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Vcla ppt ch=vector space

The document provides information about a linear algebra and vector calculus assignment for mechanical engineering students at L.D. College of Engineering in Ahmedabad, India. It includes the names of 10 students, an outline of 8 topics to be covered, and sample definitions, examples, and explanations related to those topics, such as definitions of vector spaces and subspaces, linear combinations, linear independence, and span of a set of vectors.

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L.D. College Of Engineering Ahmedabad TOPIC : VECTOR SPACES BRANCH : MECHANICAL DIVISION : B SEM : 2ND ACADEMIC YEAR : 2014-15 LINEAR ALIGEBRA AND VECTOR CALCULAS ACTIVE LEARNING ASSIGNMENT 1
NAME ENROLLMENT NO. TEJAS 140280119081 TRUPAL 140280119082 VATSAL 140280119083 JUGAL 140280119084 DHARMANSHU 140280119085 PRATIK 140280119086 -------- 140280119087 KISHAN 140280119088 MAHARSH 140280119089 NIRAV 140280119090 2
1) Real Vector Spaces 2) Sub Spaces 3) Linear combination 4) Linear independence 5) Span Of Set Of Vectors 6) Basis 7) Dimension 8)Coordinate and change of basis 9)Linear dependence and linear independence of function 3
 Definition and Examples Vector Space
Vector space is a system consisting of a set of generalized vectors and a field of scalars,having the same rules for vector addition and scalar multiplication as physical vectors and scalars. What is Vector Space? Let V be a non empty set of objects on which the operations of addition and multiplication by scalars are defined. If the following axioms are satisfied by all objects u,v,w in V and all scalars k1,k2 then V is called a vector space and the objects in V are called vectors.
 Addition conditions:- 1.If u and v are objects in V then u+v is in V. 2.u+v=v+u 3.u+(v+w) = (u+v)+w 4.There is an object 0 in V, called zero vector , such that 0+u=u+0 for all u in V. 5.For each object u in V, there exists an object -u in V called a negative of u.
6. If K1 is any scalar and u is an object in V, then k1u is in V. 7.k1(u+v) = k1u +k1v 8.If k1,k2 are scalars and u is an object in V, then (k1+k2)u = k1u+k2u. 9.k1(k2u) = (k1k2)u. 10. 1u=u .  Scalar conditions:-
Determine whether the set R+ of all positive real numbers with operations x + y = xy kx = xk. Is a vector space. Example:-
9 Definition: ),,( V : a vector space      VW W  : a non empty subset ),,( W ：a vector space (under the operations of addition and scalar multiplication defined in V)  W is a subspace of V Subspaces If W is a set of one or more vectors in a vector space V, then W is a sub space of V if and only if the following condition hold; a)If u,v are vectors in a W then u+v is in a W. b)If k is any scalar and u is any vector In a W then ku is in W.
10 Every vector space V has at least two subspaces (1)Zero vector space {0} is a subspace of V. (2) V is a subspace of V.  Ex: Subspace of R2    00,(1) 00 originhethrough tLines(2) 2 (3) R • Ex: Subspace of R3 originhethrough tPlanes(3) 3 (4) R    00,0,(1) 00 originhethrough tLines(2) If w1,w2,. . .. wr subspaces of vector space V then the intersection is this subspaces is also subspace of V.
Let W be the set of all 2×2 symmetric matrices. Show that W is a subspace of the vector space M2×2, with the standard operations of matrix addition and scalar multiplication. sapcesvector:2222  MMW Sol: )(Let 221121 AA,AAWA,A TT  )( 21212121 AAAAAAWAW,A TTT  )( kAkAkAWA,Rk TT  22ofsubspaceais  MW )( 21 WAA  )( WkA Ex : (A subspace of M2×2) 11
WBA        10 01 222 ofsubspaceanotis  MW Let W be the set of singular matrices of order 2 Show that W is not a subspace of M2×2 with the standard operations. WB,WA              10 00 00 01 Sol: Ex : (The set of singular matrices is not a subspace of M2×2) 12
13 Linear Combination • A vector V is called a linear combination of the vectors v1,v2,..., vr if it can be expressed in the form as V = k1v1 + k2v2 + ... + krvr where k1, k2, ...., kr are scalars. • Note: If r=1, then V = k1v1. This shows that a vector V is a linear combination of a single vector v1 if it is a scalar multiple of v1.
