Linear Transformations

Transcription

1 DM554 Linear and Integer Programming Lecture 8 Marco Chiarandini Department of Mathematics & Computer Science University of Southern Denmark

2 Outline

3 Resume Linear dependence and independence Determine linear dependency of a set of vertices, ie, find non-trivial lin. combination that equal zero Basis Find a basis for a linear space Find a basis for the null space, range and row space of a matrix (from its reduced echelon form) Dimension (finite, infinite) Rank-nullity theorem 3

4 Outline

5 Definition (Linear Transformation) Let V and W be two vector spaces. A function T : V W is linear if for all u, v V and all α R: 1. T (u + v) = T (u) + T (v) 2. T (αu) = αt (u) A linear transformation is a linear function between two vector spaces If V = W also known as linear operator Equivalent condition: T (αu + βv) = αt (u) + βt (v) for all 0 V, T (0) = 0 5

6 Example () vector space V = R, F 1 (x) = px for any p R x, y R, α, β R : F 1 (αx + βy) = p(αx + βy) = α(px) + β(px) = αf 1 (x) + βf 1 (y) vector space V = R, F 1 (x) = px + q for any p, q R or F 3 (x) = x 2 are not linear transformations T (x + y) T (x) + T (y) x, y R vector spaces V = R n, W = R m, m n matrix A, T (x) = Ax for x R n T (u + v) = A(u + v) = Au + Av = T (u) + T (v) T (αu) = A(αu) = αau = αt (u) 6

7 Example () vector spaces V = R n, W : f : R R. T : R n W : u 1 u 2 T (u) = T. = p u 1,u 2,...,u n = p u u n p u1,u 2,...,u n = u 1 x 1 + u 2 x 2 + u 3 x u n x n p u+v (x) = = (p u + p v )(x) p αu (x) = = αp u (x) 7

8 and Matrices any m n matrix A defines a linear transformation T : R n R m T A for every linear transformation T : R n R m there is a matrix A such that T (v) = Av A T Theorem Let T : R n R m be a linear transformation and {e 1, e 2,..., e n } denote the standard basis of R n and let A be the matrix whose columns are the vectors T (e 1 ), T (e 2 ),..., T (e n ): that is, A = [ T (e 1 ) T (e 2 )... T (e n ] Then, for every x R n, T (x) = Ax. Proof: write any vector x R n as lin. comb. of standard basis and then make the image of it. 8

9 Example T : R 3 R 3 T x y = z x + y + z x y x + 2y 3z The image of u = [1, 2, 3] T can be found by substitution: T (u) = [6, 1, 4] T. to find A T : 1 T (e 1 ) = 1 1 T (e 2 ) = 1 1 T (e 3 ) = A = [T (e 1 ) T (e 2 ) T (e n )] = T (u) = Au = [6, 1, 4] T.

10 Linear Transformation in R 2 We can visualize them! Reflection in the x axis: [ [ ] x x T : y] y A T = [ ] Stretching the plane away from the origin [ ] [ ] 2 0 x T (x) = 0 3 y 10

11 (the matrix A is correct, in the lecture, I made a mistake placing the θ angle on the other side of e 2) Rotation anticlockwise by an angle θ y 1 T (e e 2 2 ) θ T (e 1 ) θ (0, 0) e 1 1 x we search the images of the standard basis vector e 1, e 2 [ ] [ ] a d T (e 1 ) =, T (e c 1 ) = b they will be orthogonal and with length 1. [ ] [ ] a b cos θ sin θ A = = c d sin θ cos θ For π/4: A = [ ] a b = c d [ ] cos θ sin θ sin θ cos θ [ ] =

12 Identity and Zero For T : V V the linear transformation such that T (v) = v is called the identity. if V = R n, the matrix A T = I (of size n n) For T : V W the linear transformation such that T (v) = 0 is called the zero transformation. If V = R n and W = R m, the matrix A T is an m n matrix of zeros. 12

13 Composition of Let T : V W and S : W U be linear transformations. The composition of ST is again a linear transformation given by: ST (v) = S(T (v)) = S(w) = u where w = T (v) ST means do T and then do S: V T W S U if T : R n R m and S : R m R p in terms of matrices: ST (v) = S(T (v)) = S(A T v) = A S A T v note that composition is not commutative 13

14 Combinations of If S, T : V W are linear transformations between the same vector spaces, then S + T and αs, α R are linear transformations. hence also αs + βt, α, β R is 14

15 Inverse If V and W are finite-dimensional vector spaces of the same dimension, then the inverse of a lin. transf. T : V W is the lin. transf such that T 1 (T (v)) = v In R n if T 1 exists, then its matrix satisfies: T 1 (T (v)) = A T 1A T v = I v that is, T 1 exists iff (A T ) 1 exists and A T 1 = (A T ) 1 (recall that if BA = I then B = A 1 ) In R 2 for rotations: [ ] cos( θ) sin( θ) A T 1 = = sin( θ) cos( θ) [ ] cos θ sin θ sin θ cos θ 15

16 Example Is there an inverse to T : R 3 R 3 T x y = x + y + z x y z x + 2y 3z A = Since det(a) = 9 then the matrix is invertible, and T 1 is given by the matrix: A 1 = u T 1 3 u v w v = u 4 9 v w w 3 u v 2 9 w 16

17 from V to W Theorem Let V be a finite-dimensional vector space and let T be a linear transformation from V to a vector space W. Then T is completely determined by what it does to a basis of V. Proof (unique representation in V implies unique representation in T ) 17

