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LINEAR COMBINATIONS Linear Algebra and Numerical Methods MA 5356 1 / 17
Linear Combination Let V be a vector space and S be a non empty subset of V. A vector v ∈ V is called a linear combination of vectors of S if there exists a finite number of vectors u1, u2, u3, · · · , un in S and scalars a1, a2, a3, · · · , an ∈ F such that v = a1u1 + a2u2 + a3u3 + · · · + anun. Linear Algebra and Numerical Methods MA 5356 2 / 17
Linear Combination Let V be a vector space and S be a non empty subset of V. A vector v ∈ V is called a linear combination of vectors of S if there exists a finite number of vectors u1, u2, u3, · · · , un in S and scalars a1, a2, a3, · · · , an ∈ F such that v = a1u1 + a2u2 + a3u3 + · · · + anun. Note In any vector spaceV, 0 · v = 0 ∀ v ∈ V. ∴ Zero vector is a linear combination of any non empty subset of V. Linear Algebra and Numerical Methods MA 5356 2 / 17
Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Linear Algebra and Numerical Methods MA 5356 3 / 17
Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then Linear Algebra and Numerical Methods MA 5356 3 / 17
Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) Linear Algebra and Numerical Methods MA 5356 3 / 17
Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) Linear Algebra and Numerical Methods MA 5356 3 / 17
Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) Linear Algebra and Numerical Methods MA 5356 3 / 17
Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) a + 2b = 2 (1) Linear Algebra and Numerical Methods MA 5356 3 / 17
Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) a + 2b = 2 (1) 2a + b = 2 (2) Linear Algebra and Numerical Methods MA 5356 3 / 17
Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) a + 2b = 2 (1) 2a + b = 2 (2) Solve (1) and (2), we get Linear Algebra and Numerical Methods MA 5356 3 / 17
Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) a + 2b = 2 (1) 2a + b = 2 (2) Solve (1) and (2), we get a = 2 3 , Linear Algebra and Numerical Methods MA 5356 3 / 17
Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) a + 2b = 2 (1) 2a + b = 2 (2) Solve (1) and (2), we get a = 2 3 , b = 2 3 . Linear Algebra and Numerical Methods MA 5356 3 / 17
Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) a + 2b = 2 (1) 2a + b = 2 (2) Solve (1) and (2), we get a = 2 3 , b = 2 3 . ∴ (2, 2) = 2 3 (1, 2) + 2 3 (2, 1). Hence linear combination exist. Linear Algebra and Numerical Methods MA 5356 3 / 17
Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Linear Algebra and Numerical Methods MA 5356 4 / 17
Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) Linear Algebra and Numerical Methods MA 5356 4 / 17
Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) = (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c) Linear Algebra and Numerical Methods MA 5356 4 / 17
Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) = (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c) = (a + 2b + c, −3a − 4b − 5c, 2a − b + 7c) Linear Algebra and Numerical Methods MA 5356 4 / 17
Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) = (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c) = (a + 2b + c, −3a − 4b − 5c, 2a − b + 7c) a + 2b + c = 2 (3) Linear Algebra and Numerical Methods MA 5356 4 / 17
Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) = (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c) = (a + 2b + c, −3a − 4b − 5c, 2a − b + 7c) a + 2b + c = 2 (3) − 3a − 4b − 5c = −5 (4) Linear Algebra and Numerical Methods MA 5356 4 / 17
Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) = (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c) = (a + 2b + c, −3a − 4b − 5c, 2a − b + 7c) a + 2b + c = 2 (3) − 3a − 4b − 5c = −5 (4) 2a − b + 7c = 3 (5) Linear Algebra and Numerical Methods MA 5356 4 / 17
Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) = (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c) = (a + 2b + c, −3a − 4b − 5c, 2a − b + 7c) a + 2b + c = 2 (3) − 3a − 4b − 5c = −5 (4) 2a − b + 7c = 3 (5) The above equations are written as follows   1 2 1 −3 −4 −5 2 −1 7     a b c   =   2 −5 3   Linear Algebra and Numerical Methods MA 5356 4 / 17
Solve above system of equations by Cramer’s rule. Linear Algebra and Numerical Methods MA 5356 5 / 17
Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 Linear Algebra and Numerical Methods MA 5356 5 / 17
Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 = 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8) Linear Algebra and Numerical Methods MA 5356 5 / 17
Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 = 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8) = −33 + 22 + 11 Linear Algebra and Numerical Methods MA 5356 5 / 17
Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 = 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8) = −33 + 22 + 11 = 0 Linear Algebra and Numerical Methods MA 5356 5 / 17
Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 = 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8) = −33 + 22 + 11 = 0 ⇒ ∆ is inconsistent. Linear Algebra and Numerical Methods MA 5356 5 / 17
Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 = 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8) = −33 + 22 + 11 = 0 ⇒ ∆ is inconsistent. ∴ There is no solution. Linear Algebra and Numerical Methods MA 5356 5 / 17
Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 = 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8) = −33 + 22 + 11 = 0 ⇒ ∆ is inconsistent. ∴ There is no solution. Hence linear combination does not exist. Linear Algebra and Numerical Methods MA 5356 5 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) − b = −2 (6) Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) − b = −2 (6) − 2a + 5b = k Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) − b = −2 (6) − 2a + 5b = k (7) From (3) and (4), we get Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) − b = −2 (6) − 2a + 5b = k (7) From (3) and (4), we get a = −1, b = 2. Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) − b = −2 (6) − 2a + 5b = k (7) From (3) and (4), we get a = −1, b = 2. ∴ a, b exists. Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) − b = −2 (6) − 2a + 5b = k (7) From (3) and (4), we get a = −1, b = 2. ∴ a, b exists. From (5), k = 12. Linear Algebra and Numerical Methods MA 5356 6 / 17
Example 4 write down the matrix E = 3 1 1 −1 as a linear combination of the matrices A = 1 1 1 0 , B = 0 0 1 1 and C = 0 2 0 −1 Linear Algebra and Numerical Methods MA 5356 7 / 17
Example 4 write down the matrix E = 3 1 1 −1 as a linear combination of the matrices A = 1 1 1 0 , B = 0 0 1 1 and C = 0 2 0 −1 Solution: Let a, b, c ∈ R. Linear Algebra and Numerical Methods MA 5356 7 / 17
Example 4 write down the matrix E = 3 1 1 −1 as a linear combination of the matrices A = 1 1 1 0 , B = 0 0 1 1 and C = 0 2 0 −1 Solution: Let a, b, c ∈ R. Then E = aA + bB + cC Linear Algebra and Numerical Methods MA 5356 7 / 17
Example 4 write down the matrix E = 3 1 1 −1 as a linear combination of the matrices A = 1 1 1 0 , B = 0 0 1 1 and C = 0 2 0 −1 Solution: Let a, b, c ∈ R. Then E = aA + bB + cC 3 1 1 −1 = a 1 1 1 0 + b 0 0 1 1 + c 0 2 0 −1 Linear Algebra and Numerical Methods MA 5356 7 / 17
Example 4 write down the matrix E = 3 1 1 −1 as a linear combination of the matrices A = 1 1 1 0 , B = 0 0 1 1 and C = 0 2 0 −1 Solution: Let a, b, c ∈ R. Then E = aA + bB + cC 3 1 1 −1 = a 1 1 1 0 + b 0 0 1 1 + c 0 2 0 −1 = a a a 0 + 0 0 b b + 0 2c 0 −c = a a + 2c a + b b − c Linear Algebra and Numerical Methods MA 5356 7 / 17
a = 3 a + 2c = 1 a + b = 1 b − c = −1 Linear Algebra and Numerical Methods MA 5356 8 / 17
a = 3 a + 2c = 1 a + b = 1 b − c = −1 Solve above equations, we get a = 3 Linear Algebra and Numerical Methods MA 5356 8 / 17
a = 3 a + 2c = 1 a + b = 1 b − c = −1 Solve above equations, we get a = 3 b = −2 Linear Algebra and Numerical Methods MA 5356 8 / 17
a = 3 a + 2c = 1 a + b = 1 b − c = −1 Solve above equations, we get a = 3 b = −2 c = −1 Linear Algebra and Numerical Methods MA 5356 8 / 17
a = 3 a + 2c = 1 a + b = 1 b − c = −1 Solve above equations, we get a = 3 b = −2 c = −1 ∴ a, b, c exists. Linear Algebra and Numerical Methods MA 5356 8 / 17
a = 3 a + 2c = 1 a + b = 1 b − c = −1 Solve above equations, we get a = 3 b = −2 c = −1 ∴ a, b, c exists. Hence the linear combination of E is 3A − 2B − C Linear Algebra and Numerical Methods MA 5356 8 / 17
Span Let S be a nonempty subset of a vector space V. The span of S, denoted span(S), is the set consisting of all linear combinations of the vectors in S. Linear Algebra and Numerical Methods MA 5356 9 / 17
Span Let S be a nonempty subset of a vector space V. The span of S, denoted span(S), is the set consisting of all linear combinations of the vectors in S. Note For convenience, we define span(φ) = 0. Linear Algebra and Numerical Methods MA 5356 9 / 17
Span Let S be a nonempty subset of a vector space V. The span of S, denoted span(S), is the set consisting of all linear combinations of the vectors in S. Note For convenience, we define span(φ) = 0. Generate A subset S of a vector space V generates or spans V, if span(S)=V. We also say that the vectors of S generate or span V. Linear Algebra and Numerical Methods MA 5356 9 / 17
Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Linear Algebra and Numerical Methods MA 5356 10 / 17
Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Linear Algebra and Numerical Methods MA 5356 10 / 17
Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) Linear Algebra and Numerical Methods MA 5356 10 / 17
Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) Linear Algebra and Numerical Methods MA 5356 10 / 17
Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) Linear Algebra and Numerical Methods MA 5356 10 / 17
Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Linear Algebra and Numerical Methods MA 5356 10 / 17
Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Solving the above equations, we get Linear Algebra and Numerical Methods MA 5356 10 / 17
Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Solving the above equations, we get a = 1, Linear Algebra and Numerical Methods MA 5356 10 / 17
Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Solving the above equations, we get a = 1, b = −1 Linear Algebra and Numerical Methods MA 5356 10 / 17
Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Solving the above equations, we get a = 1, b = −1 Substitute in (3), 2a + b = 2(1) + (−1) = 2 − 1 = 1 Linear Algebra and Numerical Methods MA 5356 10 / 17
Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Solving the above equations, we get a = 1, b = −1 Substitute in (3), 2a + b = 2(1) + (−1) = 2 − 1 = 1 ∴ v is the linear combination of S. Linear Algebra and Numerical Methods MA 5356 10 / 17
Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Solving the above equations, we get a = 1, b = −1 Substitute in (3), 2a + b = 2(1) + (−1) = 2 − 1 = 1 ∴ v is the linear combination of S. Hence v is spanned by S. Linear Algebra and Numerical Methods MA 5356 10 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Solving above equations, we get Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Solving above equations, we get a = 2, Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Solving above equations, we get a = 2, b = −3, Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Solving above equations, we get a = 2, b = −3, c = 2 Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Solving above equations, we get a = 2, b = −3, c = 2 Substitute in (4), we get a + b + c = 2 − 3 + 2 = 1 Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Solving above equations, we get a = 2, b = −3, c = 2 Substitute in (4), we get a + b + c = 2 − 3 + 2 = 1 ∴ v is the linear combination of S. Hence v is spanned by S. Linear Algebra and Numerical Methods MA 5356 11 / 17
