Vector space

MATHS 2

1. VECTOR SPACE PRESENTED BY :-MECHANICAL ENGINEERING DIVISION-B SEM-2 YEAR-2016-17

2. Real Vector Spaces Sub Spaces Linear combination Span Of Set Of Vectors Basis Dimension Row Space, Column Space, Null Space Rank And Nullity Coordinate and change of basis CONTENTS

3. NAME ENROLLMENT NO. PATEL JAIMIN A 150280119080 PATEL KAUSHAL K 150280119081 PATEL KRUNAL S 150280119082 PATEL MEET B 150280119083 PATEL MILAN V 150280119084 PATEL MILIND B 150280119085 PATEL PRANAV P 150280119086 PATEL RAVI K 150280119087 PATEL RAVI S 150280119088 PATEL UJJWAL G 150280119089 PATIL KRUNAL R 150280119090 PRESENTED BY:-

4. Vector Space Let V be an arbitrary non empty set of objects on which two operations are defined, addition and multiplication by scalar (number). If the following axioms are satisfied by all objects u, v, w in V and all scalars k and l, then we call V a vector space and we call the objects in V vectors. If u and v are objects in V, then u + v is in V u + v = v + u u + (v + w) = (u + v) + w There is an object 0 in V, called a zero vector for V, such that 0 + u = u + 0 = u for all u in V For each u in V, there is an object –u in V, called a negative of u, such that u + (-u) = (-u) + u = 0 If k is any scalar and u is any object in V then ku is in V k( u+v ) = ku + kv ( k+l )(u) = ku + lu k( lu ) = (kl)u 1u = uz

5. EXAMPLE OF VECTOR SPACE Determine whether the set of V of all pairs of real numbers (x,y) with the operations ( = (x1+x2+1, y1+y2+1) and k(x,y) = ( kx,ky ) is a vector space. Solution let u=(x1,y1), v=(x2,y2) and w=(x3,y3) are objects in V and k1,k2 are some scalars. 1 . u+v = (x1,y1) + (x2,y2) = ( x1+x2+1 , y1+y2+1 ) since x1+x2+1, y1+y2+1 are also real numbers . Therefore, u+v is also an object in V. 2 . u+v = ( x1+x2+1, y1+y2+1) = ( x2+x1+1 , y2+y1+1 ) = v + u Therefore , vector addition is commutative.  

6. 3 . u+( v+w ) = ( x1,y1)+[(x2,y2) +(x3,y3)] = ( x1,y1)+ ( x2+x3+1, y2+y3+1) = [x1+ ( x2+x3+1)+1 , y1+(y2+y1+1)+1) = [(x1+ x2+1)+x3+1 , (y1+y2+1)+y3+1)] = ( x1+x2+1, y1+y2+1)+ (x3+y3) = ( u+v )+w Hence, vector addition is associative. 4. Let ( a,b ) be in object in V such that ( a,b )+u=u ( a,b ) +(x1,y1)=(x1,y1) (a+x1+1,b1+y1+1) = (x1,y1) a= -1 , b=-1 Hence , (-1,-1) is zero vector in V. 5. Let ( a,b ) be in object in V such that ( a,b )+u=(-1,-1) ( a,b )+(x1,y1)=(-1,-1) ( x1+a+1,y1+b+1)=(1,-1) a = -x1-2 , b = -y1-2 Hence , (-x1-2,y1-2) is the negative of u in V

7. 6 . k1 u = k1(x1,y1) = (k1x1,k1y1) Since k1x1, k1y1 are real numbers. Therefore, V is closed under scalar multiplication. 7. K1 ( u+v )= k1(x1+ x2+1, y1+y2+1) = (k1x1 + k1x2 + k1, k1y1 + k1y2 + k1) ≠ k1 u + k1 v V is not distributive under scalar multiplication. Hence, V is not a vector space

8. 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 space V has at least two subspaces Zero vector space { } is a subspace of V. (2) V is a subspace of V. Ex: Subspace of R 2 Ex : Subspace of R 3 If w 1 ,w 2 ,. . .. w r subspaces of vector space V then the intersection is this subspaces is also subspace of V .