18 If both V and W are finite dimensional vector spaces, then we can find a matrix that represents the linear transformation: suppose V has dim(v ) = n and basis B = {v 1, v 2,..., v n } and W has dim(w ) = m and basis S = {w 1, w 2,..., w m }; coordinates of v V are [v] B coordinates of T (v) W are [T (v)] S we search for a matrix A such that: [T (v)] S = A[v] B we find it by: [T (v)] S = a 1 [T (v 1 )] S + a 2 [T (v 2 )] S + + a n [T (v n )] S = [[T (v 1 )] S [T (v 2 )] S [T (v n )] S ] [v] B where [v] B = [a 1, a 2,..., a n ] T 18

19 Range and Null Space Definition (Range and null space) T : V W. The range R(T ) of T is: R(T ) = {T (v) v V } and the null space (or kernel) N(T ) of T is N(T ) = {v V T (v) = 0} the range is a subspace of W and the null space of V. Matrix case, T : R n R m R(T ) = R(A) N(T ) = N(A) Rank-nullity theorem: rank(t ) = dim(r(t )) nullity(t ) = dim(n(t )) rank(t ) + nullity(t ) = dim(v ) 19

20 Example Construct a linear transformation T : R 3 R 3 with N(T ) = t 1 2 : t R, R(T ) = xy-plane. 3 20

21 Outline

22 Coordinates Recall: Definition (Coordinates) If S = {v 1, v 2,..., v n } is a basis of a vector space V, then any vector v V can be expressed uniquely as v = α 1 v α n v n and the real numbers α 1, α 2,..., α n are the coordinates of v wrt the basis S. To denote the coordinate vector of v in the basis S we use the notation α 1 α 2 [v] S =. α n S In the standard basis the coordinates of v are precisely the components of the vector v: v = v 1 e 1 + v 2 e v n e n How to find coordinates of a vector v wrt another basis? 22

23 Transition from Standard to Basis B Definition (Transition Matrix) Let B = {v 1, v 2,..., v n } be a basis of R n. The coordinates of a vector x wrt B, a = [a, a 2,..., a n ] T = [x] B, are found by solving the linear system: a 1 v 1 + a 2 v a n v n = x that is x = [v 1 v 2 v n ]a We call P the matrix whose columns are the basis vectors: P = [v 1 v 2 v n ] Then for any vector x R n x = P[x] B transition matrix from B coords to standard coords moreover P is invertible (columns are a basis): [x] B = P 1 x transition matrix from standard coords to B coords 23

24 Example B = 1 2, 2 1, P = [v] B = det(p) = 4 0 so B is a basis of R 3 standard coordinates of v: v = = v = B = 9 3 5

25 Example (cntd) B = 1 2, 2 1, 3 2, [x] = B coordinates of vector x: = a a a either we solve Pa = x in a by Gaussian elimination or we find the inverse P 1 : 1 [x] B = P 1 x = 1 check the calculation 2 B What are the B coordinates of the basis vector? ([1, 0, 0], [0, 1, 0], [0, 0, 1])

26 Change of Basis Since T (x) = Px then T (e i ) = v i, ie, T maps standard basis vector to new basis vectors Example Rotate basis in R 2 by π/4 anticlockwise, find coordinates of a vector wrt the new basis. [ cos π A T = 4 sin ] [ ] π sin π 4 cos π = Since the matrix A T rotates {e 1, e 2 }, then A T = P and its columns tell us the coordinates of the new basis and v = P[v] B and [v] B = P 1 v. The inverse is a rotation clockwise: [ P 1 cos( π = 4 ) sin( π 4 ) ] [ cos( π sin( π 4 ) cos( π 4 ) = 4 ) sin( π 4 ) ] sin( π 4 ) cos( π 4 ) [ ] =

27 Example (cntd) Find the new coordinates of a vector x = [1, 1] T [ ] [1 ] [x] B = P 1 x = = [ ]

28 Change of basis from B to B Given a basis B of R n with transition matrix P B, and another basis B with transition matrix P B, how do we change from coords in the basis B to coords in the basis B? coordinates in B v=p B [v] B [v] B =P 1 standard coordinates B v coordinates in B [v] B Theorem = P 1 B P B[v] B M = P 1 B P B = P 1 B [v 1 v 2 If B and B are two bases of R n, with B = {v 1, v 2,..., v n }... v n ] ex7sh3 = [P 1 B v 1 P 1 B v 2... P 1 B v n] then the transition matrix from B coordinates to B coordinates is given by M = [ ] [v 1 ] B [v 2 ] B [v n ] B (the columns of M are the B coordinates of the basis B) 28

29 Example B = {[ [ ]} 1 1, 2] 1 S = {[ [ 3 5, 1] 2]} are basis of R 2, indeed the corresponding transition matrices from standard basis: [ ] [ ] P = Q = have det(p) = 3, det(q) = 1. Hence, lin. indep. vectors. We are given [ ] 4 [x] B = 1 B find its coordinates in S. 29

30 Example (cntd) 1. find first the standard coordinates of x [ ] [ ] [ ] [ ] [ ] x = 4 = = and then find S coordinates: [ ] [ ] [x] S = Q x = = [ ] S 2. use transition matrix M from B to S coordinates: v = P[v] B and v = Q[v] S [v] S = Q 1 P[v] B : [ ] [ ] [ ] M = Q P = = [x] S = [ ] [ = 5 4 1] [ ] S