Example 3 Show that the vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F3. Linear Algebra and Numerical Methods MA 5356 12 / 17
Example 3 Show that the vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F3. Solution: Let a, b, c ∈ F. Linear Algebra and Numerical Methods MA 5356 12 / 17
Example 3 Show that the vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F3. Solution: Let a, b, c ∈ F. Then a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (x, y, z) Linear Algebra and Numerical Methods MA 5356 12 / 17
Example 3 Show that the vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F3. Solution: Let a, b, c ∈ F. Then a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (x, y, z) (a, a, 0) + (b, 0, b) + (0, c, c) = (x, y, z) Linear Algebra and Numerical Methods MA 5356 12 / 17
Example 3 Show that the vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F3. Solution: Let a, b, c ∈ F. Then a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (x, y, z) (a, a, 0) + (b, 0, b) + (0, c, c) = (x, y, z) (a + b, a + c, b + c) = (x, y, z) ⇒ a + b = x (15) a + c = y (16) b + c = z (17) Linear Algebra and Numerical Methods MA 5356 12 / 17
Solving the above equations, we get Linear Algebra and Numerical Methods MA 5356 13 / 17
Solving the above equations, we get a = 1 2 (x + y − z) Linear Algebra and Numerical Methods MA 5356 13 / 17
Solving the above equations, we get a = 1 2 (x + y − z) b = 1 2 (x − y + z) Linear Algebra and Numerical Methods MA 5356 13 / 17
Solving the above equations, we get a = 1 2 (x + y − z) b = 1 2 (x − y + z) c = 1 2 (−x + y + z) Linear Algebra and Numerical Methods MA 5356 13 / 17
Solving the above equations, we get a = 1 2 (x + y − z) b = 1 2 (x − y + z) c = 1 2 (−x + y + z) Hence every vector in F3 is spanned by S. i.e., F3 = Span(S) Linear Algebra and Numerical Methods MA 5356 13 / 17
Solving the above equations, we get a = 1 2 (x + y − z) b = 1 2 (x − y + z) c = 1 2 (−x + y + z) Hence every vector in F3 is spanned by S. i.e., F3 = Span(S) ∴ The vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F3 Linear Algebra and Numerical Methods MA 5356 13 / 17
Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Linear Algebra and Numerical Methods MA 5356 14 / 17
Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Linear Algebra and Numerical Methods MA 5356 14 / 17
Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Then a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = x y z w Linear Algebra and Numerical Methods MA 5356 14 / 17
Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Then a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = x y z w a 0 0 0 + 0 b 0 0 + 0 0 c 0 + 0 0 0 d = x y z w Linear Algebra and Numerical Methods MA 5356 14 / 17
Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Then a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = x y z w a 0 0 0 + 0 b 0 0 + 0 0 c 0 + 0 0 0 d = x y z w a b c d = x y z w Linear Algebra and Numerical Methods MA 5356 14 / 17
Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Then a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = x y z w a 0 0 0 + 0 b 0 0 + 0 0 c 0 + 0 0 0 d = x y z w a b c d = x y z w x 1 0 0 0 + y 0 1 0 0 + y 0 0 1 0 + w 0 0 0 1 = x y z w Linear Algebra and Numerical Methods MA 5356 14 / 17
Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Then a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = x y z w a 0 0 0 + 0 b 0 0 + 0 0 c 0 + 0 0 0 d = x y z w a b c d = x y z w x 1 0 0 0 + y 0 1 0 0 + y 0 0 1 0 + w 0 0 0 1 = x y z w ∴ M2×2(F) = span 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 Linear Algebra and Numerical Methods MA 5356 14 / 17
Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Then a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = x y z w a 0 0 0 + 0 b 0 0 + 0 0 c 0 + 0 0 0 d = x y z w a b c d = x y z w x 1 0 0 0 + y 0 1 0 0 + y 0 0 1 0 + w 0 0 0 1 = x y z w ∴ M2×2(F) = span 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 ∴ Given vectors span V. Linear Algebra and Numerical Methods MA 5356 14 / 17
Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Linear Algebra and Numerical Methods MA 5356 15 / 17
Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. Linear Algebra and Numerical Methods MA 5356 15 / 17
Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. ∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in any subspace of V) Linear Algebra and Numerical Methods MA 5356 15 / 17
Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. ∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in any subspace of V) Let S 6= φ, then S contains a vector z. Linear Algebra and Numerical Methods MA 5356 15 / 17
Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. ∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in any subspace of V) Let S 6= φ, then S contains a vector z. Now, 0z = 0 Linear Algebra and Numerical Methods MA 5356 15 / 17
Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. ∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in any subspace of V) Let S 6= φ, then S contains a vector z. Now, 0z = 0 ∈ span(S). Linear Algebra and Numerical Methods MA 5356 15 / 17
Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. ∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in any subspace of V) Let S 6= φ, then S contains a vector z. Now, 0z = 0 ∈ span(S). Let x, y ∈ span(S). Linear Algebra and Numerical Methods MA 5356 15 / 17
Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. ∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in any subspace of V) Let S 6= φ, then S contains a vector z. Now, 0z = 0 ∈ span(S). Let x, y ∈ span(S). Then there exist vectors u1, u2, ..., um, v1, v2, ..., vn ∈ S and scalars a1, a2, ..., am, b1, b2, ..., bn such that x = a1u1 + a2u2 + + amum y = b1v1 + b2v2 + + bnvn . Linear Algebra and Numerical Methods MA 5356 15 / 17
Then Linear Algebra and Numerical Methods MA 5356 16 / 17
Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn Linear Algebra and Numerical Methods MA 5356 16 / 17
Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn ∈ span(S) Linear Algebra and Numerical Methods MA 5356 16 / 17
Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn ∈ span(S) For any scalar c, cx = c(a1u1 + a2u2 + + amum) Linear Algebra and Numerical Methods MA 5356 16 / 17
Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn ∈ span(S) For any scalar c, cx = c(a1u1 + a2u2 + + amum) = (ca1)u1 + (ca2)u2 + + (cam)um Linear Algebra and Numerical Methods MA 5356 16 / 17
Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn ∈ span(S) For any scalar c, cx = c(a1u1 + a2u2 + + amum) = (ca1)u1 + (ca2)u2 + + (cam)um ∈ span(S) Linear Algebra and Numerical Methods MA 5356 16 / 17
Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn ∈ span(S) For any scalar c, cx = c(a1u1 + a2u2 + + amum) = (ca1)u1 + (ca2)u2 + + (cam)um ∈ span(S) ∴ x + y and cx are in span(S). Linear Algebra and Numerical Methods MA 5356 16 / 17
Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn ∈ span(S) For any scalar c, cx = c(a1u1 + a2u2 + + amum) = (ca1)u1 + (ca2)u2 + + (cam)um ∈ span(S) ∴ x + y and cx are in span(S). Thus span(S) is a subspace of V. Linear Algebra and Numerical Methods MA 5356 16 / 17
Let W denote any subspace of V that contains S. i.e., S ⊆ W Linear Algebra and Numerical Methods MA 5356 17 / 17
Let W denote any subspace of V that contains S. i.e., S ⊆ W To prove: span(S) ⊆ W Linear Algebra and Numerical Methods MA 5356 17 / 17
Let W denote any subspace of V that contains S. i.e., S ⊆ W To prove: span(S) ⊆ W Let w ∈ span(S). Linear Algebra and Numerical Methods MA 5356 17 / 17
Let W denote any subspace of V that contains S. i.e., S ⊆ W To prove: span(S) ⊆ W Let w ∈ span(S). Then w = c1w1 + c2w2 + ... + ckwk for some vectors w1, w2, ..., wk ∈ S and some scalars c1, c2, ..., ck. Linear Algebra and Numerical Methods MA 5356 17 / 17
Let W denote any subspace of V that contains S. i.e., S ⊆ W To prove: span(S) ⊆ W Let w ∈ span(S). Then w = c1w1 + c2w2 + ... + ckwk for some vectors w1, w2, ..., wk ∈ S and some scalars c1, c2, ..., ck. Since S ⊆ W, we have w1, w2, ..., wk ∈ W. We know that If W is a subspace of a vector space V and w1, w2, ..., wn ∈ W, then a1w1 + a2w2 + ... + anwn ∈ W for any scalars a1, a2, ..., an. Linear Algebra and Numerical Methods MA 5356 17 / 17
Let W denote any subspace of V that contains S. i.e., S ⊆ W To prove: span(S) ⊆ W Let w ∈ span(S). Then w = c1w1 + c2w2 + ... + ckwk for some vectors w1, w2, ..., wk ∈ S and some scalars c1, c2, ..., ck. Since S ⊆ W, we have w1, w2, ..., wk ∈ W. We know that If W is a subspace of a vector space V and w1, w2, ..., wn ∈ W, then a1w1 + a2w2 + ... + anwn ∈ W for any scalars a1, a2, ..., an. By the above fact, we have w = c1w1 + c2w2 + ... + ckwk ∈ W Linear Algebra and Numerical Methods MA 5356 17 / 17
Let W denote any subspace of V that contains S. i.e., S ⊆ W To prove: span(S) ⊆ W Let w ∈ span(S). Then w = c1w1 + c2w2 + ... + ckwk for some vectors w1, w2, ..., wk ∈ S and some scalars c1, c2, ..., ck. Since S ⊆ W, we have w1, w2, ..., wk ∈ W. We know that If W is a subspace of a vector space V and w1, w2, ..., wn ∈ W, then a1w1 + a2w2 + ... + anwn ∈ W for any scalars a1, a2, ..., an. By the above fact, we have w = c1w1 + c2w2 + ... + ckwk ∈ W ∴ span(S) ⊆ W Hence proved. Linear Algebra and Numerical Methods MA 5356 17 / 17

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Linear Combinations.pdfhkuhcvurcyyr6thtu

  • 1. LINEAR COMBINATIONS Linear Algebra and Numerical Methods MA 5356 1 / 17
  • 2. Linear Combination Let V be a vector space and S be a non empty subset of V. A vector v ∈ V is called a linear combination of vectors of S if there exists a finite number of vectors u1, u2, u3, · · · , un in S and scalars a1, a2, a3, · · · , an ∈ F such that v = a1u1 + a2u2 + a3u3 + · · · + anun. Linear Algebra and Numerical Methods MA 5356 2 / 17
  • 3. Linear Combination Let V be a vector space and S be a non empty subset of V. A vector v ∈ V is called a linear combination of vectors of S if there exists a finite number of vectors u1, u2, u3, · · · , un in S and scalars a1, a2, a3, · · · , an ∈ F such that v = a1u1 + a2u2 + a3u3 + · · · + anun. Note In any vector spaceV, 0 · v = 0 ∀ v ∈ V. ∴ Zero vector is a linear combination of any non empty subset of V. Linear Algebra and Numerical Methods MA 5356 2 / 17
  • 4. Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Linear Algebra and Numerical Methods MA 5356 3 / 17
  • 5. Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then Linear Algebra and Numerical Methods MA 5356 3 / 17
  • 6. Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) Linear Algebra and Numerical Methods MA 5356 3 / 17
  • 7. Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) Linear Algebra and Numerical Methods MA 5356 3 / 17
  • 8. Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) Linear Algebra and Numerical Methods MA 5356 3 / 17
  • 9. Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) a + 2b = 2 (1) Linear Algebra and Numerical Methods MA 5356 3 / 17
  • 10. Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) a + 2b = 2 (1) 2a + b = 2 (2) Linear Algebra and Numerical Methods MA 5356 3 / 17
