Examples 6 1 Rotation R by θ about the e 3 z axis choose the standard basis for

Examples 6 1 rotation r by θ about the e 3 z axis

This preview shows page 13 - 17 out of 26 pages.

Examples 6. 1. Rotation R by θ about the e 3 = z –axis; choose the standard basis for R 3 R e 1 = cos θ e 1 + sin θ e 2 = r 11 e 1 + r 21 e 2 + r 31 e 3 R e 2 = - sin θ e 1 + cos θ e 2 = r 12 e 1 + r 22 e 2 + r 32 e 3 R e 3 = e 3 = r 13 e 1 + r 23 e 2 + r 33 e 3 Reading off gives the matrix (note the transpose!): R = R z , θ = cos θ - sin θ 0 sin θ cos θ 0 0 0 1 (1.7) 2. Similarly, rotations by θ about the x and y axes are R = R x , θ = 1 0 0 0 cos θ - sin θ 0 sin θ cos θ , R = R y , θ = cos θ 0 - sin θ 0 1 0 sin θ 0 cos θ (1.8) 3. For V = polynomials of degree 3 , choose basis { e 1 , e 2 , e 3 , e 4 } = { 1, x , x 2 , x 3 } Take
Image of page 13

Subscribe to view the full document.

14 CHAPTER 1. LINEAR ALGEBRA (APPROX. 9 LECTURES) linear operator D = d dx : V V . D e 1 = 0 = 4 i =1 d i 1 e i D e 2 = 1 = 4 i =1 d i 2 e i D e 3 = 2 x = 4 i =1 d i 3 e i D e 4 = 3 x 2 = 4 i =1 d i 4 e i From the first equation we can read off the first column of D : d 11 = d 21 = d 31 = d 41 = 0 Continuing, gives the matrix D = [ d ij ] = 0 1 0 0 0 0 2 0 0 0 0 3 0 0 0 0 One can summarize these steps in a memorable way: D [1, x , x 2 , x 3 ] = [0, 1, 2 x , 3 x 3 ] = [1, x , x 2 , x 3 ] D In other words, DU = U D . Now check that D does the job. For example, P ( x ) = 1 + 2 x 3 has coordinates (1, 0, 0, 2) T . Apply D to this column vector: 0 1 0 0 0 0 2 0 0 0 0 3 0 0 0 0 1 0 0 2 = 0 0 6 0 which gives the coordinates for P ( x ) = 6 x 2 . Question: What happens when we compose linear operators? Answer: Let A and B be linear operators V V : then C = B ◦ A : V V is also a linear operator. Given a basis { u i } n i =1 for V : how do their matrices A , B , C relate? Well, C u i = B ( A u i ) = B ( m j =1 a ji u j ) = m j =1 a ji ( B u j ) = m j =1 a ji ( p k =1 b kj u k ) = p k =1 ( m j =1 a ji b kj ) u k = p k =1 c ki u k
Image of page 14
1.4. LINEAR OPERATORS 15 from which we find matrix multiplication again: c ki = m j =1 b kj a ji or C = BA Examples 7. 1. Figure 1.1 shows several compositions of linear operators. The impor- tant fact about the matrix A of A given the basis { u i } is that the defining equations (1.6) are equivalent to A ◦ U = U ◦ A and A = U ◦ A ◦ U - 1 . (1.9) 2. The general rotation in R 3 can be written R = R z , ψ R x , θ R z , φ (1.10) φ , θ , ψ are called the three “Euler angles”. Note again that matrix multiplication is not commutative! 3. D 2 is the second derivative matrix. Question: Given A : V V and two bases { u i } and { v i } for V , how does the u -matrix A compare to the v -matrix A ? Answer: This is called change of basis . In Assignment 2 you are asked to show that A = C u , v A C - 1 u , v where C u , v = V - 1 U converts u -coordinates of any vector b V into its v -coordinates. Question: How do we get a matrix for an operator A : V W ? Answer: We need two bases: a basis { u i } i =1: N for V and a basis { w i } i =1: M for W . Relative to these two bases, the matrix of A has components A = ( a ij ) i =1: M , j =1: N defined by the equations A u i = M j =1 a ji w j , for all i = 1 : N Things go much the same as the special case when W = V and { u i } = { w i } . In summary, one finds that equation (1.9) extends to A ◦ U = W ◦ A , A = W ◦ A ◦ U - 1 , where U is the coordinate operator for the u -basis and W is the coordinate operator for the w -basis.
Image of page 15

Subscribe to view the full document.

16 CHAPTER 1. LINEAR ALGEBRA (APPROX. 9 LECTURES) 1.5 Invertible Linear Operators
Image of page 16
Image of page 17
  • Winter '10
  • kovarik

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern

Ask Expert Tutors You can ask You can ask ( soon) You can ask (will expire )
Answers in as fast as 15 minutes