Engineering Mathematics-1 Unit V Notes PDF

Engineering Mathematics- Mathematics-1 Unit V Vector Spaces Definition: Let V be a nonempty set of vectors with two binary operations: vector addition (+) and scalar multiplication (....

Engineering Mathematics- Mathematics-1 Unit V Vector Spaces Definition: Let V be a nonempty set of vectors with two binary operations: vector addition (+) and scalar multiplication (.). Then (V, +,.) is said to be a vector space over a field F (may set of real number, set of complex numbers) if i. (V, +) is an Abelian group i.e., vector addition (+) satisfies following properties (a) Closure property: For any two vectors, x, y ∈ V ⇒ x + y ∈ V (b) Associative Law Hold: For any three vectors, x, y , z ∈ V ⇒ ( x + y ) + z = x + ( y + z ) (c) Existence Existence of Identity: For any vector, x ∈ V , ∃ (there exist) a vector e ∈V such that x+e = x ⇒e=0 This e is called Identity element of V. (d) Existence of Inverse: For any vector, x ∈ V , ∃ (there exist) a vector y ∈ V such that x + y = e (Identity) ⇒ x+ y =0 ⇒ y = −x This y is called Inverse of element x. (e) Commutative Law Hold: : For any two vectors, x, y ∈ V ⇒ x + y = y + x ii. Scalar multiplication satisfies following properties: (a) For any vector x ∈ V , 1.x = x (b) For any two arbitrary vectors x, y ∈ V , α ( x + y ) = α.x + α. y , ∀α ∈ F (c) For any arbitrary vector x ∈ V , (α + β ).x = α.x + β.x , ∀α , β ∈ F (d) For any arbitrary vector x ∈ V , (αβ ).x = α ( β x ) , ∀α , β ∈ F Example 1: Show that ‘set of real numbers with addition and multiplication that is, ( M , +,.) form a vector space. Solution: To show M is vector space; first of all we have to show that 1. ( M , + ) is an Abelian group- (a) Since for any two vectors, x, y ∈ M ⇒ x + y ∈ M. Therefore Closure property hold. (b) Since for any three vectors, x, y , z ∈ M ⇒ ( x + y ) + z = x + ( y + z ) , therefore Associative Law hold. (c) Let x ∈ M and e be the identity element of V then x+e = x ⇒ e = 0∈M This shows the existence of Identity. Dr. T. N. Mishra Page 1 of 6  Engineering Mathematics- Mathematics-1 (d) Let x ∈ M and y be the inverse element of x then x+ y =0 ⇒ y = −x ∈ M This shows the existence of Inverse. (e) Since for any two vectors, x, y ∈ M ⇒ x + y = y + x. Thus commutative Law holds. Thus from (a)-(e), ( M , + ) is an Abelian group. 2. Now we have to following properties scalar multiplication (.) (a) 1.x = x, ∀x ∈ M (b) For any two arbitrary vectors x, y ∈ M , α ( x + y ) = α.x + α. y , ∀α ∈ M, here F is also real field M. (c) For any arbitrary vector x ∈ M , (α + β ).x = α.x + β.x , ∀α , β ∈ M (d) For any arbitrary vector x ∈ M , (αβ ).x = α ( β x ) Thus, from (1) and (2), we say that ( M , +,.) is a vector space. Example 2: Show that ‘set of all real matrices of order n’, that is M n = { A : A is the real square matrix of order n}. Then show that ( M n , +,.) forms a vector space with binary operations matrix addition and scalar multiplication. Solution: To show ( M n , +) is an Abelian group. (a) Since for any two square matrices, A, B ∈ M n ⇒ A + B ∈ M n. Therefore Closure property hold. (b) Since for any three square matrices, A, B, C ∈ M n ⇒ ( A + B) + C = A + ( B + C ) , therefore Associative Law hold. (c) Let A ∈ M n and I be the identity matrix of M n then A+ I = A ⇒ I = 0∈ Mn This shows the existence of Identity. (d) Let A ∈ M