  • 11. Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) a + 2b = 2 (1) 2a + b = 2 (2) Solve (1) and (2), we get Linear Algebra and Numerical Methods MA 5356 3 / 17
  • 12. Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) a + 2b = 2 (1) 2a + b = 2 (2) Solve (1) and (2), we get a = 2 3 , Linear Algebra and Numerical Methods MA 5356 3 / 17
  • 13. Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) a + 2b = 2 (1) 2a + b = 2 (2) Solve (1) and (2), we get a = 2 3 , b = 2 3 . Linear Algebra and Numerical Methods MA 5356 3 / 17
  • 14. Example 1 Let V = R2 ; S = {(1, 2), (2, 1)}. Then, check v = (2, 2) ∈ V a linear combination of S or not? Solution: Let a, b ∈ R. Then (2, 2) = a(1, 2) + b(2, 1) = (a, 2a) + (2b, b) = (a + 2b, 2a + b) a + 2b = 2 (1) 2a + b = 2 (2) Solve (1) and (2), we get a = 2 3 , b = 2 3 . ∴ (2, 2) = 2 3 (1, 2) + 2 3 (2, 1). Hence linear combination exist. Linear Algebra and Numerical Methods MA 5356 3 / 17
  • 15. Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Linear Algebra and Numerical Methods MA 5356 4 / 17
  • 16. Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) Linear Algebra and Numerical Methods MA 5356 4 / 17
  • 17. Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) = (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c) Linear Algebra and Numerical Methods MA 5356 4 / 17
  • 18. Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) = (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c) = (a + 2b + c, −3a − 4b − 5c, 2a − b + 7c) Linear Algebra and Numerical Methods MA 5356 4 / 17
  • 19. Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) = (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c) = (a + 2b + c, −3a − 4b − 5c, 2a − b + 7c) a + 2b + c = 2 (3) Linear Algebra and Numerical Methods MA 5356 4 / 17
  • 20. Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) = (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c) = (a + 2b + c, −3a − 4b − 5c, 2a − b + 7c) a + 2b + c = 2 (3) − 3a − 4b − 5c = −5 (4) Linear Algebra and Numerical Methods MA 5356 4 / 17
  • 21. Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) = (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c) = (a + 2b + c, −3a − 4b − 5c, 2a − b + 7c) a + 2b + c = 2 (3) − 3a − 4b − 5c = −5 (4) 2a − b + 7c = 3 (5) Linear Algebra and Numerical Methods MA 5356 4 / 17
  • 22. Example 2 Check whether the vector v = (2, −5, 3) is a linear combination of the vectors e1 = (1, −3, 2), e2 = (2, −4, −1) and e3 = (1, −5, 7) or not? Solution: Let a, b, c ∈ R. Then (2, −5, 3) = ae1 + be2 + ce3 = a(1, −3, 2) + b(2, −4, −1) + c(1, −5, 7) = (a, −3a, 2a) + (2b, −4b, −b) + (c, −5c, 7c) = (a + 2b + c, −3a − 4b − 5c, 2a − b + 7c) a + 2b + c = 2 (3) − 3a − 4b − 5c = −5 (4) 2a − b + 7c = 3 (5) The above equations are written as follows   1 2 1 −3 −4 −5 2 −1 7     a b c   =   2 −5 3   Linear Algebra and Numerical Methods MA 5356 4 / 17
  • 23. Solve above system of equations by Cramer’s rule. Linear Algebra and Numerical Methods MA 5356 5 / 17
  • 24. Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 Linear Algebra and Numerical Methods MA 5356 5 / 17
  • 25. Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 = 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8) Linear Algebra and Numerical Methods MA 5356 5 / 17
  • 26. Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 = 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8) = −33 + 22 + 11 Linear Algebra and Numerical Methods MA 5356 5 / 17
  • 27. Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 = 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8) = −33 + 22 + 11 = 0 Linear Algebra and Numerical Methods MA 5356 5 / 17
  • 28. Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 = 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8) = −33 + 22 + 11 = 0 ⇒ ∆ is inconsistent. Linear Algebra and Numerical Methods MA 5356 5 / 17
  • 29. Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 = 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8) = −33 + 22 + 11 = 0 ⇒ ∆ is inconsistent. ∴ There is no solution. Linear Algebra and Numerical Methods MA 5356 5 / 17
  • 30. Solve above system of equations by Cramer’s rule. Let ∆ = 1 2 1 −3 −4 −5 2 −1 7 = 1(−28 − 5) − 2(−21 + 10) + 1(3 + 8) = −33 + 22 + 11 = 0 ⇒ ∆ is inconsistent. ∴ There is no solution. Hence linear combination does not exist. Linear Algebra and Numerical Methods MA 5356 5 / 17
  • 31. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 32. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 33. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 34. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 35. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 36. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 37. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 38. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) − b = −2 (6) Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 39. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) − b = −2 (6) − 2a + 5b = k Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 40. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) − b = −2 (6) − 2a + 5b = k (7) From (3) and (4), we get Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 41. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) − b = −2 (6) − 2a + 5b = k (7) From (3) and (4), we get a = −1, b = 2. Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 42. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) − b = −2 (6) − 2a + 5b = k (7) From (3) and (4), we get a = −1, b = 2. ∴ a, b exists. Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 43. Example 3 For which values of k will the vector v = (1, −2, k) in R3 be a linear combination of u = (3, 0, −2) and w = (2, −1, 5). Solution: Let a, b ∈ R. Then (1, −2, k) = au + bw = a(3, 0, −2) + b(2, −1, 5) = (3a, 0, −2a) + (2b, −b, 5b) = (3a + 2b, −b, −2a + 5b) 3a + 2b = 1 (5) − b = −2 (6) − 2a + 5b = k (7) From (3) and (4), we get a = −1, b = 2. ∴ a, b exists. From (5), k = 12. Linear Algebra and Numerical Methods MA 5356 6 / 17
