Math 307 Spring, 2005 Hentzel Time: 10:00 to 10:50 MWF Room: 205 Carver Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: hentzel@iastate.edu http://orion.math.iastate.edu/hentzel/class.307.05 Text: Linear Algebra and its Applications, Third Edition David C. Lay Monday, March 7 Section 4.3 Page 243 Problem 11, 21, 22, 38 Main Idea: All that independence and spanning stuff explains the solutions of differential equations. Did you recognize the Wronskian? Key Words: Linearly independence, Spanning, Basis, Goal: Get experience thinking of x x c1 e Cos[x] + c2 e Sin[x] as just more of the same stuff as we were studying up till now. Previous Assignment Page 236 Problem 25 A denotes an mxn matrix. (a) The null space of A is the solution set of the equation AX = 0 True (b) The null space of an mxn matrix is in R^m False, it is as long and the width of A, not the depth of A. (c) The column space of A is the range of the mapping X -> AX True (d) If the equation AX = B is consistent, then Column Space of A is R^m. False. If it is consistent for all B, then Column Space of A is R^m. But if it is consistent for some B, it only says that B itself is in Column Space of A. (e) The kernel of a linear transformation is a vector space. True. (f) The column space of A is the set of all vectors that can be written as AX for some X. True. ----------------------------------------- Page 236 Problem 26 (a) A null space is a vector space. True (b) The column space of an mxn matrix is in R^m True, it is the same length as each of the columns of A. (c) The column space of A is the set of all solutions of AX = B False. These solutions are not necessarily a vector space. (d) The null space of A is the kernel of the mapping X -> AX. True. (e) The range of a linear transformation is a vector space. True. (f) The set of all solutions of a homogeneous linear differential equation is the kernel of a linear transformation. True -------------------------------------------------------- Page 236 Problem 33 Let M2x2 be the vector space of all 2x2 matrices, and define T:M2x2 -> M2x2 by T(A) = A+A^T where A = | a b | | c d | (a) Show that T is a linear transformation. T(A+B) = A+B + (A+B)^T = A+A^T + B+B^T = T(A)+T(B) T(cA) = cA + (cA)^T = cA + cA^T = c (A+A^T) = c T(A) (b) Let B be any element of M2x2 such that B^% = B. Find an A in M2x2 such that T(A) = B. T(1/2 A) = 1/2 A + (1/2 A)^T = 1/2 A + 1/2 A = A (c) Show that the range of T is the set of B in M2x2 with the property that B^T = B. We already showed that if B=B^T then B is in the range of T. We have to show that everything in the range of T satisfies B = B^T. T(A) = A+A^T and (A+A^T)^T = A^T + (A^T)^T = A^T + A = A+A^T. (d) Describe the kernel of T. Those matrices where A^T = -A. So symmetric spots across the diagonal add to zero. -------------------------------------------------------- Page 236 Problem 36 Given T:V----> W as in Exercise 35, and given a subspace Z of W, let U be the set of all X in V such that T(X) is in Z. Show that U is a subspace of V. V W ------------------- ----------------------- | | | | | __ | | _________ | | / \ | T \\ | | | | | | U | | =========>| | Z | | | \__/ | // | |_______| | | | | | |_________________| |_____________________| let u,u' be elements of U. Then T(u+u') = T(u) + T(u') is in Z because Z is closed under addition. T(cu) = c T(u) is in Z because Z is closed under scalar multiplication. Therefore U is closed under addition and scalar multiplication and is a subspace. ------------------------------------------------------------- Essentially, we use V ,V , ..., V and call them vectors. While 1 2 n before we thought of them as columns of a matrix. Now we think of them as more general objects. E.G. as polynomials, or things 3 x like x e Sin[x] A set of vectors V , V , ... V is said to be linearly independent 1 2 n ===================== if the only solution to c V + c V + ... c V = 0 1 1 2 2 n n is the trivial solution, c = c = ... = c = 0. 1 2 n 2 2 Show that Sin(x) , Cos(x) and 1 are not linearly independent. The set V , V , ..., V is said to be linearly dependent if there 1 2 n ================== is a non trivial solution to c V + c V + ... c V = 0 1 1 2 2 n n This a trivial theorem, but truly has a myriad of uses. ################################################################### Theorem: If V , V , ..., V are linearly dependent and V =/= 0, 1 2 n 1 then some V is a linear combination of the preceding and i can be removed without changing the span. ################################################################### Proof: Suppose that c V + c V + ... c V = 0 1 1 2 2 n n and not all of the c are zero. Let c be the last nonzero coefficient. i m Then c V + c V + ... c V = 0 and solving for V gives 1 1 2 2 m m m V = -(c1/cm) V -(c2/cm) V ... -(cm-1/cm) V m 1 2 m-1 So V is one vector that is a linear combination of the previous ones. m There may be others. If we remove a vector, it is not possible that the span gets larger, but it is possible for the span to be smaller. However, if a V + a V + ..... a V + ..... a V is some element of the 1 1 2 2 m m n n span of V , V , ..., V , .... V , then we can replace V by 1 2 m n m V = -(c1/cm) V -(c2/cm) V ... -(cm-1/cm) V m 1 2 m-1 giving a V + a V + ..... a V + ..... a V = 1 1 2 2 m m n n / \ a V + a V + ..... a |-(c1/cm) V -(c2/cm) V ...-(cm-1/cm) V |+ ..... a V 1 1 2 2 m\ 1 2 m-1/ n n and by reshuffling terms and collecting coefficients, this same element is in the span of omit \ / \ / V1, V2, ..., Vm, .... Vn /\ / \ Definition: Let V be a vector space and H a subspace of V. A set of vectors V , V , ... V in H is a basis for H if 1 2 p (a) V , V , ... V are linearly independent 1 2 p and (b) V , V , ... V span H. 1 2 p ######################################################## Theorem: Any spanning set can be reduced to a Basis. ######################################################## Proof; Start with any spanning set of V. As long as they are linearly independent, we can remove a vector that does not change the span. After doing this, the span is still V. Since the set is finite, we cannot do this foreever. Therefore at some time we reach an independent set which is then a basis for V. ########################################################### Theorem: Any linearly independent set can be augmented to a basis of V. ########################################################### Suppose that W , W , ..., W is an independent set. 1 2 r Let V , V , ...., V be any spanning set of V. Then 1 2 n W , W , ... W , V , V , ... V is still a spanning set of V. 1 2 r 1 2 n If we reduce this set to a basis, we will never remove a W because no W is a linear combination of the previous because the W's are linearly independent. The the resulting basis will contain all of the original W's and is thus an augmenting of the original W's to be a basis of V. ############################################################ Theorem: Any two bases of a vector space V have the same number of elements. ############################################################ Proof: Suppose that V , V ...., V and 1 2 r W , W , ..., W are two bases of V. 1 2 s We start adding the W's to the V's one a a time. Each time the set becomes linearly dependent, and each time it can be reduced to a basis without removing any of the W's. So at least one V has to be removed. This means that there are at least as many V's as W's. By symmetry there must also be at least as many W's as V's. Thus the number of V's and W's is equal.