Example:  Every vector v = (a, b, c) in R3 is expressible as a linear combination of the standard basis vectors i = (1,0,0), j = (0,1,0), k=(0,0,1) since v = (a,b,c) = a(1,0,0) + b(0,1,0) + c(0,0,1)
15 Linear Combination • Example: Consider the vectors u=(1,2,-1) and v=(6,4,2) in R3. Show that w=(9,2,7) is a linear combination of u and v and that w’=(4,-1,8) is not a linear combination of u and v. vuw vuw 23so,2,3 72 242 96 )2,42,6()7,2,9( )2,4,6()1,2,1()7,2,9( 21 21 21 21 212121 21 21        kk kk kk kk kkkkkk kk kk
16 Linear Combination System of equations is inconsistent, so no such scalar k1 and k2 exist. w’ is not a linear combination of u and v. 822 142 46 )22,42,6()8,1,4( )2,4,6()1,2,1()8,1,4( 21 21 21 212121 21 21       kk kk kk kkkkkk kk kk vuw
dependent.linearlycalledisthen zeros),allnot(i.e.,solutionnontrivialahasequationtheIf(2) t.independenlinearlycalledisthen )0(solutiontrivialonly thehasequationtheIf(1) 21 S S ccc k     0vvv vvv   kk k ccc S   2211 21 ,,, : a set of vectors in a vector space V Linear Independent (L.I.) and Linear Dependent (L.D.): Definition: 17 Theorem A set S with two or more vectors is (a) Linearly dependent if and only if at least one of the vectors in S is expressible as a linear combination of the other vectors in S. (b) Linearly independent if and only if no vector in S is expressible as a linear combination of the other vectors in S.
18 tindependenlinearlyis(1)  dependent.linearlyis(2) SS 0   tindependenlinearlyis(3) v0v  21(4) SS  dependentlinearlyisdependentlinearlyis 21 SS  tindependenlinearlyistindependenlinearlyis 12 SS  Notes
      10,2,,21,0,,32,1, S 023 02 02 321 21 31    ccc cc cc  0vvv 332211 ccc Sol: Determine whether the following set of vectors in R 3 is L.I. or L.D.             0123 0012 0201   nEliminatioJordan-Gauss           0100 0010 0001  solutiontrivialonly the0321  ccc tindependenlinearlyisS v1 v2 v3 Ex : Testing for linearly independent 19
SPAN
What is the Spanning Set? Let S = {v1, v2,…, vr } be a set of vectors in a vector space V, then there exists a subspace W of V consisting of all linear combinations of the vectors in S. W is called the space spanned by v1, v2,…, vr. Alternatively, we say that the vectors v1, v2,…, vr span W. Thus, W = span(S) = span {v1, v2,…, vr } and the set S is the spanning set of the subspace W. In short, if every vector in V can be expressed as a linear combinations of the vectors in S, then S is the spanning set of the vector space V.