  • 44. Example 4 write down the matrix E = 3 1 1 −1 as a linear combination of the matrices A = 1 1 1 0 , B = 0 0 1 1 and C = 0 2 0 −1 Linear Algebra and Numerical Methods MA 5356 7 / 17
  • 45. Example 4 write down the matrix E = 3 1 1 −1 as a linear combination of the matrices A = 1 1 1 0 , B = 0 0 1 1 and C = 0 2 0 −1 Solution: Let a, b, c ∈ R. Linear Algebra and Numerical Methods MA 5356 7 / 17
  • 46. Example 4 write down the matrix E = 3 1 1 −1 as a linear combination of the matrices A = 1 1 1 0 , B = 0 0 1 1 and C = 0 2 0 −1 Solution: Let a, b, c ∈ R. Then E = aA + bB + cC Linear Algebra and Numerical Methods MA 5356 7 / 17
  • 47. Example 4 write down the matrix E = 3 1 1 −1 as a linear combination of the matrices A = 1 1 1 0 , B = 0 0 1 1 and C = 0 2 0 −1 Solution: Let a, b, c ∈ R. Then E = aA + bB + cC 3 1 1 −1 = a 1 1 1 0 + b 0 0 1 1 + c 0 2 0 −1 Linear Algebra and Numerical Methods MA 5356 7 / 17
  • 48. Example 4 write down the matrix E = 3 1 1 −1 as a linear combination of the matrices A = 1 1 1 0 , B = 0 0 1 1 and C = 0 2 0 −1 Solution: Let a, b, c ∈ R. Then E = aA + bB + cC 3 1 1 −1 = a 1 1 1 0 + b 0 0 1 1 + c 0 2 0 −1 = a a a 0 + 0 0 b b + 0 2c 0 −c = a a + 2c a + b b − c Linear Algebra and Numerical Methods MA 5356 7 / 17
  • 49. a = 3 a + 2c = 1 a + b = 1 b − c = −1 Linear Algebra and Numerical Methods MA 5356 8 / 17
  • 50. a = 3 a + 2c = 1 a + b = 1 b − c = −1 Solve above equations, we get a = 3 Linear Algebra and Numerical Methods MA 5356 8 / 17
  • 51. a = 3 a + 2c = 1 a + b = 1 b − c = −1 Solve above equations, we get a = 3 b = −2 Linear Algebra and Numerical Methods MA 5356 8 / 17
  • 52. a = 3 a + 2c = 1 a + b = 1 b − c = −1 Solve above equations, we get a = 3 b = −2 c = −1 Linear Algebra and Numerical Methods MA 5356 8 / 17
  • 53. a = 3 a + 2c = 1 a + b = 1 b − c = −1 Solve above equations, we get a = 3 b = −2 c = −1 ∴ a, b, c exists. Linear Algebra and Numerical Methods MA 5356 8 / 17
  • 54. a = 3 a + 2c = 1 a + b = 1 b − c = −1 Solve above equations, we get a = 3 b = −2 c = −1 ∴ a, b, c exists. Hence the linear combination of E is 3A − 2B − C Linear Algebra and Numerical Methods MA 5356 8 / 17
  • 55. Span Let S be a nonempty subset of a vector space V. The span of S, denoted span(S), is the set consisting of all linear combinations of the vectors in S. Linear Algebra and Numerical Methods MA 5356 9 / 17
  • 56. Span Let S be a nonempty subset of a vector space V. The span of S, denoted span(S), is the set consisting of all linear combinations of the vectors in S. Note For convenience, we define span(φ) = 0. Linear Algebra and Numerical Methods MA 5356 9 / 17
  • 57. Span Let S be a nonempty subset of a vector space V. The span of S, denoted span(S), is the set consisting of all linear combinations of the vectors in S. Note For convenience, we define span(φ) = 0. Generate A subset S of a vector space V generates or spans V, if span(S)=V. We also say that the vectors of S generate or span V. Linear Algebra and Numerical Methods MA 5356 9 / 17
  • 58. Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Linear Algebra and Numerical Methods MA 5356 10 / 17
  • 59. Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Linear Algebra and Numerical Methods MA 5356 10 / 17
  • 60. Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) Linear Algebra and Numerical Methods MA 5356 10 / 17
  • 61. Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) Linear Algebra and Numerical Methods MA 5356 10 / 17
  • 62. Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) Linear Algebra and Numerical Methods MA 5356 10 / 17
  • 63. Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Linear Algebra and Numerical Methods MA 5356 10 / 17
  • 64. Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Solving the above equations, we get Linear Algebra and Numerical Methods MA 5356 10 / 17
  • 65. Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Solving the above equations, we get a = 1, Linear Algebra and Numerical Methods MA 5356 10 / 17
  • 66. Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Solving the above equations, we get a = 1, b = −1 Linear Algebra and Numerical Methods MA 5356 10 / 17
  • 67. Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Solving the above equations, we get a = 1, b = −1 Substitute in (3), 2a + b = 2(1) + (−1) = 2 − 1 = 1 Linear Algebra and Numerical Methods MA 5356 10 / 17
  • 68. Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Solving the above equations, we get a = 1, b = −1 Substitute in (3), 2a + b = 2(1) + (−1) = 2 − 1 = 1 ∴ v is the linear combination of S. Linear Algebra and Numerical Methods MA 5356 10 / 17
  • 69. Example 1 Determine whether the vector v = (2, −1, 1) is in the span of S = {(1, 0, 2), (−1, 1, 1)}. Solution: Let a, b ∈ F. Then a(1, 0, 2) + b(−1, 1, 1) = (2, −1, 1) (a, 0, 2a) + (−b, b, b) = (2, −1, 1) (a − b, b, 2a + b) = (2, −1, 1) ⇒ a − b = 2 (8) b = −1 (9) 2a + b = 1 (10) Solving the above equations, we get a = 1, b = −1 Substitute in (3), 2a + b = 2(1) + (−1) = 2 − 1 = 1 ∴ v is the linear combination of S. Hence v is spanned by S. Linear Algebra and Numerical Methods MA 5356 10 / 17
