1 Definitions

    2 Spanning Sets and Linear Independence

    3 Bases and Dimension

    4 Change of Basis

    5 Inner Product Spaces

    6 Gram-Schmidt Orthogonalisation

    VECTOR SPACE ITC 2 / 32

  • Vector Spaces

    Let K denote the set of real numbers or the set of complex numbers.

    Definition 1

    Let V be a nonempty set (whose elements are called vectors) on whichare defined an addition operation (+) and a scalar multiplicationoperation (.) with scalars in K. We call V a vector space over K,provided the following ten conditions are satisfied.

    1 u, v V : u + v V2 u, v V : u + v = v + u3 u, v, w V : (u + v) + w =

    u + (v + w)4 0 V, v V : v + 0 = 05 v V, v V :

    v + (v) = 06 k K, v V : kv V

    7 k K, u, v V :k(u + v) = ku + kv

    8 k1, k2 K, v V :(k1 + k2)v = k1v + k2v

    9 k1, k2 K, v V :(k1k2)v = k1(k2v)

    10 v V : 1v = v.

    VECTOR SPACE ITC 3 / 32

  • Vector Spaces

    In this whole chapter, V will be denoted as a vector space over K.Otherwise, it must be mentioned.

    Theorem 2

    Let V be a vector space over K. We have1 The zero vector is unique.

    2 0v = 0 for all v V .3 k0 = 0 for all scalars k K.4 The additive inverse of each element of V is unique.

    5 For v V : v = (1)v.6 kv = 0 k = 0 or v = 0.

    VECTOR SPACE ITC 4 / 32

  • Subpaces

    Definition 3

    Let 6= S V . If S is a vector space under the same operations ofaddition and scalar multiplication as used in V , then we say that S is asubspace of V .

    Theorem 4

    Let S V . Then S is a subspace of V iff1 S 6= ,2 S is closed under addition and scalar multiplication.

    VECTOR SPACE ITC 5 / 32

  • Subpaces

    Theorem 5

    Let 6= S V . Then S is a subspace of V iff:1 0 S,2 u, v S, k K : u + kv S.

    Theorem 6

    If S1, S2, . . . , Sp are subspaces of V , then S1 S2 Sp is also asubspace of V .

    VECTOR SPACE ITC 6 / 32

  • Contents

    1 Definitions

    2 Spanning Sets and Linear Independence

    3 Bases and Dimension

    4 Change of Basis

    5 Inner Product Spaces

    6 Gram-Schmidt Orthogonalisation

    VECTOR SPACE ITC 7 / 32

  • Subpaces

    Definition 7

    Let v1, v2, . . . , vp V , and k1, k2, . . . , kp K.1 The expression of the form k1v1 + k2v2 + + kpvp is called a

    linear combination of v1, v2, . . . , vp.

    2 The set of all linear combinations of v1, v2, . . . , vp is called thespan of v1, v2, . . . , vp, and denoted by Span{v1, v2, . . . , vp}. Thatis,

    Span{v1, v2, . . . , vp} = {k1v1 + + kpvp : ki K, i = 1, 2, . . . , p}

    3 A subspace S of V is said to be spanned by v1, v2, . . . , vp if

    S = Span{v1, v2, . . . , vp}.

    In this case, {v1, v2, . . . , vp} is called a spanning set for S.VECTOR SPACE ITC 8 / 32

  • Subpaces

    Definition 8

    The vectors v1, v2, . . . , vp V are said to be linearly independent ifthe equation

    k1v1 + k2v2 + + kpvp = 0

    has a unique solution k1 = = kp = 0. Otherwise, v1, v2, . . . , vp aresaid to be linearly dependent.

    Theorem 9

    Let v1, v2, . . . , vp V . Then1 Span{v1, v2, . . . , vp} is a subspace of V and the smallest subspace

    containing {v1, v2, . . . , vp}.2 A vector v Span{v1, v2, . . . , vp} can be written uniquely as a

    linear combination of v1, v2, . . . , vp if and only if v1, v2, . . . , vp arelinearly independent.

    VECTOR SPACE ITC 9 / 32

  • Subpaces

    Theorem 10

    The set {v1, v2, . . . , vp} V is linearly dependent if and only if at leastone of the vectors of the set can be expressed as a linearly combinationof the others.

    VECTOR SPACE ITC 10 / 32

  • Subpaces

    Definition 11

    Let A = (aij) be an m n matrix with entries in K. The row space ofA denoted r(A) is a subspace of Kn spanned by row vectors of A andthe column space of A denoted c(A) is a subspace of Km spanned bycolumn vectors of A.

    Theorem 12

    Two matrices A and B are row-equivalent matrices, then r(A) = r(B).

    Theorem 13

    The set of nonzero row vectors in any row-echelon form of a matrix islinearly independent.