How to Find the Space Spanned by a Set of Vectors? S = {u, v, w } = {(1,1,2),(-1,3,0),(0,1,2)} is a set of vectors in the vector space ℜ³, and Is Or can we solve for any x? Yes, if A-1 exists. Find det(A) to see if there is a unique solution? If we let W be the subspace of ℜ³ consisting of all linear combinations of the vectors in S, then x ∈ W for any x ∈ ℜ³. Thus, W = span(S) = ℜ³. The span of any subset of a vector space is a subspace span S is the smallest vector space containing all members of S. (x1, x2 , x 3 )  r x W ? x  A r k
23 Ex: A spanning set for R3   sapns)1,0,2(),2,1,0(),3,2,1(setthat theShow 3 RS  .and,,ofncombinatiolinearaasbecanin ),,(vectorarbitraryanwhetherdeterminemustWe 321 3 321 vvv u R uuu Sol: 332211 3 vvvuu cccR  3321 221 131 23 2 2 uccc ucc ucc    .and,,ofvaluesallforconsistentis systemthiswhethergdeterminintoreducesthusproblemThe 321 uuu 0 123 012 201   A u.everyforsolutiononeexactlyhasbx  A 3 )( RSspan 
Consistency of a system of linear equations:  Theorem: The set of equation Ax=B are consistence if and only if the coefficient matrix A and augmented matrix [A|B] have the same rank.
Conditions for consistency of non homogeneous linear equation Ax=B : If rank of [A|B]=rank of (A)=no. of variables , the equation are consistent and have unique solution. If rank of [A|B]=rank of(A)<no. of variables, the equation are consistent and have infinite solution. If rank of [A|B]≠rank of (A),the equation are inconsistent and have no solution.
Condition for consistency of homogeneous linear equation Ax=0 : x=0 is always solution . This solution in which each x1=0,x2=0,x3=0…..xn=0 is called null solution or the trivial solution.  If rank of (A)=number of variable ,the system has only trivial solution. If rank of (A)<no. of variable ,the system has an infinite non-trivial solution.
27 Basis • Definition: V：a vector space Generating Sets Bases Linearly Independent Sets  S is called a basis for V S ={v1, v2, …, vn}V • S spans V (i.e., span(S) = V ) • S is linearly independent (1) Ø is a basis for {0} (2) the standard basis for R3: {i, j, k} i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)  Notes:
28 (3) the standard basis for R n : {e1, e2, …, en} e1=(1,0,…,0), e2=(0,1,…,0), en=(0,0,…,1) Ex: R4 {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)} Ex: matrix space:                               10 00 , 01 00 , 00 10 , 00 01 22 (4) the standard basis for mn matrix space: { Eij | 1im , 1jn } (5) the standard basis for Pn(x): {1, x, x2, …, xn} Ex: P3(x) {1, x, x2, x3}
29 THEOREMS Uniqueness of basis representation If S= {v1,v2,…,vn} is a basis for a vector space V, then every vector in V can be written in one and only one way as a linear combination of vectors in S. If S= {v1,v2,…,vn} is a basis for a vector space V, then every set containing more than n vectors in V is linearly dependent. Bases and linear dependence
30 If a vector space V has one basis with n vectors, then every basis for V has n vectors. (All bases for a finite-dimensional vector space has the same number of vectors.) Number of vectors in a basis An INDEPENDENT set of vectors that SPANS a vector space V is called a BASIS for V.
31 Dimension  Definition: The dimension of a finite dimensional vector space V is defined to be the number of vectors in a basis for V. V: a vector space S: a basis for V Finite dimensional A vector space V is called finite dimensional, if it has a basis consisting of a finite number of elements Infinite dimensional If a vector space V is not finite dimensional,then it is called infinite dimensional. • Dimension of vector space V is denoted by dim(V).
32 Theorems for dimention THEOREM 1 All bases for a finite-dimensional vector space have the same number of vectors. THEOREM 2 Let V be a finite-dimensional vector space, and let be any basis. (a) If a set has more than n vectors, then it is linearly dependent. (b) If a set has fewer than n vectors, then it does not span V.