n and B be the inverse element of A then A+ B = 0 ⇒ B = −A∈ Mn This shows the existence of Inverse. (e) Since for any two vectors, A, B ∈ M n ⇒ A + B = B + A. Thus commutative Law holds. Thus from (a)-(e), ( M n , +) is an Abelian group. Dr. T. N. Mishra Page 2 of 6  Engineering Mathematics- Mathematics-1 2. Now we have to following properties scalar multiplication (.) (a) 1. A = A, ∀A ∈ M n (b) For any two arbitrary matrices A, B ∈ M n , α ( A + B ) = α A + α B , ∀α ∈ M, here F is also real field M. (c) For any arbitrary matrix A ∈ M n , (α + β ). A = α. A + β. A , ∀α , β ∈ M (d) For any arbitrary matrix A ∈ M n , (αβ ). A = α ( β A), ∀α , β ∈ M Thus, from (1) and (2), we say that ( M n , +,.) is a vector space. 3: Show that ‘set of all real polynomials of degree ≤ n’, that is Pn = { P : is a polynomial of degree Example 3: ≤ n}. Then show that ( Pn , +,.) forms a vector space with binary operations polynomial addition and scalar multiplication. Solution: 1. To show ( Pn , +) is an Abelian group. (a) Since for any two polynomials of degree ≤ n, P, Q ∈ Pn ⇒ P + Q ∈ Pn. Therefore Closure property hold. (b) Since for any three polynomials of degree ≤ n, P, Q, R ∈ Pn ⇒ ( P + Q) + R = P + (Q + R) , therefore Associative Law hold. (c) Let P ∈ Pn and I be the identity of Pn then P+I = P ⇒ I = 0 ∈ Pn This shows the existence of Identity. (d) Let P ∈ Pn and Q be the inverse element of P then P+Q = 0 ⇒ Q = − P ∈ Pn This shows the existence of Inverse. (e) Since for any two polynomials, P, Q ∈ Pn ⇒ P + Q = Q + P. Thus, commutative Law holds. Thus, from (a)-(e), ( Pn , +) is an Abelian group. 2. Now we have to show following properties of scalar multiplication (.) (a) 1.P = P, ∀P ∈ Pn (b) For any two arbitrary polynomials, P, Q ∈ Pn , α ( P + Q ) = α P + α Q , ∀α ∈ M, here F is also real field M. (c) For any arbitrary polynomial, P ∈ Pn , (α + β ).P = α.P + β.P , ∀α , β ∈ M (d) For any arbitrary polynomial, P ∈ Pn , (αβ ).P = α ( β P ), ∀α , β ∈ M Thus, from (1) and (2), we say that ( Pn , +,.) is a vector space. Dr. T. N. Mishra Page 3 of 6  Engineering Mathematics- Mathematics-1 Linear Transformation: Let V1 and V2 be two vector spaces and T : V1 → V2 be a map, then T is said to be a linear transformation, if T (α x + β y ) = α T ( x ) + β T ( y ), ∀α , β ∈ F Example 1: Let V be the set of polynomials in variable ‘t’ and D : V → V be a derivative map defined by df D( f ) =. Then show that D is a linear transformation. dt Solution: Let f1 and f 2 be two polynomials in V then consider D(α f1 + β f 2 ) = d (α f1 + β f 2 ) α d ( f1 ) β d ( f 2 ) = + dt dt dt = α D( f1 ) + β D( f 2 ) showing that D is a linear transformation. Example 2: Let V be the set of polynomials in variable ‘t’. and I : V → V be a integration map defined by I ( f ) = ∫ f (t ) dt. Then show that I is a linear transformation. Solution: Let f1 and f 2 be two polynomials in V then consider I (α f1 + β f 2 ) = ∫ {α f (t ) + β f (t )}dt 1 2 = α ∫ f (t )dt + β ∫ f (t )dt 1 2 = α I ( f1 ) + β I ( f 2 ) showing that I is a linear transformation. Example 3: Let R 3 = {( x, y , z ) : x, y , z ∈ R} be a vector space over real field and T : R 3 → R 3 be a map defined by T ( x, y , z ) = ( x, y , 0) then show that T is a linear transformation. Solution: Let X = ( x1 , y1 , z1 ) ∈ R 3 and Y = ( x2 , y2 , z2 ) ∈ R 3 then consider (α X + β Y ) = α ( x1 , y1 , z1 ) + β ( x2 , y2 , z2 ) = (α x1 , α y1 , α z1 ) + ( β x2 , β y2 , β z2 ) = (α x1 + β x2 , α y1 + β y2 , α z1 + β z2 ) -----------------------------(1) Now consider, T (α X + β Y ) = T (α x1 + β x2 , α y1 + β y2 , α z1 + β z2 ) (from equation (1)) Dr. T. N. Mishra Page 4 of 6  Engineering Mathematics- Mathematics-1 = (α x1 + β x2 , α y1 + β y2 , 0) (by the given map) = (α x1 , α y1 , 0) + ( β x2 , β y2 , 0) = α ( x1 , y1 , 0) + β ( x2 , y2 ,0) -----------------------------------(2) Since T ( X ) = T ( x1 , y1 , z1 ) = ( x1 , y1 , 0) and T (Y ) = T ( x2 , y2 , z2 ) = ( x2 , y2 , 0) , so from (2) T (α X + β Y ) = α T ( X ) + β T (Y ) This shows that T is a linear transformation. Example 4: Let R n = {( x1 , x2 , x3..........xn −1 , xn ) : x1 , x2 , x3......xn ∈ R} be a vector space over real field and T : R n → R n be a map defined by T ( x1 , x2 , x3..........xn −1 , xn ) = ( x1 , x2 , x3..........xn −1 , 0) then show that T is a linear transformation. Solution: Same as above Example 3, only let X = ( x1 , x2 , x3..........xn −1 , xn ) ∈ R n and Y = ( y1 , y2 , y3.......... yn −1 , yn ) ∈ R n and consider (α X + β Y ) = α ( x1 , x2 , x3..........xn −1 , xn ) + β ( y1 , y2 , y3.......... yn −1 , yn ) (α x1 , α x2 , α x3..........α xn −1 , α xn ) + ( β y1 , β y2 , β y3..........β yn −1 , β yn ) = (α x1 + β y1 , α x2 + β y2 , α x3 + β y3...........α xn−1 + β yn−1 , α xn + β yn ) -----------------(1) Now consider, T (α X + β Y ) = T (α x1 + β y1 , α x2 + β y2 , α x3 + β y3...........α xn −1 + β yn −1 , α xn + β yn ) (from equation (1)) = (α x1 + β y1 , α x2 + β y2 , α x3 + β y3...........α xn −1 + β yn−1 , 0) (by the given Map) = (α x1 , α x2 , α x3...........α xn −1 , 0) + ( β y1 , β y2 , β y3...........β yn −1 , 0) = α ( x1 , x2 , x3...........xn−1 , 0) + β ( y1 , y2 , y3........... yn −1 , 0) = α ( x1 , x2 , x3...........xn−1 , 0) + β ( y1 , y2 , y3........... yn −1 , 0) -----------------------(2) Since T ( X ) = T ( x1 , x2 , x3.....xn ) = ( x1 , x2 , x3.....0) and T (Y ) = T ( y1 , y2 , y3..... yn ) = ( y1 , y2 , y3.....0) , so from (2) T (α X + β Y ) = α T ( X ) + β T (Y ) This shows that T is a linear transformation. Example 5: Let R 3 = {( x, y , z ) : x, y , z ∈ R} be a vector space over real field and T : R 3 → R 2 be a map defined by T ( x, y , z ) = ( x + 1, y ) then check whether T is a linear transformation on not. Solution: To check given map is linear or not, you have to check the image of identity of R 3 i.e. Dr. T. N. Mishra Page 5 of 6  Engineering Mathematics- Mathematics-1 T (0, 0, 0) = (1, 0) (by the definition of Map) Since (1, 0, 0) ≠ (0, 0) (the identity of R 2 ) Therefore T is not a linear transformation. Example 6: Show that following are not a linear transformations (a) Let R 3 = {( x, y , z ) : x, y , z ∈ R} be a vector space over real field and T : R 3 → R 3 be a map defined by T ( x, y , z ) = (1, y, z ) (b) Let R 3 = {( x, y , z ) : x, y , z ∈ R} be a vector space over real field and T : R 3 → R 3 be a map defined by T ( x, y , z ) = ( x,1, z ) (c) Let R 3 = {( x, y , z ) : x, y , z ∈ R} be a vector space over real field and T : R 3 → R 3 be a map defined by T ( x, y , z ) = ( x 2 , y , z ) (d) Let R 3 = {( x, y , z ) : x, y , z ∈ R} be a vector space over real field and T : R 3 → R 3 be a map defined by T ( x, y , z ) = (1, y, z ) Dr. T. N. Mishra Page 6 of 6