    VECTOR SPACE ITC 11 / 32

  • Contents

    1 Definitions

    2 Spanning Sets and Linear Independence

    3 Bases and Dimension

    4 Change of Basis

    5 Inner Product Spaces

    6 Gram-Schmidt Orthogonalisation

    VECTOR SPACE ITC 12 / 32

  • Bases and Dimension

    Definition 14

    A set {v1, v2, . . . , vn} V is called a basis for V if:1 v1, v2, . . . , vn are linearly independent,

    2 Span{v1, v2, . . . , vn} = V .

    Theorem 15

    If Span{v1, v2, . . . , vn} = V , then any collection of m vectors in V ,where m > n, is linearly dependent.

    Theorem 16

    If {v1, v2, . . . , vn} and {u1, u2, . . . , um} are both bases for V , thenm = n.

    VECTOR SPACE ITC 13 / 32

  • Bases and Dimension

    Definition 17

    The dimension of a vector space V is the number of elements in thebasis. That is, if B = {v1, v2, . . . , vn} is a basis for V , then dimensionof V equals to n. We write,

    dimV = n.

    V is said to be finite-dimensional if there is a finite set of vectorsthat spans V ; otherwise, we say that V is infinite-dimensional.

    VECTOR SPACE ITC 14 / 32

  • Bases and Dimension

    Theorem 18

    Let B = {v1, v2, . . . , vn} V , and dimV = n. The following statementsare equivalent.

    1 B is a basis for V .

    2 B is linearly independent.

    3 B spans V .

    Theorem 19

    If S is a subspace of V and if dimV = n, then dimS n.

    VECTOR SPACE ITC 15 / 32

  • Contents

    1 Definitions

    2 Spanning Sets and Linear Independence

    3 Bases and Dimension

    4 Change of Basis

    5 Inner Product Spaces

    6 Gram-Schmidt Orthogonalisation

    VECTOR SPACE ITC 16 / 32

  • Bases and Dimension

    Theorem 20

    Let B = {v1, v2, . . . , vn} be a basis for V . Then every vector v V canbe written uniquely as a linear combination of v1, v2, . . . , vn. That is,! c1, c2, . . . , cn, such that

    v = c1v1 + c2v2 + + cnvn. (1)

    VECTOR SPACE ITC 17 / 32

  • Bases and Dimension

    Definition 21

    The unique n-tuple (c1, c2, . . . , cn) Kn defined in (1) is called thecoordonates of v relative to the ordered basisB = {v1, v2, . . . , vn}. We denote

    [v]B =

    c1c2...cn

    VECTOR SPACE ITC 18 / 32

  • Bases and Dimension

    Theorem 22

    Let B be an ordered basis for a finite dimensional vector space V . Letu, v V and c K. Then,

    [u + cv]B = [u]B + c[v]B.

    Definition 23

    Let B = {v1, v2, . . . , vn} and C = {w1, w2, . . . , wn} be two orderedbases for a vector space V . We call the n n matrix defined by

    PCB = ([v1]C [vn]C)

    the change-of-basis matrix or transition matrix from orderedbasis B to ordered basis C.

    VECTOR SPACE ITC 19 / 32

  • Bases and Dimension

    Theorem 24

    Let A,B and C be three ordered bases for a finite dimensional vectorspace V and v V . Then

    [v]C = PCB[v]B, PCBPBC = I, and PCA = PCBPBA.

    Theorem 25

    The set of nonzero row vectors in any row-echelon form of an m nmatrix A is a basis for the row space r(A).

    VECTOR SPACE ITC 20 / 32

  • Bases and Dimension

    Definition 26

    Let S1, S2, . . . , Sp and S be subspaces of V .

    1 The sum of S1, S2, . . . , Sp is defined by

    S1 + + Sp = {s1 + + sp : si Si, i = 1, 2, . . . , p}

    2 S is the direct sum of S1, S2, . . . , Sp denoted by

    S = S1 S2 Sp

    if for each s S can be written uniquely as

    s = s1 + s2 + + sp

    where si Si, i = 1, 2, . . . , p.

    VECTOR SPACE ITC 21 / 32

  • Bases and Dimension

    Theorem 27

    Let S1, S2, . . . , Sp be subspaces of V . Then S1 + S2 + + Sp is also asubspace of V .

    Theorem 28

    Let S1, S2 be two subspaces of a finite dimensional vector space V .Then

    dim(S1 + S2) = dimS1 + dimS2 dim(S1 S2)

    VECTOR SPACE ITC 22 / 32

  • Bases and Dimension

    Theorem 29

    Let S1, S2, . . . , Sp be subspaces of V . The following properties areequivalent.