33 Dimensions of Some Familiar Vector Spaces (1) Vector space Rn  basis {e1 , e2 ,  , en} (2) Vector space Mm  basis {Eij | 1im , 1jn} (3) Vector space Pn(x)  basis {1, x, x2,  , xn} (4) Vector space P(x)  basis {1, x, x2, }  dim(Rn) = n  dim(Mmn)=mn  dim(Pn(x)) = n+1  dim(P(x)) = 
34 Dimension of a Solution Space EXAMPLE
35 Coordinates and change of basis • Coordinate representation relative to a basis Let B = {v1, v2, …, vn} be an ordered basis for a vector space V and let x be a vector in V such that .2211 nnccc vvvx   The scalars c1, c2, …, cn are called the coordinates of x relative to the basis B. The coordinate matrix (or coordinate vector) of x relative to B is the column matrix in Rn whose components are the coordinates of x.                n B c c c  2 1 x
Find the coordinate matrix of x=(1, 2, –1) in R3 relative to the (nonstandard) basis B ' = {u1, u2, u3}={(1, 0, 1), (0, – 1, 2), (2, 3, – 5)} Sol: 2100 8010 5001 1521 2310 1201 E.G.J.                         )5,3,2()2,1,0()1,0,1()1,2,1( 321332211  cccccc uuux                                       1 2 1 521 310 201 i.e. 152 23 12 3 2 1 321 32 31 c c c ccc cc cc              2 8 5 ][ B x Finding a coordinate matrix relative to a nonstandard basis
LINEAR DEPENDENCE AND INDEPENDENCE OF FUNCTIONS If 𝒇 𝟏=𝒇 𝟏 𝒙 , 𝒇 𝟐=𝒇 𝟐 𝒙 ,…, 𝒇 𝒏=𝒇 𝒏 𝒙 Are 𝒏 − 𝟏 times differentiable Functions on the interval −∞, ∞ Then the wronskian of these function is……..
W= 𝒇 𝟏 𝒙 𝒇 𝟐 𝒙 … 𝒇 𝒏 𝒙 𝒇 𝟏 | (𝒙) 𝒇 𝟐 | (𝒙) … 𝒇 𝒏 | (𝒙) ⋮ ⋮ ⋮ 𝒇 𝟏 𝒏−𝟏 (𝒙) 𝒇 𝟐 𝒏−𝟏 (𝒙) … 𝒇 𝒏 𝒏−𝟏 (𝒙) THEOREM: if the wronskian of (n-1) times differentiable functions on the interval −∞, ∞ is not identically zero on this interval then these functions are linearly independent.
• NOTE: If the Wronskian of the functions is identically zero on the interval −∞, ∞ , then no conclusion can be made about the linear dependence or independence of the functions.
EXAMPLES • Which of following set of the function F −∞, ∞ are linearly independent ? (1) x,sinx the wronskian of the functions is W= 𝑥 𝑠𝑖𝑛𝑥 1 𝑐𝑜𝑠𝑥 = xcosx-sinx Since, the function is not zero for all values of x in the interval −∞, ∞ , the given function are linearly independent.
(2) 6,3𝑠𝑖𝑛2 𝑥,2𝑐𝑜𝑠2 𝑥 the wronskian of the function is W= 6 3𝑠𝑖𝑛2 𝑥 2𝑐𝑜𝑠2 𝑥 0 6𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 −4𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥 0 6𝑐𝑜𝑠2𝑥 −4𝑐𝑜𝑠2𝑥 = 0 No conclusion can be made about the linear independence of the functions.
6=6𝑠𝑖𝑛2 𝑥+6𝑐𝑜𝑠2 𝑥 =2(3𝑠𝑖𝑛2 𝑥)+3(2𝑐𝑜𝑠2 𝑥) this shows that 6 can be expressed as a linear combination to given functions, hence the given functions are linearly dependent. NOTE : appropriate can be used directly to show linear dependence without using Wronskian.
43

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Vcla ppt ch=vector space

  • 1.
    L.D. College OfEngineering Ahmedabad TOPIC : VECTOR SPACES BRANCH : MECHANICAL DIVISION : B SEM : 2ND ACADEMIC YEAR : 2014-15 LINEAR ALIGEBRA AND VECTOR CALCULAS ACTIVE LEARNING ASSIGNMENT 1
  • 2.
