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Vector space and subspace Math 112, week 8 Goals: • Vector space, subspace, span. • Null space, column space. • Linearly independent, bases. Suggested Textbook Readings: Sections §4.1, 4.2, 4.3 Week 8: Vector space and subspace Review of R2 : u1 v1 Addition: ~u = , ~v = , ~u + ~v = u2 v2 u1 Multiplication by scalars: ~u = , c is a scalar, c~u = u2 Show that the following properties are satisfied: 1. The sum ~u + ~v is in R2 . 2. ~u + ~v = ~v + ~u. 3. (~u + ~v ) + w ~ = ~u + (~v + w). ~ 4. The zero vector satisfies: ~u + ~0 = ~u. 5. The vector −~u is also in R2 , and ~u + (−~u) = ~0. 6. The scalar multiple c~u is in R2 . 7. c(~u + ~v ) = c~u + c~v . 8. (c + d)~u = c~u + d~u. 9. c(d~u) = (cd)~u. 10. 1~u = ~u. These ten properties are known as the Axioms of vector space. 2 Week 8: Vector space and subspace 3 Definition (Vector space): A vector space is a non-empty set V of objects, called vectors, on which are defined two operations, called addition and multiplication by scalars (real numbers), subject to the ten axioms listed below: 1. The sum of ~u, ~v , denoted by ~u + ~v , is in V . 2. ~u + ~v = ~v + ~u. 3. (~u + ~v ) + w ~ = ~u + (~v + w). ~ 4. There is a zero vector ~0 in V such that ~u + ~0 = ~u. 5. For each ~u in V , there is a vector −~u in V such that ~u + (−~u) = ~0. 6. The scalar multiple c~u is in V . 7. c(~u + ~v ) = c~u + c~v . 8. (c + d)~u = c~u + d~u. 9. c(d~u) = (cd)~u. 10. 1~u = ~u. The spaces Rn are premier examples of vector spaces. Week 8: Vector space and subspace 4 Examples of vector spaces. 1. Let S be the space of all infinite sequence of numbers: {yk } = (y0 , y1 , y2 , · · · ) If {zk } is another element of S, define The sum {yk } + {zk } = The scalar multiplication c{yk } = 2. The set Pn consists of all polynomials of degree at most n: p~(t) = a0 + a1 t + a2 t2 + · · · + an tn If ~q(t) = b0 + b1 t + b2 t2 + · · · + bn tn is another element in Pn , define The sum p~ + ~q: The scalar multiplication c~p: 3. Let P be the set of all polynomials with real coefficients, with sum and scalar multiplication as above. Then it is a vector space. Week 8: Vector space and subspace Definition (Subspace) A subspace of a vector space V is a subset H of V such that: a. The zero vector of V is in H. b. H is closed under vector addition. That is, c. H is closed under multiplication by scalars. That is, Note: Properties of a. b. c. guarantee that H is itself a vector space. Examples: 1. Zero subspace: The set H = {~0} is a subspace of V . 2. The set s H = { t : s and t are real} 0 is a subspace of R3 . 3. R2 is not a subspace of R3 . 5 Week 8: Vector space and subspace 6 4. A plane in R3 which is not through the origin is not a subspace of R3 . 5. A line in R2 which is not through the origin is not a subspace of R2 . Example 1: Given v~1 and ~v2 in a vector space V , let H = Span{~v1 , ~v2 }. Show that H is a subspace of V . Week 8: Vector space and subspace 7 Theorem: If ~v1 , · · · , ~vk are in a vector space V , then Span{~v1 , · · · , ~vk } is a subspace of V . a − 3b b−a , Example 2: Let H be the set of all vectors of the form a b where a, b are arbitrary real numbers. Show that H is a subspace of R4 . Example 3: Let W be the union of the first and third quadrants x in the xy-plane. That is, W = { : xy ≥ 0}. Is W a subspace of R2 ? y Week 8: Vector space and subspace 8 Definition (Null space). The null space of an m × n matrix A, written as NulA, is the set of all solutions of A~x = ~0. In set notation, NulA = {~x : ~x ∈ Rn and A~x = ~0} 3 1 −2 −2 , and let ~u = Example 4: Let A = 1, −5 11 3 1 is ~u in the null space of A? Theorem: The null space of an m × n matrix A is a subspace of Rn . Equivalently, the solution set of the linear system A~x = ~0 with m equations and n variables is a subspace of Rn . Proof. Week 8: Vector space and subspace 9 An explicit description of Nul A Example 5: Find a spanning set for the null space of the matrix −3 6 −1 1 −7 A = 1 −2 2 3 −1 2 −4 5 8 −4 Note: 1. The spanning set produced by the method in Example 5 is automatically linearly independent. 