Vector Space Examples Math 203 Spring 2014 Myers Example: S = set of all doubly infinite sequences of real numbers = {{y k } : k Z, y k R} Define {y k } + {z k } = {y k + z k } and c{y k } = {cy k }. Example: y k = 2 k, z k = cos(k(π/2)), c = 2 {y k } = (..., 1 4, 1 2, 1, 2, 4,...) {z k } = (..., cos( π), cos( π/2), cos(0), cos(π/2), cos(π),...) = (..., 1, 0, 1, 0, 1,...) {y k } + {z k } = (..., 3 4, 1 2, 2, 2, 3,...) 1 3 {y k} = (..., 1 12, 1 6, 1 3, 2 3, 4 3,...) Claim: The set S of all doubly infinite sequences of real numbers is a vector space. Proof: Let {u k }, {v k }, and {w k } be any three doubly infinite sequences in S, and let c and d be any two scalars. 1. {u k } + {v k } is in S since {u k + v k } is a doubly infinite sequence. 2. {u k } + {v k } = {v k } + {u k } because: {u k } + {v k } = {u k + v k } definition of addition in S = {v k + u k } addition is commutative in R = {v k } + {u k } definition of addition in S 3. ({u k } + {v k }) + {w k } = {u k } + ({v k } + {w k }) because: ({u k } + {v k }) + {w k } = {u k + v k } + {w k } definition of addition in S = {(u k + v k ) + w k } definition of addition in S = {u k + (v k + w k )} addition is associative in R = {u k } + {v k + w k } definition of addition in S = {u k } + ({v k } + {w k }) definition of addition in S 4. Let {0} = {..., 0, 0, 0, 0, 0,...} denote the doubly infinite sequence of zeros. Then {u k } + {0} = {u k } because: {u k } + {0} = {u k + 0} definition of addition in S = {u k } property of real numbers
5. Let {u k } = ( 1){u k }. Then {u k } + ( {u k }) = {0} because: {u k } + ( {u k }) = {u k } + ( 1){u k } definition of {u k } = {u k } + {( 1)u k } definition of scalar multiplication in S = {u k } + { u k } property of real numbers = {u k + ( u k )} definition of addition in S = {0} property of real numbers 6. c{u k } is in S since {cu k } is a doubly infinite sequence. 7. c({u k } + {v k }) = c{u k } + c{v k } because: c({u k } + {v k }) = c{u k + v k } definition of addition in S = {c(u k + v k )} definition of scalar multiplication in S = {cu k + cv k } distributive property of R = {cu k } + {cv k } definition of addition in S = c{u k } + c{v k } definition of scalar multiplication in S 8. (c + d){u k } = c{u k } + d{u k } because: (c + d){u k } = {(c + d)u k } definition of scalar multiplication in S = {cu k + du k } distributive property of R = {cu k } + {du k } definition of addition in S = c{u k } + d{u k } definition of scalar multiplication in S 9. c(d{u k }) = (cd){u k } because: c(d{u k }) = c{du k } definition of scalar multiplication in S = {c(du k )} definition of scalar multiplication in S = {(cd)u k } multiplication is associative in R = (cd){u k } definition of scalar multiplication in S 10. 1{u k } = {u k } because: 1{u k } = {1u k } definition of scalar multiplication in S = {u k } property of real numbers Since all ten vector space axioms hold, we can conclude that S is a vector space.
Homework Exercise 1: P n = set of all polynomials of degree at most n with real coefficients = { a 0 + a 1 t + a 2 t 2 +... + a n t n : a 1, a 2,..., a n R } Given any two polynomials p(t) = a 0 + a 1 t + a 2 t 2 +... + a n t n and q(t) = b 0 + b 1 t + b 2 t 2 +... + b n t n and any scalar c, define p(t) + q(t) = (a 0 + b 0 ) + (a 1 + b 1 )t + (a 2 + b 2 )t 2 +... + (a n + b n )t n and c(p(t)) = (ca 0 ) + (ca 1 )t + (ca 2 )t 2 +... + (ca n )t n. Example: p(t) = 1 + 2t + 3t 2, q(t) = 2 + 4t, c = 2 p(t) + q(t) = (1 + 2) + (2 + 4)t + (3 + 0)t 2 = 3 + 6t + 3t 2 2(p(t)) = (2 1) + (2 2)t + (2 3)t 2 = 2 + 4t + 6t 2 Claim: The set P n of all polynomials of degree at most n with real coefficients is a vector space. Proof: Let u(t) = u 0 + u 1 t + u 2 t 2 +... + u n t n, v(t) = v 0 + v 1 t + v 2 t 2 +... + v n t n, and w(t) = w 0 + w 1 t + w 2 t 2 +... + w n t n be any three polynomials in P n, and let c and d be any two scalars. Homework: Fill in the details of the proof on a separate piece of paper. (For help see Example 4 on page 192.) 1. u(t) + v(t) is in P n since???. 2. u(t) + v(t) = v(t) + u(t) because:??? 3. (u(t) + v(t)) + w(t) = u(t) + (v(t) + w(t)) because:??? 4. Let 0 denote the polynomial in P n with all coefficients equal to zero. Then u(t) + 0 = u(t) because:??? 5. Let u(t) = ( 1)u(t). Then u(t) + ( u(t)) = 0 because:??? 6. cu(t) is in P n since???. 7. c(u(t) + v(t)) = cu(t) + cv(t) because:??? 8. (c + d)u(t) = cu(t) + du(t) because:??? 9. c(du(t)) = (cd)u(t) because:??? 10. 1u(t) = u(t) because:??? Since all ten vector space axioms hold, we can conclude that P n is a vector space.
Homework Exercise 2: F = set of all real-valued functions defined on a set D = {f : D R} Given any two functions f and g, and any scalar c, define f + g to be the function whose value at the point t in D is f(t) + g(t), and define cf to be the function whose value at the point t in D is c(f(t)). Example: D = (0, ), f(t) = t 2, g(t) = ln(t), c = 1 2 (f + g) : (0, ) R is defined by (f + g)(t) = t 2 + ln(t) ( 1 2 f) : (0, ) R is defined by ( 1 2 f) (t) = 1 2 t2 Note: to say f = g means f(t) = g(t) for all t in D. Claim: The set F of all real-valued functions defined on a set D is a vector space. Proof: Let f, g, and h be any three functions in F, and let c and d be any two scalars. Homework: Fill in the details of the proof on a separate piece of paper. (For help see Example 5 on page 192.) 1. f + g is in F since???. 2. f + g = g + f because:??? 3. (f + g) + h = f + (g + h) because:??? 4. Let 0 denote the function whose value is zero for all t in D. Then f + 0 = f because:??? 5. Let f = ( 1)f. Then f + ( f) = 0 because:??? 6. cf is in F since???. 7. c(f + g) = cf + cg because:??? 8. (c + d)f = cf + df because:??? 9. c(df) = (cd)f because:??? 10. 1f = f because:??? Since all ten vector space axioms hold, we can conclude that F is a vector space.
Homework Exercise 3: {[ a11 a M 2 2 = set of all 2 2 matrices with real entries = 12 a 21 a 22 ] } : a 11, a 12, a 21, a 22 R Homework: Show that the set M 2 2 of all 2 2 matrices with real entries is a vector space.