    NAME ENROLLMENT NO. TEJAS140280119081 TRUPAL 140280119082 VATSAL 140280119083 JUGAL 140280119084 DHARMANSHU 140280119085 PRATIK 140280119086 -------- 140280119087 KISHAN 140280119088 MAHARSH 140280119089 NIRAV 140280119090 2
  • 3.
    1) Real VectorSpaces 2) Sub Spaces 3) Linear combination 4) Linear independence 5) Span Of Set Of Vectors 6) Basis 7) Dimension 8)Coordinate and change of basis 9)Linear dependence and linear independence of function 3
  • 4.
     Definition andExamples Vector Space
  • 5.
    Vector space isa system consisting of a set of generalized vectors and a field of scalars,having the same rules for vector addition and scalar multiplication as physical vectors and scalars. What is Vector Space? Let V be a non empty set of objects on which the operations of addition and multiplication by scalars are defined. If the following axioms are satisfied by all objects u,v,w in V and all scalars k1,k2 then V is called a vector space and the objects in V are called vectors.
  • 6.
     Addition conditions:- 1.Ifu and v are objects in V then u+v is in V. 2.u+v=v+u 3.u+(v+w) = (u+v)+w 4.There is an object 0 in V, called zero vector , such that 0+u=u+0 for all u in V. 5.For each object u in V, there exists an object -u in V called a negative of u.
  • 7.
    6. If K1is any scalar and u is an object in V, then k1u is in V. 7.k1(u+v) = k1u +k1v 8.If k1,k2 are scalars and u is an object in V, then (k1+k2)u = k1u+k2u. 9.k1(k2u) = (k1k2)u. 10. 1u=u .  Scalar conditions:-
  • 8.
    Determine whether theset R+ of all positive real numbers with operations x + y = xy kx = xk. Is a vector space. Example:-
  • 9.
    9 Definition: ),,( V :a vector space      VW W  : a non empty subset ),,( W ：a vector space (under the operations of addition and scalar multiplication defined in V)  W is a subspace of V Subspaces If W is a set of one or more vectors in a vector space V, then W is a sub space of V if and only if the following condition hold; a)If u,v are vectors in a W then u+v is in a W. b)If k is any scalar and u is any vector In a W then ku is in W.
  • 10.
    10 Every vector spaceV has at least two subspaces (1)Zero vector space {0} is a subspace of V. (2) V is a subspace of V.  Ex: Subspace of R2    00,(1) 00 originhethrough tLines(2) 2 (3) R • Ex: Subspace of R3 originhethrough tPlanes(3) 3 (4) R    00,0,(1) 00 originhethrough tLines(2) If w1,w2,. . .. wr subspaces of vector space V then the intersection is this subspaces is also subspace of V.
  • 11.
    Let W bethe set of all 2×2 symmetric matrices. Show that W is a subspace of the vector space M2×2, with the standard operations of matrix addition and scalar multiplication. sapcesvector:2222  MMW Sol: )(Let 221121 AA,AAWA,A TT  )( 21212121 AAAAAAWAW,A TTT  )( kAkAkAWA,Rk TT  22ofsubspaceais  MW )( 21 WAA  )( WkA Ex : (A subspace of M2×2) 11
  • 12.
    WBA        10 01 222 ofsubspaceanotis MW Let W be the set of singular matrices of order 2 Show that W is not a subspace of M2×2 with the standard operations. WB,WA              10 00 00 01 Sol: Ex : (The set of singular matrices is not a subspace of M2×2) 12
  • 13.
    13 Linear Combination • Avector V is called a linear combination of the vectors v1,v2,..., vr if it can be expressed in the form as V = k1v1 + k2v2 + ... + krvr where k1, k2, ...., kr are scalars. • Note: If r=1, then V = k1v1. This shows that a vector V is a linear combination of a single vector v1 if it is a scalar multiple of v1.
