Lecture 11: Vector space and subspace

Transcription

1 Lecture : Vector space and subspace Vector space. R n space Definition.. The space R n consists of all column vector v with n real components, i.e. R n = { v : v = [v,v 2,...,v n ] T, v j R,j =,2,...,n }, (.) Simple cases for n =,2,3 = {all n-by- column vectors with real entries}. (.2). The one-dimensional space R is a line (like the x-axis). 2. The two-dimensional space R 2 is the whole xy-plane. Each vector gives the x and y coordinates of a point in the plane v = (x,y) T. 3. The vectors in R 3 correspond to the points v = (x,y,z) T in three-dimensional space. Important property: Suppose V is a vector space. If v,w V, any linear combination c v+c 2 w V. Definition.2. Let V be any set of elements called vectors, in which two operations are defined: (i) addition of vectors and (ii) multiplication of vectors by scalars. Then, V is called a vector space provided that for any x,y,z V and scalars c,c,c 2, the following rules (axioms) must be obeyed:. x+y = y+x; 2. x+(y+z) = (x+y)+z; 3. There is a unique zero vector such that x+0 = x for all x; 4. For each x, there is a unique vector x such that x+( x) = 0; 5. Scalar times x equals to x; 6. (c c 2 )x = c (c 2 x); Copy right reserved by Yingwei Wang

2 7. c(x+y) = cx+cy; 8. (c +c 2 )x = c x+c 2 x. Remark.. It is easy to verify that the R n defined in Definition. satisfies the eight rules in Definition.2. The above eight conditions are required of every vector space V. For any set S, in order to prove this set to be a space, we need to check the eight rules shown above..2 Subspace Definition.3. Let W be a nonempty subset of the vector space V. Then W is called a subspace of V if and only if W itself is also a vector space. Given a vector space V, let W be a non-empty subset of V, denoted as W V. In other words, every w W, we have w V. Question: How to check whether W is a subspace of V or not? Answer: We can check either the eight rules shown in Definition.2 or the three rules shown below: (0) 0 W; () Suppose v,w W, then v+w W; (2) Suppose v W, then cv W. In other words, the subspace W should be closed under addition v + w and multiplication cv. The rules () and (2) can be combined into a single requirement: the rule for subspaces: A subspace containing v and w must contain all linear combinations cv+dw. Several examples:. R is a subspace of R R is a subspace of R R 2 is a subspace of R Two trivial subspaces of R n is {0} and R n itself. 2 Copy right reserved by Yingwei Wang

3 .3 Solution subspaces Consider an m-by-n matrix A and the following homogeneous linear system Ax = 0. (.3) Let us collect all of the solutions to (.3) together: W = {x : Ax = 0} (.4) = {all solutions to Ax = 0}. (.5) We can double check the conditions (0), () and (2) shown in previous subsection: (0) 0 W since A0 = 0. () Suppose v,w W, then v+w W. A(v+w) = Av+Aw = 0. (2) Suppose v W, then cv W. A(cv) = cav = c0 = 0. Later, we will see that the subspace W defined in (.4) or (.5) is called null space of the matrix A. Remark.2. The solution set of the following non-homogeneous linear system Ax = b, b 0, (.6) is NOT a space. In order to disprove the subspace, we only need to pick up one from the conditions (0), () or (2) shown in previous subsection. Let us check the condition (2). Suppose v is a solution to Ax = b, then A(2v) = 2Av = 2b b. 3 Copy right reserved by Yingwei Wang

4 2 Linear combination Consider a set U = {u,u 2,...,u K } R n. Note that the total number of vectors in U is K and each vector u k, for k =,2,...,K, is an n-by- column vector. Definition2.. Thevector w is calledalinear combination ofthe vectors u,u 2,...,u K provided that there exist scalars c,c 2,...,c K such that w = c u +c 2 u c K u K. (2.) Question: Given a vector w R n and a set of vectors U = {u,u 2,...,u K } R n, how to find the coefficients c,c 2,...,c K such that (2.) holds? Answer: We can solve c from the following linear system Uc = w, (2.2) where the matrix U = [u,u 2,...,u K ] and the column vector c = [c,c 2,...,c K ] T. [ ] [ ] [ ] 2 Example. Let w = and u 7 =,u 2 2 =. Find c,c 2 such that (2.) holds. Solution: By (2.), we know that c u +c 2 u 2 = w, [ ] [ ] [ ] 2 c +c 2 2 =, 7 [ ][ ] [ ] 2 c =, 2 7 which is exactly the same linear system shown in Example in Lecture note 7. The solution is c = 3,c 2 = Example 2. Let w = and u =,u 2 =,u 3 =. Find c,c 2,c such that (2.) holds. Solution: By (2.), we know that c 2 c u +c 2 u 2 +c 3 u 3 = w, 4 Copy right reserved by Yingwei Wang

