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

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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 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 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. Subscribe to view the full document.

16 CHAPTER 1. LINEAR ALGEBRA (APPROX. 9 LECTURES) 1.5 Invertible Linear Operators  • Winter '10
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