  • 14.
    Example:  Every vectorv = (a, b, c) in R3 is expressible as a linear combination of the standard basis vectors i = (1,0,0), j = (0,1,0), k=(0,0,1) since v = (a,b,c) = a(1,0,0) + b(0,1,0) + c(0,0,1)
  • 15.
    15 Linear Combination • Example:Consider the vectors u=(1,2,-1) and v=(6,4,2) in R3. Show that w=(9,2,7) is a linear combination of u and v and that w’=(4,-1,8) is not a linear combination of u and v. vuw vuw 23so,2,3 72 242 96 )2,42,6()7,2,9( )2,4,6()1,2,1()7,2,9( 21 21 21 21 212121 21 21        kk kk kk kk kkkkkk kk kk
  • 16.
    16 Linear Combination System ofequations is inconsistent, so no such scalar k1 and k2 exist. w’ is not a linear combination of u and v. 822 142 46 )22,42,6()8,1,4( )2,4,6()1,2,1()8,1,4( 21 21 21 212121 21 21       kk kk kk kkkkkk kk kk vuw
  • 17.
    dependent.linearlycalledisthen zeros),allnot(i.e.,solutionnontrivialahasequationtheIf(2) t.independenlinearlycalledisthen )0(solutiontrivialonly thehasequationtheIf(1) 21 S S ccck     0vvv vvv   kk k ccc S   2211 21 ,,, : a set of vectors in a vector space V Linear Independent (L.I.) and Linear Dependent (L.D.): Definition: 17 Theorem A set S with two or more vectors is (a) Linearly dependent if and only if at least one of the vectors in S is expressible as a linear combination of the other vectors in S. (b) Linearly independent if and only if no vector in S is expressible as a linear combination of the other vectors in S.
  • 18.
    18 tindependenlinearlyis(1)  dependent.linearlyis(2) SS0   tindependenlinearlyis(3) v0v  21(4) SS  dependentlinearlyisdependentlinearlyis 21 SS  tindependenlinearlyistindependenlinearlyis 12 SS  Notes
  • 19.
         10,2,,21,0,,32,1, S 023 02 02 321 21 31    ccc cc cc  0vvv 332211 ccc Sol: Determine whether the following set of vectors in R 3 is L.I. or L.D.             0123 0012 0201   nEliminatioJordan-Gauss           0100 0010 0001  solutiontrivialonly the0321  ccc tindependenlinearlyisS v1 v2 v3 Ex : Testing for linearly independent 19
  • 20.
  • 21.
    What is theSpanning Set? Let S = {v1, v2,…, vr } be a set of vectors in a vector space V, then there exists a subspace W of V consisting of all linear combinations of the vectors in S. W is called the space spanned by v1, v2,…, vr. Alternatively, we say that the vectors v1, v2,…, vr span W. Thus, W = span(S) = span {v1, v2,…, vr } and the set S is the spanning set of the subspace W. In short, if every vector in V can be expressed as a linear combinations of the vectors in S, then S is the spanning set of the vector space V.
  • 22.
    How to Findthe Space Spanned by a Set of Vectors? S = {u, v, w } = {(1,1,2),(-1,3,0),(0,1,2)} is a set of vectors in the vector space ℜ³, and Is Or can we solve for any x? Yes, if A-1 exists. Find det(A) to see if there is a unique solution? If we let W be the subspace of ℜ³ consisting of all linear combinations of the vectors in S, then x ∈ W for any x ∈ ℜ³. Thus, W = span(S) = ℜ³. The span of any subset of a vector space is a subspace span S is the smallest vector space containing all members of S. (x1, x2 , x 3 )  r x W ? x  A r k
  • 23.
