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 2 Review of R 2 : Addition: u = u 1, v = v 1, u + v = u 2 v 2 Multiplication by scalars: u = u 1, c is a scalar, c u = Show that the following properties are satisfied: 1. The sum u + v is in R 2. 2. u + v = v + u. 3. ( u + v) + w = u + ( v + w). 4. The zero vector satisfies: u + = u. 5. The vector u is also in R 2, and u + ( u) =. 6. The scalar multiple c u is in R 2. 7. c( u + v) = c u + c v. 8. (c + d) u = c u + d u. 9. c(d u) = (cd) u. 1. 1 u = u. u 2 These ten properties are known as the Axioms of vector space.
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 in V such that u + = u. 5. For each u in V, there is a vector u in V such that u + ( u) =. 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. 1. 1 u = u. The spaces R n 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: {y k } = (y, y 1, y 2, ) If {z k } is another element of S, define The sum {y k } + {z k } = The scalar multiplication c{y k } = 2. The set P n consists of all polynomials of degree at most n: p(t) = a + a 1 t + a 2 t 2 + + a n t n If q(t) = b + b 1 t + b 2 t 2 + + b n t n is another element in P n, 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 5 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 = { } is a subspace of V. 2. The set s H = { t : s and t are real} is a subspace of R 3. 3. R 2 is not a subspace of R 3.
Week 8: Vector space and subspace 6 4. A plane in R 3 which is not through the origin is not a subspace of R 3. 5. A line in R 2 which is not through the origin is not a subspace of R 2. Example 1: Given v 1 and v 2 in a vector space V, let H = Span{ v 1, v 2 }. Show that H is a subspace of V.
Week 8: Vector space and subspace 7 Theorem: If v 1,, v k are in a vector space V, then Span{ v 1,, v k } 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 R 4. Example 3: Let W be the union of the first and third quadrants in the xy-plane. That is, W = { x : xy }. Is W a subspace of R 2? 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 =. In set notation, NulA = { x : x R n and A x = } Example 4: Let A = 1 2 2 3, and let u = 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 R n. Equivalently, the solution set of the linear system A x = with m equations and n variables is a subspace of R n. 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 =.
Week 8: Vector space and subspace 1 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 = [ a 1 a n ], then ColA = Span{ a 1,, a n } Theorem: The column space of an m n matrix is a subspace of R m. 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 R k, what is k? b. If Nul A is a subspace of R k, 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 12 Example 9: Let 2 4 2 1 3 A = 2 5 7 3, u = 2 1 3 7 8 6 3 and v = 1 3 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 Col A 1. Nul A is a subspace of R n 1. Col A is a subspace of R m 2. Nul A is implicitly defined; 2. Col A is explicitly defined. (A x = ) 3. Any vector v in Nul A has 3. Any vector v in Col A has the property that A v =. 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 = { } if and only if 5. Col A = R m if and only if A x = has only the trivial A x = b has a solution for every b solution. every b in R m.
Week 8: Vector space and subspace 14 Definition (Linearly independent): A set of vectors { v 1,, v k } in vector space V is said to be linearly independent if the vector equation c 1 v 1 + c 2 v 2 + + c k v k = has only the trivial solution: c 1 =,, c k =. (Linearly dependent): The set { v 1,, v k } is linearly dependent if the vector equation has a non-trivial solution. Theorem: A set of vectors { v 1,, v k } is linearly dependent if and only if at least one of the vectors is a linear combination of the others. Example 1: Determine if the following sets of vectors are linearly independent. 4 6 1. { 2, 3 } 6 9 2. { 5, 2, 1, 1 } 1 8 3 4 3 6 3. { 5,, 5 } 1 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 = { b 1,, b k } 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 { b 1,, b k } Example 11: Let A be an n n invertible matrix, then the columns of A form a basis for R n. Example 12: Determine if { e 1,, e n } is a basis for R n : 1 1 e 1 =, e 2 =,, e n =... 1
Week 8: Vector space and subspace 16 3 4 2 Example 13: Let v 1 =, v 2 = 1, v 3 = 1 6 7 5 Determine if { v 1, v 2, v 3 } is a basis for R 3. 2 6 Example 14: v 1 = 2, v 2 = 2, v 3 = 16, and H = Span{ v 1, v 2, v 3 }. 1 5 Note that v 3 = 5 v 1 + 3 v 2, and show that Span{ v 1, v 2, v 3 } = Span{ v 1, v 2 }. Then find a basis for H.
Week 8: Vector space and subspace 17 The Spanning Set Theorem: Let S = { v 1,, v k } be a set in V, and let H = Span{ v 1,, v k }, a. If one of the vectors in S, say v k, is a linear combination of the remaining vectors in S, then the set formed from S by removing v k still spans H. b. If H { }, 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 4 2 B = [ b 1 b 1 1 5 ] = 1
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 4 2 1 3 12 1 5 5 A = [ a 1 a 5 ] = 2 8 1 3 2 5 2 2 8 8