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![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.](https://image.slidesharecdn.com/150490106037-160505193442/85/150490106037-23-320.jpg)
![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.](https://image.slidesharecdn.com/150490106037-160505193442/85/150490106037-24-320.jpg)
![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](https://image.slidesharecdn.com/150490106037-160505193442/85/150490106037-35-320.jpg)
Uploaded byHarekrishna Jariwala
150490106037
This document discusses vector spaces and related concepts such as subspaces, linear combinations, linear independence, spanning sets, bases, and dimension. It begins by defining a vector space and providing examples. It then covers subspaces and shows that every vector space has at least two subspaces: the zero vector space and the entire vector space. The document also discusses linear combinations, linear independence, spanning sets, bases, and notes some key properties such as the uniqueness of the basis representation in a vector space.
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150490106037
- 1. Presented By 150490106037 –Harekrishna Jariwala 150490106027 – Niraj Devaiya SNPIT & RC UMRAKH VECTOR SPACE
- 2. CONTENTS 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 2
- 3. Definition andExamples Vector Space
- 4. 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.
- 5. 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.
- 6. 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:-
- 7. Determine whether theset R+ of all positive real numbers with operations x + y = xy kx = xk. Is a vector space. Example:-
- 8. 8 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.
- 9. 9 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.
- 10. 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) 10
- 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) 11
- 12. 12 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.
- 13. 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)
- 14. 14 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
- 15. 15 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
- 16. 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: 16 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.
- 17. 17 tindependenlinearlyis(1) dependent.linearlyis(2) SS0 tindependenlinearlyis(3) v0v 21(4) SS dependentlinearlyisdependentlinearlyis 21 SS tindependenlinearlyistindependenlinearlyis 12 SS Notes
- 18. 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 18
- 19.
- 20. 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.
- 21. 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 ? r x A r k
- 22. 22 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
- 23. 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.
- 24. 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.
- 25. 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.
- 26. 26 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:
- 27. 27 (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 mn matrix space: { Eij | 1im , 1jn } (5) the standard basis for Pn(x): {1, x, x2, …, xn} Ex: P3(x) {1, x, x2, x3}
- 28. 28 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
- 29. 29 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.
- 30. 30 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).
- 31. 31 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.
- 32. 32 Dimensions of SomeFamiliar Vector Spaces (1) Vector space Rn basis {e1 , e2 , , en} (2) Vector space Mm basis {Eij | 1im , 1jn} (3) Vector space Pn(x) basis {1, x, x2, , xn} (4) Vector space P(x) basis {1, x, x2, } dim(Rn) = n dim(Mmn)=mn dim(Pn(x)) = n+1 dim(P(x)) =
- 33. 33 Dimension of aSolution Space EXAMPLE
- 34. 34 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
- 35. 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
- 36. 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……..
- 37. 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.
- 38. • 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.
- 39. 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.
- 40. (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.
- 41. 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.
- 42.