10. Let W be the set of all 2×2 symmetric matrices. Show that W is a subspace of the vector space M 2×2 , with the standard operations of matrix addition and scalar multiplication. 10 Sol: Ex : (A subspace of M 2×2 )

11. SPAN - The set of all the vectors that are the linear combination of the vectors in the set S = {v 1 ,v 2….. v r } is called span of S and is denoted by Span {v 1 ,v 2….. v r } - If S = {v 1 ,v 2….. v r } is a set of vector in a vector space v then, (1)The span is a subspace of v (2)The span S is the smallest subspace of v that contains the set S. - If S 1 And S 2 are two sets of vectors in v then, Span S 1 = Span S 2 if and only if each vector in S 1 is a linear combination of those in S 1 And S 2 and vice versa

12. Method to check span of v 1.choose an arbitary vector b in v. 2. express b as alinear combination of v 1 ,v 2 ,….., v r b=k 1 v 1 +k 2 v 2 +…..+ k r v r 3. if the system of the equation in above equation is consistent for all choice of b then vectors v 1 ,v 2 ,….., v r span v.if it is inconsistent for some choice of b, vectors do not span v.

13. Definition Linear combination If S = (w1,w2,w3, . . . . , wr) is a nonempty set of vectors in a vector space V, then: (a) The set W of all possible linear combinations of the vectors in S is a subspace of V. (b) The set W in part (a) is the “smallest” subspace of V that contains all of the vectors in S in the sense that any other subspace that contains those vectors contains W.

14. 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) = a i + b j + c k Example: Consider the vectors u =(1,2,-1) and v =(6,4,2) in R 3 . 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 .

15. V 1 =(1,2,3) v 2 =(0,0,2) v3= (-1,0,1) Prove W=(1,-2,2) is not a linear combination of v 1 ,v 2 ,v 3 w = c 1 v 1 + c 2 v 2 + c 3 v 3 Finding a linear combination Sol :

16. Basis Definition: V ： a vector space Generating Sets Bases Linearly Independent Sets  S is called a basis for V S ={ v 1 , v 2 , … , v n }  V S spans V (i.e., span ( S ) = V ) S is linearly independent (1) Ø is a basis for { } (2) the standard basis for R 3 : { i , j , k } i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) Notes:

17. (3) the standard basis for R n : { e 1 , e 2 , …, e n } e 1 =(1,0,…,0), e 2 =(0,1,…,0), e n =(0,0,…,1 ) Ex: R 4 {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)} Ex : matrix space : (4) the standard basis for m  n matrix space: { E ij | 1  i  m , 1 j  n } (5) the standard basis for P n ( x ): {1, x , x 2 , …, x n } Ex: P 3 ( x ) {1, x , x 2 , x 3 }

18. Example Let V  =  P 3  and let S  =  {1, t, t 2 , t 3 }.  Show that S is a basis for V.   Solution We must show both linear independence and span.     Linear Independence: Let          c 1 (1) + c 2 (t) + c 3 (t 2 ) + c 4 (t 3 )  =  0 Then since a polynomial is zero if and only if its coefficients are all zero we have         c 1   =  c 2   =  c 3   =  c 4   =  0 Hence S is a linearly independent set of vectors in V.

19. Span A general vector in P 3  is given by          a + bt + ct 2  + dt 3   We need to find constants c 1 , c 2 , c 3 , c 4  such that          c 1 (1) + c 2 (t) + c 3 (t 2 ) + c 4 (t 3 )  =  a + bt + ct 2  + dt 3   We just let                 c 1   =  a,   c 2   =  b,   c 3   =  c,   c 4   =  d Hence S spans V.   We can conclude that S is a basis for V.   In general the basis {1, t, t 2 , ... , t n } is called the   STANDARD BASIS  for P n .  