2. The number of vectors in the spanning set for Nul A equals the number of free variables in the equation A~x = ~0. Week 8: Vector space and subspace 10 Definition (Column space). The column space of an m × n matrix A, written as Col A, is the set of all linear combinations of the columns of A. If A = [~a1 · · · ~an ], then ColA = Span{~a1 , · · · , ~an } Theorem: The column space of an m × n matrix is a subspace of Rm . Example 6: Find a matrix A such that W = ColA. 6a − b W = { a + b : a, b ∈ R} −7a Week 8: Vector space and subspace 11 The contrast between Null A and Col A. 2 4 −2 1 Example 7: A = −2 −5 7 3 3 7 −8 6 a. If Col A is a subspace of Rk , what is k? b. If Nul A is a subspace of Rk , what is k? 2 4 −2 1 Example 8: A = −2 −5 7 3 3 7 −8 6 a. Find a nonzero vector in Col A. b. Find a nonzero vector in Nul A. Week 8: Vector space and subspace Example 2 A = −2 3 12 9: Let −2 1 −5 7 3, 7 −8 6 4 3 3 −2 ~u = −1 and ~v = −1 3 0 a. Determine if ~u is in Nul A. Could ~u be in Col A? b. Determine if ~v is in Col A. Could ~v be in Nul A? Week 8: Vector space and subspace 13 Contrast between Nul A and Col A for an m × n matrix A. Nul A 1. Nul A is a subspace of Rn Col A 1. Col A is a subspace of Rm 2. Nul A is implicitly defined; 2. Col A is explicitly defined. (A~x = ~0) 3. Any vector ~v in Nul A has 3. Any vector ~v in Col A has the property that A~v = ~0. the property that A~x = ~v is consistent. 4. Given a specific vector ~v , 4. Given a specific vector ~v , it is easy to tell if ~v is it may take time to tell if ~v is in Nul A. in Col A. 5. Nul A = {~0} if and only if A~x = ~0 has only the trivial 5. Col A = Rm if and only if A~x = ~b has a solution for every ~b solution. every ~b in Rm . Week 8: Vector space and subspace 14 Definition (Linearly independent): A set of vectors {~v1 , · · · , ~vk } in vector space V is said to be linearly independent if the vector equation c1~v1 + c2 v~2 + · · · + ck~vk = ~0 has only the trivial solution: c1 = 0, · · · , ck = 0. (Linearly dependent): The set {~v1 , · · · , ~vk } is linearly dependent if the vector equation has a non-trivial solution. Theorem: A set of vectors {~v1 , · · · , ~vk } is linearly dependent if and only if at least one of the vectors is a linear combination of the others. Example 10: Determine if the following sets of vectors are linearly independent. 4 6 1. {−2 , −3} 9 6 −1 1 2 5 2. { , , , } 4 3 1 8 3 0 −6 3. { 5 , 0 , 5 } −1 0 4 1 2 3 4. {2 , 3 , 5} 3 4 7 Week 8: Vector space and subspace 15 Definition (Basis): Let H be a subspace of a vector space V . The vectors B = {~b1 , · · · , ~bk } in V is a basis for H if a. B is a linearly independent set, and b. the subspace spanned by B coincides with H; that is, H = Span { ~b1 , · · · , ~bk } Example 11: Let A be an n × n invertible matrix, then the columns of A form a basis for Rn . Example 12: Determine if {~e1 , · · · , ~en } is a basis for Rn : 1 0 ~e1 = .. , . 0 0 1 ~e2 = .. , . 0 ··· , 0 0 ~en = .. . 1 Week 8: Vector space and subspace 16 −2 −4 3 Example 13: Let ~v1 = 0 , ~v2 = 1 , ~v3 = 1 5 7 −6 Determine if {~v1 , ~v2 , ~v3 } is a basis for R3 . 0 2 6 Example 14: ~v1 = 2 , ~v2 = 2 , ~v3 = 16 , and H = Span{~v1 , ~v2 , ~v3 }. −1 0 −5 Note that ~v3 = 5~v1 + 3~v2 , and show that Span{~v1 , ~v2 , ~v3 } = Span{~v1 , ~v2 }. Then find a basis for H. Week 8: Vector space and subspace 17 The Spanning Set Theorem: Let S = {~v1 , · · · , ~vk } be a set in V , and let H = Span{~v1 , · · · , ~vk }, a. If one of the vectors in S, say ~vk , is a linear combination of the remaining vectors in S, then the set formed from S by removing ~vk still spans H. b. If H 6= {~0}, some subset of S is a basis for H. Proof. Week 8: Vector space and subspace 18 Two views of a basis. • A basis is a spanning set that is as small as possible. • A basis is also a linearly independent set that is as large as possible. Bases for Nul A. See the method in Example 5, page 9. Bases for Col A. Example 15: Find a basis for ColB, where 1 0 B = [~b1 · · · ~b5 ] = 0 0 4 0 2 0 0 1 −1 0 0 0 0 1 0 0 0 0 Week 8: Vector space and subspace 19 What about a matrix A that is not in RREF? How can we find a basis for Col A? Example 16: Find a basis for ColA, where 1 3 A = [~a1 · · · ~a5 ] = 2 5 4 0 2 −1 12 1 5 5 8 1 3 2 20 2 8 8