  • 70. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 71. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 72. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 73. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 74. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 75. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 76. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 77. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Solving above equations, we get Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 78. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Solving above equations, we get a = 2, Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 79. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Solving above equations, we get a = 2, b = −3, Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 80. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Solving above equations, we get a = 2, b = −3, c = 2 Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 81. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Solving above equations, we get a = 2, b = −3, c = 2 Substitute in (4), we get a + b + c = 2 − 3 + 2 = 1 Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 82. Example 2 Determine whether the vector v = 2x3 − x2 + x + 3 is in the span of S = {x3 + x2 + x + 1, x2 + x + 1, x + 1}. Solution: Let a, b, c ∈ F. Then a(x3 + x2 + x + 1) + b(x2 + x + 1) + c(x + 1) = v (ax3 + ax2 + ax + a) + (bx2 + bx + b) + (cx + c) = 2x3 − x2 + x + 3 ax3 + (a + b)x2 + (a + b + c)x + (a + b + c) = 2x3 − x2 + x + 3 Comparing the coefficients of x3, x2, x, constant, we get a = 2 (11) a + b = −1 (12) a + b + c = 1 (13) a + b + c = 3 (14) Solving above equations, we get a = 2, b = −3, c = 2 Substitute in (4), we get a + b + c = 2 − 3 + 2 = 1 ∴ v is the linear combination of S. Hence v is spanned by S. Linear Algebra and Numerical Methods MA 5356 11 / 17
  • 83. Example 3 Show that the vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F3. Linear Algebra and Numerical Methods MA 5356 12 / 17
  • 84. Example 3 Show that the vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F3. Solution: Let a, b, c ∈ F. Linear Algebra and Numerical Methods MA 5356 12 / 17
  • 85. Example 3 Show that the vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F3. Solution: Let a, b, c ∈ F. Then a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (x, y, z) Linear Algebra and Numerical Methods MA 5356 12 / 17
  • 86. Example 3 Show that the vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F3. Solution: Let a, b, c ∈ F. Then a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (x, y, z) (a, a, 0) + (b, 0, b) + (0, c, c) = (x, y, z) Linear Algebra and Numerical Methods MA 5356 12 / 17
  • 87. Example 3 Show that the vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F3. Solution: Let a, b, c ∈ F. Then a(1, 1, 0) + b(1, 0, 1) + c(0, 1, 1) = (x, y, z) (a, a, 0) + (b, 0, b) + (0, c, c) = (x, y, z) (a + b, a + c, b + c) = (x, y, z) ⇒ a + b = x (15) a + c = y (16) b + c = z (17) Linear Algebra and Numerical Methods MA 5356 12 / 17
  • 88. Solving the above equations, we get Linear Algebra and Numerical Methods MA 5356 13 / 17
  • 89. Solving the above equations, we get a = 1 2 (x + y − z) Linear Algebra and Numerical Methods MA 5356 13 / 17
  • 90. Solving the above equations, we get a = 1 2 (x + y − z) b = 1 2 (x − y + z) Linear Algebra and Numerical Methods MA 5356 13 / 17
  • 91. Solving the above equations, we get a = 1 2 (x + y − z) b = 1 2 (x − y + z) c = 1 2 (−x + y + z) Linear Algebra and Numerical Methods MA 5356 13 / 17
  • 92. Solving the above equations, we get a = 1 2 (x + y − z) b = 1 2 (x − y + z) c = 1 2 (−x + y + z) Hence every vector in F3 is spanned by S. i.e., F3 = Span(S) Linear Algebra and Numerical Methods MA 5356 13 / 17
  • 93. Solving the above equations, we get a = 1 2 (x + y − z) b = 1 2 (x − y + z) c = 1 2 (−x + y + z) Hence every vector in F3 is spanned by S. i.e., F3 = Span(S) ∴ The vectors (1, 1, 0), (1, 0, 1) and (0, 1, 1) generate F3 Linear Algebra and Numerical Methods MA 5356 13 / 17
  • 94. Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Linear Algebra and Numerical Methods MA 5356 14 / 17
  • 95. Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Linear Algebra and Numerical Methods MA 5356 14 / 17
  • 96. Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Then a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = x y z w Linear Algebra and Numerical Methods MA 5356 14 / 17
  • 97. Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Then a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = x y z w a 0 0 0 + 0 b 0 0 + 0 0 c 0 + 0 0 0 d = x y z w Linear Algebra and Numerical Methods MA 5356 14 / 17
  • 98. Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Then a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = x y z w a 0 0 0 + 0 b 0 0 + 0 0 c 0 + 0 0 0 d = x y z w a b c d = x y z w Linear Algebra and Numerical Methods MA 5356 14 / 17
  • 99. Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Then a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = x y z w a 0 0 0 + 0 b 0 0 + 0 0 c 0 + 0 0 0 d = x y z w a b c d = x y z w x 1 0 0 0 + y 0 1 0 0 + y 0 0 1 0 + w 0 0 0 1 = x y z w Linear Algebra and Numerical Methods MA 5356 14 / 17
  • 100. Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Then a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = x y z w a 0 0 0 + 0 b 0 0 + 0 0 c 0 + 0 0 0 d = x y z w a b c d = x y z w x 1 0 0 0 + y 0 1 0 0 + y 0 0 1 0 + w 0 0 0 1 = x y z w ∴ M2×2(F) = span 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 Linear Algebra and Numerical Methods MA 5356 14 / 17
  • 101. Example 4 Show that the matrices 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 generate M2×2(F) Solution: Let a, b, c, d ∈ F. Then a 1 0 0 0 + b 0 1 0 0 + c 0 0 1 0 + d 0 0 0 1 = x y z w a 0 0 0 + 0 b 0 0 + 0 0 c 0 + 0 0 0 d = x y z w a b c d = x y z w x 1 0 0 0 + y 0 1 0 0 + y 0 0 1 0 + w 0 0 0 1 = x y z w ∴ M2×2(F) = span 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 ∴ Given vectors span V. Linear Algebra and Numerical Methods MA 5356 14 / 17