    23 Ex: A spanningset for R3   sapns)1,0,2(),2,1,0(),3,2,1(setthat theShow 3 RS  .and,,ofncombinatiolinearaasbecanin ),,(vectorarbitraryanwhetherdeterminemustWe 321 3 321 vvv u R uuu Sol: 332211 3 vvvuu cccR  3321 221 131 23 2 2 uccc ucc ucc    .and,,ofvaluesallforconsistentis systemthiswhethergdeterminintoreducesthusproblemThe 321 uuu 0 123 012 201   A u.everyforsolutiononeexactlyhasbx  A 3 )( RSspan 
  • 24.
    Consistency of asystem of linear equations:  Theorem: The set of equation Ax=B are consistence if and only if the coefficient matrix A and augmented matrix [A|B] have the same rank.
  • 25.
    Conditions for consistencyof non homogeneous linear equation Ax=B : If rank of [A|B]=rank of (A)=no. of variables , the equation are consistent and have unique solution. If rank of [A|B]=rank of(A)<no. of variables, the equation are consistent and have infinite solution. If rank of [A|B]≠rank of (A),the equation are inconsistent and have no solution.
  • 26.
    Condition for consistencyof homogeneous linear equation Ax=0 : x=0 is always solution . This solution in which each x1=0,x2=0,x3=0…..xn=0 is called null solution or the trivial solution.  If rank of (A)=number of variable ,the system has only trivial solution. If rank of (A)<no. of variable ,the system has an infinite non-trivial solution.
  • 27.
    27 Basis • Definition: V：a vectorspace Generating Sets Bases Linearly Independent Sets  S is called a basis for V S ={v1, v2, …, vn}V • S spans V (i.e., span(S) = V ) • S is linearly independent (1) Ø is a basis for {0} (2) the standard basis for R3: {i, j, k} i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1)  Notes:
  • 28.
    28 (3) the standardbasis for R n : {e1, e2, …, en} e1=(1,0,…,0), e2=(0,1,…,0), en=(0,0,…,1) Ex: R4 {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)} Ex: matrix space:                               10 00 , 01 00 , 00 10 , 00 01 22 (4) the standard basis for mn matrix space: { Eij | 1im , 1jn } (5) the standard basis for Pn(x): {1, x, x2, …, xn} Ex: P3(x) {1, x, x2, x3}
  • 29.
    29 THEOREMS Uniqueness of basisrepresentation If S= {v1,v2,…,vn} is a basis for a vector space V, then every vector in V can be written in one and only one way as a linear combination of vectors in S. If S= {v1,v2,…,vn} is a basis for a vector space V, then every set containing more than n vectors in V is linearly dependent. Bases and linear dependence
  • 30.
    30 If a vectorspace V has one basis with n vectors, then every basis for V has n vectors. (All bases for a finite-dimensional vector space has the same number of vectors.) Number of vectors in a basis An INDEPENDENT set of vectors that SPANS a vector space V is called a BASIS for V.
  • 31.
    31 Dimension  Definition: The dimensionof a finite dimensional vector space V is defined to be the number of vectors in a basis for V. V: a vector space S: a basis for V Finite dimensional A vector space V is called finite dimensional, if it has a basis consisting of a finite number of elements Infinite dimensional If a vector space V is not finite dimensional,then it is called infinite dimensional. • Dimension of vector space V is denoted by dim(V).
  • 32.
    32 Theorems for dimention THEOREM1 All bases for a finite-dimensional vector space have the same number of vectors. THEOREM 2 Let V be a finite-dimensional vector space, and let be any basis. (a) If a set has more than n vectors, then it is linearly dependent. (b) If a set has fewer than n vectors, then it does not span V.
  • 33.
    33 Dimensions of SomeFamiliar Vector Spaces (1) Vector space Rn  basis {e1 , e2 ,  , en} (2) Vector space Mm  basis {Eij | 1im , 1jn} (3) Vector space Pn(x)  basis {1, x, x2,  , xn} (4) Vector space P(x)  basis {1, x, x2, }  dim(Rn) = n  dim(Mmn)=mn  dim(Pn(x)) = n+1  dim(P(x)) = 
  • 34.