5 3 3 c 4 c 2 =, c 3 8 which is exactly the same linear system shown in Example in Lecture note 8. There is no solution to this linear system so we can not find c,c 2,c 3 such that (2.) holds. See the Examples, 2, 3 in Chapter 4.3 and Example 5 in Chapter 4.4 of your textbook. 3 Null space of a matrix The main goal of Chapter 4 is to give a big picture of linear algebra for any given matrix A m n, we will consider the so called four fundamental subspaces related to the matrix A: row space, column space, null space and left null space. Let us start from the null space of a matrix. 3. Definition Definition 3.. The nullspace of A consists of all solutions to Av = 0, denoted as Null(A). Theorem 3.. The solution set V consisting of v R n satisfying Av = 0 is a subspace of R n, where A is an m-by-n matrix. Proof. Check the conditions of subspace:. 0 V since A0 = Suppose u,v V, which means it implies that cu+dv V since Au = 0, Av = 0, A(cu+dv) = 0, for any scalars c,d. By previous section, we can conclude that V is a subspace of R n. 5 Copy right reserved by Yingwei Wang

6 3.2 Another way to solve Ax = 0 How to find the linearly independent special solutions (non-trivial solutions, nonzero solutions) to the linear system Ax = 0. Example 3. Find the special solutions to Solution: We know that x = 2x 2. x 2x 2 = 0. (3.) Method I: Let the free variable x 2 = t, then the leading variable x = 2t. So the solution is ( ) ( ) ( ) x 2t 2 x = = = t. t x 2 It implies that all of ( the ) solution to the problem (3.) can be written as a constant 2 t times the vector. ( ) 2 Method II: We can choose x = 2,x 2 = which means the special solution is x =. The null space of the coefficient matrix A = [ 2 ] { ( )} 2 is Null(A) = c. ( ) 2 In next lecture, we will see that the vector x = is called the base of the space Null(A). Example 4. Find the special solutions to { x 2x 2 = 0, 3x 6x 2 = 0. Solution: Perform the Gaussian elimination to the coefficient matrix [ ] 2 A = 3 6 [ ] ( 3)R +R 2 R Copy right reserved by Yingwei Wang

7 Now the linear system becomes x = 2x( 2.) Similarly, we can choose x = 2,x 2 = 2 which means the special solution is x =. The null space of the coefficient matrix [ ] { ( )} 2 2 A = is Null(A) = c. 3 6 Example 5. Find the special solutions to Solution: It implies that x +2x 2 x 3 = 0. x = 2x 2 +x 3. Recall that x is called pivot variable, and x 2,x 3 are called free variables. We have two methods to find all of the solutions. Method I: Choose the free variables x 2 = t,x 3 = s and we can get the pivot variable x = 2t+s. Then the solution can be written as 2t+s 2t s 2 x = t = t + 0 = t +s 0. s 0 s 0 Method II: (a) If we choose x 2 = 0,x 3 =, then x = and v = 0 ; 2 (b) If we choose x 2 =,x 3 = 0, then x = 2 and v 2 =. 0 Now we find two linearly independent special solutions v = 0,v 2 = So the complete solutions to the linear system x +2x 2 x 3 = 0 is 2 x = c 0 +c Copy right reserved by Yingwei Wang

8 The null space of the coefficient matrix A = [ 2 ] is 2 Null(A) = c 0 +c 2. 0 Basically, the v and v 2 are linearly independent means that one can not be written as the linear combination of another one. Besides, the choice of v,v 2 are not unique. We will see the definition next lecture. Example 6. Find the special solutions to x +2x 2 x 3 = 0, 2x +4x 2 2x 3 = 0, 3x +6x 2 3x 3 = 0. Solution: Perform the Gaussian elimination to the coefficient matrix 2 A = ( 2)R +R 2 R ( 3)R +R 3 R Repeat the procedure of previous example, we can conclude that the nullspace of A = is Null(A) = c 0 +c See example 5 in Chapter 4.2 of your textbook. 3.3 Special cases of null space Consider the null space of identity matrix I n and zero matrix 0 n.. The null space of identity matrix is zero space, i.e., Null(I n ) = {0}; 2. The null space of zero matrix is the whole space, i.e., Null(0 n ) = R n. 8 Copy right reserved by Yingwei Wang

9 More remarks about n-by-n square matrix A n n :. If A is invertible (or nonsingular, or det(a) 0), then Null(A) = {0}. 2. If A is not invertible (or singular, or det(a) = 0), then in general {0} Null(A) R n. For instant, consider the matrix A in Example 6. 9 Copy right reserved by Yingwei Wang