  • 102. Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Linear Algebra and Numerical Methods MA 5356 15 / 17
  • 103. Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. Linear Algebra and Numerical Methods MA 5356 15 / 17
  • 104. Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. ∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in any subspace of V) Linear Algebra and Numerical Methods MA 5356 15 / 17
  • 105. Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. ∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in any subspace of V) Let S 6= φ, then S contains a vector z. Linear Algebra and Numerical Methods MA 5356 15 / 17
  • 106. Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. ∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in any subspace of V) Let S 6= φ, then S contains a vector z. Now, 0z = 0 Linear Algebra and Numerical Methods MA 5356 15 / 17
  • 107. Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. ∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in any subspace of V) Let S 6= φ, then S contains a vector z. Now, 0z = 0 ∈ span(S). Linear Algebra and Numerical Methods MA 5356 15 / 17
  • 108. Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. ∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in any subspace of V) Let S 6= φ, then S contains a vector z. Now, 0z = 0 ∈ span(S). Let x, y ∈ span(S). Linear Algebra and Numerical Methods MA 5356 15 / 17
  • 109. Theorem The span of any subset S of a vector space V is a subspace of V. Moreover, any subspace of V that contains S must also contain the span of S. Proof: Let S = φ, then span(φ) = {0}. ∴ span(φ) is a subspace of V (∵ {0} is a subspace that is contained in any subspace of V) Let S 6= φ, then S contains a vector z. Now, 0z = 0 ∈ span(S). Let x, y ∈ span(S). Then there exist vectors u1, u2, ..., um, v1, v2, ..., vn ∈ S and scalars a1, a2, ..., am, b1, b2, ..., bn such that x = a1u1 + a2u2 + + amum y = b1v1 + b2v2 + + bnvn . Linear Algebra and Numerical Methods MA 5356 15 / 17
  • 110. Then Linear Algebra and Numerical Methods MA 5356 16 / 17
  • 111. Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn Linear Algebra and Numerical Methods MA 5356 16 / 17
  • 112. Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn ∈ span(S) Linear Algebra and Numerical Methods MA 5356 16 / 17
  • 113. Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn ∈ span(S) For any scalar c, cx = c(a1u1 + a2u2 + + amum) Linear Algebra and Numerical Methods MA 5356 16 / 17
  • 114. Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn ∈ span(S) For any scalar c, cx = c(a1u1 + a2u2 + + amum) = (ca1)u1 + (ca2)u2 + + (cam)um Linear Algebra and Numerical Methods MA 5356 16 / 17
  • 115. Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn ∈ span(S) For any scalar c, cx = c(a1u1 + a2u2 + + amum) = (ca1)u1 + (ca2)u2 + + (cam)um ∈ span(S) Linear Algebra and Numerical Methods MA 5356 16 / 17
  • 116. Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn ∈ span(S) For any scalar c, cx = c(a1u1 + a2u2 + + amum) = (ca1)u1 + (ca2)u2 + + (cam)um ∈ span(S) ∴ x + y and cx are in span(S). Linear Algebra and Numerical Methods MA 5356 16 / 17
  • 117. Then x + y = a1u1 + a2u2 + + amum + b1v1 + b2v2 + + bnvn ∈ span(S) For any scalar c, cx = c(a1u1 + a2u2 + + amum) = (ca1)u1 + (ca2)u2 + + (cam)um ∈ span(S) ∴ x + y and cx are in span(S). Thus span(S) is a subspace of V. Linear Algebra and Numerical Methods MA 5356 16 / 17
  • 118. Let W denote any subspace of V that contains S. i.e., S ⊆ W Linear Algebra and Numerical Methods MA 5356 17 / 17
  • 119. Let W denote any subspace of V that contains S. i.e., S ⊆ W To prove: span(S) ⊆ W Linear Algebra and Numerical Methods MA 5356 17 / 17
  • 120. Let W denote any subspace of V that contains S. i.e., S ⊆ W To prove: span(S) ⊆ W Let w ∈ span(S). Linear Algebra and Numerical Methods MA 5356 17 / 17
  • 121. Let W denote any subspace of V that contains S. i.e., S ⊆ W To prove: span(S) ⊆ W Let w ∈ span(S). Then w = c1w1 + c2w2 + ... + ckwk for some vectors w1, w2, ..., wk ∈ S and some scalars c1, c2, ..., ck. Linear Algebra and Numerical Methods MA 5356 17 / 17
  • 122. Let W denote any subspace of V that contains S. i.e., S ⊆ W To prove: span(S) ⊆ W Let w ∈ span(S). Then w = c1w1 + c2w2 + ... + ckwk for some vectors w1, w2, ..., wk ∈ S and some scalars c1, c2, ..., ck. Since S ⊆ W, we have w1, w2, ..., wk ∈ W. We know that If W is a subspace of a vector space V and w1, w2, ..., wn ∈ W, then a1w1 + a2w2 + ... + anwn ∈ W for any scalars a1, a2, ..., an. Linear Algebra and Numerical Methods MA 5356 17 / 17
  • 123. Let W denote any subspace of V that contains S. i.e., S ⊆ W To prove: span(S) ⊆ W Let w ∈ span(S). Then w = c1w1 + c2w2 + ... + ckwk for some vectors w1, w2, ..., wk ∈ S and some scalars c1, c2, ..., ck. Since S ⊆ W, we have w1, w2, ..., wk ∈ W. We know that If W is a subspace of a vector space V and w1, w2, ..., wn ∈ W, then a1w1 + a2w2 + ... + anwn ∈ W for any scalars a1, a2, ..., an. By the above fact, we have w = c1w1 + c2w2 + ... + ckwk ∈ W Linear Algebra and Numerical Methods MA 5356 17 / 17
  • 124. Let W denote any subspace of V that contains S. i.e., S ⊆ W To prove: span(S) ⊆ W Let w ∈ span(S). Then w = c1w1 + c2w2 + ... + ckwk for some vectors w1, w2, ..., wk ∈ S and some scalars c1, c2, ..., ck. Since S ⊆ W, we have w1, w2, ..., wk ∈ W. We know that If W is a subspace of a vector space V and w1, w2, ..., wn ∈ W, then a1w1 + a2w2 + ... + anwn ∈ W for any scalars a1, a2, ..., an. By the above fact, we have w = c1w1 + c2w2 + ... + ckwk ∈ W ∴ span(S) ⊆ W Hence proved. Linear Algebra and Numerical Methods MA 5356 17 / 17