    34 Dimension of aSolution Space EXAMPLE
  • 35.
    35 Coordinates and changeof basis • Coordinate representation relative to a basis Let B = {v1, v2, …, vn} be an ordered basis for a vector space V and let x be a vector in V such that .2211 nnccc vvvx   The scalars c1, c2, …, cn are called the coordinates of x relative to the basis B. The coordinate matrix (or coordinate vector) of x relative to B is the column matrix in Rn whose components are the coordinates of x.                n B c c c  2 1 x
  • 36.
    Find the coordinatematrix of x=(1, 2, –1) in R3 relative to the (nonstandard) basis B ' = {u1, u2, u3}={(1, 0, 1), (0, – 1, 2), (2, 3, – 5)} Sol: 2100 8010 5001 1521 2310 1201 E.G.J.                         )5,3,2()2,1,0()1,0,1()1,2,1( 321332211  cccccc uuux                                       1 2 1 521 310 201 i.e. 152 23 12 3 2 1 321 32 31 c c c ccc cc cc              2 8 5 ][ B x Finding a coordinate matrix relative to a nonstandard basis
  • 37.
    LINEAR DEPENDENCE AND INDEPENDENCEOF FUNCTIONS If 𝒇 𝟏=𝒇 𝟏 𝒙 , 𝒇 𝟐=𝒇 𝟐 𝒙 ,…, 𝒇 𝒏=𝒇 𝒏 𝒙 Are 𝒏 − 𝟏 times differentiable Functions on the interval −∞, ∞ Then the wronskian of these function is……..
  • 38.
    W= 𝒇 𝟏 𝒙𝒇 𝟐 𝒙 … 𝒇 𝒏 𝒙 𝒇 𝟏 | (𝒙) 𝒇 𝟐 | (𝒙) … 𝒇 𝒏 | (𝒙) ⋮ ⋮ ⋮ 𝒇 𝟏 𝒏−𝟏 (𝒙) 𝒇 𝟐 𝒏−𝟏 (𝒙) … 𝒇 𝒏 𝒏−𝟏 (𝒙) THEOREM: if the wronskian of (n-1) times differentiable functions on the interval −∞, ∞ is not identically zero on this interval then these functions are linearly independent.
  • 39.
    • NOTE: Ifthe Wronskian of the functions is identically zero on the interval −∞, ∞ , then no conclusion can be made about the linear dependence or independence of the functions.
  • 40.
    EXAMPLES • Which offollowing set of the function F −∞, ∞ are linearly independent ? (1) x,sinx the wronskian of the functions is W= 𝑥 𝑠𝑖𝑛𝑥 1 𝑐𝑜𝑠𝑥 = xcosx-sinx Since, the function is not zero for all values of x in the interval −∞, ∞ , the given function are linearly independent.
  • 41.
    (2) 6,3𝑠𝑖𝑛2 𝑥,2𝑐𝑜𝑠2𝑥 the wronskian of the function is W= 6 3𝑠𝑖𝑛2 𝑥 2𝑐𝑜𝑠2 𝑥 0 6𝑠𝑖𝑛𝑥𝑐𝑜𝑠𝑥 −4𝑐𝑜𝑠𝑥𝑠𝑖𝑛𝑥 0 6𝑐𝑜𝑠2𝑥 −4𝑐𝑜𝑠2𝑥 = 0 No conclusion can be made about the linear independence of the functions.
  • 42.
    6=6𝑠𝑖𝑛2 𝑥+6𝑐𝑜𝑠2 𝑥 =2(3𝑠𝑖𝑛2𝑥)+3(2𝑐𝑜𝑠2 𝑥) this shows that 6 can be expressed as a linear combination to given functions, hence the given functions are linearly dependent. NOTE : appropriate can be used directly to show linear dependence without using Wronskian.
  • 43.