    1 The sum of S1, S2, . . . , Sp is a direct sum

    2 Each s S can be written uniquely as s = s1 + s2 + + sp, wheresi Si, i = 1, 2, . . . , p.

    3 For all i {1, 2, . . . , p}, Si

    1jp

    j 6=i

    Sj

    = {0}4 For all i {1, 2, . . . , p}, Si

    ( 1ji1

    Sj

    )= {0}

    5 For all (s1, s2, . . . , sp) (S1 {0}) (Sp {0}), s1, s2, . . . , spare linearly independent (Suppose that the subspaces S1, S2, . . . , Spare all different from {0}).VECTOR SPACE ITC 23 / 32

  • Bases and Dimension

    Theorem 30

    Let V be a finite-dimensional vector space of n dimensions over K andS1, . . . , Sp be subspaces of V . Let dj = dimSj and

    {bj1, . . . , bjdj

    }be a

    basis for Sj. Then the following statements are equivalent.

    1 S1 + + Sp = S1 Sp.2{b11, . . . , b1d1 , . . . , bp1, . . . , bpdp

    }is a basis for S1 + + Sp.

    3 dim (S1 + + Sp) = dim (S1) + + dim (Sp)

    VECTOR SPACE ITC 24 / 32

  • Contents

    1 Definitions

    2 Spanning Sets and Linear Independence

    3 Bases and Dimension

    4 Change of Basis

    5 Inner Product Spaces

    6 Gram-Schmidt Orthogonalisation

    VECTOR SPACE ITC 25 / 32

  • Inner Product Spaces

    Definition 31

    A map ., . : V V K is called an inner product in V , if itsatisfies the following properties. For all vectors u, v, w V and for allscalars k K,

    1 u, u 0, and u, u = 0 u = 0.2 u, v = v, u.3 ku, v = k u, v .4 u + v, w = u,w+ v, w .

    VECTOR SPACE ITC 26 / 32

  • Inner Product Spaces

    Definition 32

    Any vector space V that is equipped with an inner product u, vis called an inner product space.

    The norm of u is defined in term of the inner product by

    u =u, u.

    The distance between u and v is defined to be the number

    d(u, v) = u v.

    VECTOR SPACE ITC 27 / 32

  • Inner Product Spaces

    Definition 33

    Let V be an inner product space over K.1 Given two vectors u, v 6= 0. The scalar projection of u onto v is

    defined by k = u,vv and the vector projection of u onto v is

    defined by p = k(

    1vv

    )= u,vv2 v.

    2 Two vectors u, v in V is said to be orthogonal if u, v = 0.3 A set of vectors {v1, v2, . . . , vp} in V is called an orthogonal set

    if vi, vj = 0 for i 6= j.4 A vector v V is called a unit vector if v = 1.5 An orthogonal set of unit vectors is called an orthonormal set of

    vectors.

    6 Let dimV = n. A basis {v1, v2, . . . , vn} is called an orthogonalbasis if it is an orthogonal set. It is called an orthonormal basisif it is an orthonormal set.

    VECTOR SPACE ITC 28 / 32

  • Inner Product Spaces

    Theorem 34

    Let V be an inner product space over K. Let u, v V and k K.Then,

    1 kv = |k|v2 u + v u+ v3 | u, v | uv4 u + v2 + u v2 = 2

    (u2 + v2

    )

    VECTOR SPACE ITC 29 / 32

  • Contents

    1 Definitions

    2 Spanning Sets and Linear Independence

    3 Bases and Dimension

    4 Change of Basis

    5 Inner Product Spaces

    6 Gram-Schmidt Orthogonalisation

    VECTOR SPACE ITC 30 / 32

  • Gram-Schmidt Orthogonalisation

    Theorem 35

    If {v1, v2, . . . , vp} of nonzero vectors in an inner product space Vis an orthogonal set then it is linearly independent.

    If {v1, v2, . . . , vn} is an orthogonal basis for a finite-dimensionalinner product space V , then for every vector v V ,

    v =v, v1v12

    v1 +v, v2v22

    v2 + +v, vnvn2

    vn.

    If {v1, v2, . . . , vn} is an orthonormal basis for a finite-dimensionalinner product space V , then for every vector v V ,

    v = v, v1 v1 + v, v2 v2 + + v, vn vn

    VECTOR SPACE ITC 31 / 32

  • Gram-Schmidt Orthogonalisation

    Theorem 36 (Gram-Schmidt Orthogonalisation Process)

    Let {x1, x2, . . . , xp} be an independent set of vectors in afinite-dimensional inner product space V . Then an orthogonal basis fora subspace of V spanned by these vectors is {v1, v2, . . . , vp} where

    v1 = x1, vi = xi i1j=1

    xi, vjvj2

    vj ,

    i = 2, 3, . . . , p.

    VECTOR SPACE ITC 32 / 32

    DefinitionsSpanning Sets and Linear IndependenceBases and DimensionChange of BasisInner Product SpacesGram-Schmidt Orthogonalisation