20. Change of basis problem You were given the coordinates of a vector relative to one basis B and were asked to find the coordinates relative to another basis B ' . Transition matrix from B ' to B : where is called the transition matrix from B ' to B If [ v ] B is the coordinate matrix of v relative to B [ v ] B ‘ is the coordinate matrix of v relative to B '

21. If P is the transition matrix from a basis B ' to a basis B in R n , then (1) P is invertible (2) The transition matrix from B to B ' is P –1 THEOREM 1 The inverse of a transition matrix THEOREM 2 Let B= { v 1 , v 2 , … , v n } and B ' = { u 1 , u 2 , … , u n } be two bases for R n . Then the transition matrix P –1 from B to B ' can be found by using Gauss-Jordan elimination on the n ×2 n matrix as follows. Transition matrix from B to B '

22. B ={(–3, 2), (4,–2)} and B ' ={(–1, 2), (2,–2)} are two bases for R 2 (a) Find the transition matrix from B ' to B . (b) (c) Find the transition matrix from B to B ' . Ex : (Finding a transition matrix) Sol: G.J.E. B B ' I P ( the transition matrix from B ' to B ) (a) Sol: G.J.E. B B ' (a)

23. (b) G.J.E. B ' B I P -1 ( the transition matrix from B to B ' ) Check: (c)

24. 25 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).

25. 26 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 .

26. 27 Dimensions of Some Familiar Vector Spaces (1) Vector space R n  basis { e 1 , e 2 ,  , e n } (2) Vector space M m   basis { E ij | 1 i  m , 1 j  n } (3) Vector space P n ( x )  basis {1, x , x 2 ,  , x n } (4) Vector space P ( x )  basis {1, x , x 2 , }  dim( R n ) = n  dim( M m n )= mn  dim( P n ( x )) = n +1  dim( P ( x )) = 

27. Coordinates Coordinate representation relative to a basis Let B = { v 1 , v 2 , …, v n } be an ordered basis for a vector space V and let x be a vector in V such that The scalars c 1 , c 2 , …, c n 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 R n whose components are the coordinates of x .

28. Find the coordinate matrix of x =(1, 2, – 1) in R 3 relative to the (nonstandard) basis B ' = { u 1 , u 2 , u 3 }={(1, 0, 1), (0, – 1, 2), (2, 3, – 5)} Sol: Finding a coordinate matrix relative to a nonstandard basis

29. Row space: The row space of A is the subspace of R n spanned by the row vectors of A. Column space: The column space of A is the subspace of R m spanned by the column vectors of A. Null space: The null space of A is the set of all solutions of A x = and it is a subspace of R n . Let A be an m × n matrix Row space ,column space and null space The null space of A is also called the solution space of the homogeneous system A x = .

30. Find a basis of row space of A = Sol : A= B = Finding a basis for a row space EXAMPLE A basis for RS ( A ) = {the nonzero row vectors of B } (Thm 3) = { w 1 , w 2 , w 3 } = {(1, 3, 1, 3), (0, 1, 1, 0), (0, 0, 0, 1)}

31. Finding a basis for the column space of a matrix S ol: EXAMPLE:

32. CS ( A )= RS ( A T ) (A basis for the column space of A ) A Basis For CS ( A ) = a basis for RS ( A T ) = {the nonzero vectors of B } = { w 1 , w 2 , w 3 }

33. Rank and Nullity If A is an m  n matrix, then the row space and the column space of A have the same dimension. dim( RS ( A )) = dim( CS ( A )) THEOREM The dimension of the row (or column) space of a matrix A is called the rank of A and is denoted by rank( A ). rank( A ) = dim( RS ( A )) = dim( CS ( A )) Rank : Nullity: The dimension of the null space of A is called the nullity of A . nullity( A ) = dim( NS ( A ))

34. THEOREM If A is an m  n matrix of rank r , then the dimension of the solution space of A x = is n – r . That is n = rank( A ) + nullity( A ) Notes: (1) rank( A ) : The number of leading variables in the solution of A x = . (The number of nonzero rows in the row-echelon form of A ) (2) nullity ( A ) : The number of parameters in the general solution of A x = .

35. 36 Rank and nullity of a matrix Let the column vectors of the matrix A be denoted by a 1 , a 2 , a 3 , a 4 , and a 5 . a 1 a 2 a 3 a 4 a 5 EXAMPLE Find the Rank and nullity of the matrix. Sol: Let B be the reduced row-echelon form of A. a 1 a 2 a 3 a 4 a 5 b 1 b 2 b 3 b 4 b 5 G.E. rank( A ) = 3 (the number of nonzero rows in B ) Nullity(A) = n – rank(A) = 5 – 2 = 3

36. THANK YOU GUIDED BY:- MR. JAYDEV PATEL