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Math 307 Spring, 2003 Hentzel Time: 1:10-2:00 MWF Room: 1324 Howe Hall Instructor: Irvin Roy Hentzel Office 432 Carver Phone 515-294-8141 E-mail: [email protected] http://www.math.iastate.edu/hentzel/class.307.ICN Text: Linear Algebra With Applications, Second Edition Otto Bretscher Wednesday, March 5, Chapter 4.3 Problems 4,10,50 Page 173 Main Idea: Just look at the coefficients. Key Words: Representation of a vector with respect to a basis. VB PCB PCB TBB PBC Goal: Set up the constructions symbolically before you try to do them with the actual matrices. Previous Assignment Monday, March 3 Chapter 4.2 Problems 6,14,16 Page 164 Page 164 Problem 6 Find out which of the transformations in Exercises 1 through 25 are linear. For those that are linear, determine whether they are isomorphisms. T(A) = BA where B = | 1 2 | |36| The fact that T is linear is a direct result of the fact that matrix multiplication is linear. We can write out a few details as follows. T(X+Y) = B(X+Y) = BX + BY = T(X) + T(Y) T(cX) = B(cX) = c BX = c T(X) So T is linear. To show that T is not invertible, we show that the kernel is not zero. Since T( |-2 0| ) = | 1 2 | |-2 0 | = | 0 0 | | 1 0| |36||10| |00| T has a non trivial kernel so T will not be invertible. Page 164 Problem 14 T(a+bt+ct2) = a-bt+ct2. What T does is to change the sign of the linear term. The additive part of the proof that T is linear: T((a+bt+ct2) + (a'+b't+c't2)) = T((a+a') + (b+b')t + (c+c')t2) = (a+a') -(b+b')t + (c+c')t2 = a-bt+ct2 + a'-b't+c't2 = T(a+bt+ct2) + T(a'+b't+c't2). The scalar multiplication part of the proof that T is linear. T( d(a+bt+ct2) ) = T( da + dbt + dct2) = da -dbt +dct2 = d(a-bt+ct2) = d T(a+bt+ct2) To show that T is invertible, we will actually find the inverse of T. Since T(a-bt+ct2) = a+bt+ct2 the inverse of T exists and is T itself. Page 164 Problem 16 T(f(t)) = t f '(t). We first show the additive part of linearity. T( f(t)+g(t) ) = t ( f(t)+g(t) )' = t ( f '(t) + g '(t) ) = t f '(t) + t g '(t) = T( f(t)) + T(g(t)). We next show the scalar multiplication part of linearity. T( c f(t) ) = t (c f(t))' = c t f '(t) = c T( f(t) ). We know that T is not invertible because it has non-zero elements in its kernel. T kills off all of the constant functions. New Material. The numbers in a vector have to be interpreted through the appropriate basis. We keep track of the particular basis that is being used by giving a subscript on the vector. (1) If we have a Basis B = [B1 B2 ... Bn] we can express something in terms of the basis by giving the coefficients of the B's. | c1 | | c2 | | c3 | V = [B1 B2 ... Bn] | . | = c1 B1 + c2 B2 + ... + cn Bn. | . | | cn | The coefficient vector VB is called the representation of V with respect to the basis [B1 B2 ... Bn]. VB = | c1 | | c2 | | c3 | | . | | . | | cn | For example: (a) Represent the polynomial x4 + 3 x + 2 with respect to the basis [ x4 x3 x2 x 1 ]. |1| |0| answer: | 0 | |3| |2| (b) Represent x Sin[x] + 3 Cos[x] with respect to the basis [ x Sin[x], x Cos[x], Sin[x], Cos[x] ]. |1| |0| answer: | 0 | . |3| (2) If you choose a different basis for whatever reason, you have to change the representation of the vector. This change is easily done by using a matrix called the change of basis matrix. | a11 a12 ... a1n | | a21 a22 ... a2n | [B1 B2 ... Bn] | . . . | = [C1 C2 ... Cn] | . . . | | an1 an2 ... ann | You create this matrix column by column. Column i will be the coefficients needed to make Ci a linear combination of the B’s. The matrix is called PBC To get this straight, derive what you are doing. | a11 a12 ... a1n | | x1 | | x1 | | a21 a22 ... a2n | | x2 | | x2 | [B1 B2 ... Bn] | . . . | | . | = [C1 C2 ... Cn] | . | | . . . || . | | . | | an1 an2 ... ann | | xn | | xn | We multiply the expression by the vector X. This gives us B A X = C X. On the right hand side, X is a representation of a vector in the C basis. On the left hand side AX is the representation of the vector in the B basis. So X B = A XC . Thus A is P BC For example. (a) Find the change of basis matrix for [ 1 x x2 x3 ] [ 1 x x2 and [ x3 x2 x 1] |0001| x3 ] | 0 0 1 0 | = [ x3 x2 x 1] |0100| |1000| (b) Find the change of basis matrix for [ ex, e -x] and [ Sinh(x), Cosh(x) ]. ex – e -x Sinh(x) = -------------2 ex + e -x Cosh(x) = --------------2 [ e x e -x] | ½ ½ | = [ Sinh(x) Cosh(x) ]. |-½ ½ | (c) Find the change of basis matrix which rotates the axis sending (1,0) into (5/13, 12/13) and (0,1) into (-12/13, 5/13) | 1 0 | | 5/13 -12/13 | = | 5/13 -12/13 | | 0 1 | | 12/13 5/13 | |12/13 5/13 | P BB' What does (3,5) in the B' system correspond to in the B system? | 5/13 -12/13 | | 3 | = | -45/13 | |12/13 5/13 | | 5 | | 61/13 | P BB’ VB’ VB What does the equation 2 x y = 4 become in the B' basis. [xy]|0 1 ||x| =4 |1 0 ||y| [ x' y‘ ] | 5/13 12/13 | | 0 1 | | 5/13 -12/13 | | x' | = 4 |-12/13 5/13 | | 1 0 | |12/13 5/13 | | y' | [ x’ y’ ] | 120 /169 | -119 /169 - 119/169 | | x’ | = 4 -120/169 | | y’ | (120/169) (x')2 -238/169 x' y' -120/169 (y')2 = 4 (3) The matrix of a linear transformation is written with respect to a particular basis. If you change basis, you have to change the matrix of the linear transformation. Suppose T(V) = W and you know how to compute this linear transformation in the B basis. Then using matrix multiplication we know that: T BB VB = WB Now we want to represent T in the C Basis. We first change VC to VB and then use TBB to get WB and then change WB to WC VOILA: PCB TBB PBC VC |____________| TCC = WC For Example: Write the matrix for differentiation using the basis [ ex, e -x ] and [ Sinh(x), Cosh(x) ] ex ex e-x 1 0 e-x 0 -1 Sinh(x) Sinh(x) | 0 Cosh(x) | 1 Cosh(x) 1 | 0 | Using the change of basis matrix. x [ e e x ] | 1/2 1/2 | = [ Sinh[ x ] Cosh[ x ] ] B |-1/2 1/2 | C PBC PCB TBB PBC VC = WC | 1 -1 | | 1 0 | | 1/2 1/2 | | 1 1 | | 0 -1 | |-1/2 1/2 | | 1 1 | | 1/2 1/2 | = | 0 1 | | 1 -1 | |-1/2 1/2 | | 1 0 | Page 175 Problem 61. Let V be the linear space of all functions of the form c1Cos[t] + c2Sin[t] + c3t Cos[t] + c t Sin[t]. Solve this differential equation: f’’ + f = Cos[t] We are looking to solve (D2+I) f(t) = Cos[t]. | D=| | | 0 -1 0 0 1 1 0 0 0 0 0 -1 0| 1| 1| 0| We write out the matrix of differentiation with respect to the basis: Cos[t] Sin[t] t Cos[t] t Sin[t] Cos[t] 0 1 1 0 Sin[t] -1 0 0 1 t Cos[t] 0 0 0 1 t Sin[t] 0 0 -1 0 | D=| | | 0 -1 0 0 | D2 = | | | 1 1 0 0 0 0 0 -1 0| 1| 1| 0| -1 0 0 0 -1 -2 0 0 -1 0 0 0 2 0 0 -1 | | | | | D2 + I = | | | 0 0 0 0 0 0 0 -2 0 0 0 0 2 0 0 0 | | | | We wish to solve (D2+ I) f(t) = Cos[t] This is just a typical linear equation once the basis has been chosen. | | | | 0 0 0 0 0 0 0 -2 0 0 0 0 2 0 0 0 | | c1 | | | c2 | | | c3 | | | c4 | = | | | | 1 0 0 0 | | | | The Row Canonical Form is c1=a c2=b c3 c4 RHS | 0 0 1 0 0 | | 0 0 0 1 ½ | | 0 0 0 0 0 | | 0 0 0 0 0 | | c1 | | 0| | 1 | | 0 | | c2 | = | 0 |+ a | 0 | + b | 1 | | c3 | | 0| | 0 | | 0 | | c4 | |½| | 0 | | 0 | f(t) = a Cos[t] + b Sin[t] + ½ t Sin[t] check: f’(t) = -a Sin[t] + b Cos[t] + ½ Sin[t] + ½ t Cos[t] f”(t) = -a Cos[t] –b Sin[t] + ½ Cos[t] + ½ Cos[t] -1/2 t Sin[t] f”(t) + f(t) = Cos[t]. It checks. Graph your solution(s). (The differential equation f” + f = Cos[t] describes a forced undamped oscillator. In this example, we observe the phenomenon of resonance.