MATH 304 Linear Algebra Lecture 15: Linear transformations (continued). Range and kernel. Matrix transformations.

Transcription

1 MATH 304 Linear Algebra Lecture 15: Linear transformations (continued). Range and kernel. Matrix transformations.

2 Linear mapping = linear transformation = linear function Definition. Given vector spaces V 1 and V 2, a mapping L : V 1 V 2 is linear if L(x + y) = L(x) + L(y), L(rx) = rl(x) for any x,y V 1 and r R.

3 Properties of linear mappings Let L : V 1 V 2 be a linear mapping. L(r 1 v r k v k ) = r 1 L(v 1 ) + + r k L(v k ) for all k 1, v 1,...,v k V 1, and r 1,...,r k R. L(r 1 v 1 + r 2 v 2 ) = L(r 1 v 1 ) + L(r 2 v 2 ) = r 1 L(v 1 ) + r 2 L(v 2 ), L(r 1 v 1 + r 2 v 2 + r 3 v 3 ) = L(r 1 v 1 + r 2 v 2 ) + L(r 3 v 3 ) = = r 1 L(v 1 ) + r 2 L(v 2 ) + r 3 L(v 3 ), and so on. L(0 1 ) = 0 2, where 0 1 and 0 2 are zero vectors in V 1 and V 2, respectively. L(0 1 ) = L(00 1 ) = 0L(0 1 ) = 0 2. L( v) = L(v) for any v V 1. L( v) = L(( 1)v) = ( 1)L(v) = L(v).

4 Examples of linear mappings Scaling L : V V, L(v) = sv, where s R. L(x + y) = s(x + y) = sx + sy = L(x) + L(y), L(rx) = s(rx) = r(sx) = rl(x). Dot product with a fixed vector l : R n R, l(v) = v v 0, where v 0 R n. l(x + y) = (x + y) v 0 = x v 0 + y v 0 = l(x) + l(y), l(rx) = (rx) v 0 = r(x v 0 ) = rl(x). Cross product with a fixed vector L : R 3 R 3, L(v) = v v 0, where v 0 R 3. Multiplication by a fixed matrix L : R n R m, L(v) = Av, where A is an m n matrix and all vectors are column vectors.

5 Linear mappings of functional vector spaces Evaluation at a fixed point l : F(R) R, l(f ) = f (a), where a R. Multiplication by a fixed function L : F(R) F(R), L(f ) = gf, where g F(R). Differentiation D : C 1 (R) C(R), L(f ) = f. D(f + g) = (f + g) = f + g = D(f ) + D(g), D(rf ) = (rf ) = rf = rd(f ). Integration over a finite interval l : C(R) R, l(f ) = a, b R, a < b. b a f (x) dx, where

6 Properties of linear mappings If a linear mapping L : V W is invertible then the inverse mapping L 1 : W V is also linear. If L : V W and M : W X are linear mappings then the composition M L : V X is also linear. If L 1 : V W and L 2 : V W are linear mappings then the sum L 1 + L 2 is also linear.

7 Linear differential operators an ordinary differential operator L : C (R) C d 2 (R), L = g 0 dx + g d 2 1 dx + g 2, where g 0, g 1, g 2 are smooth functions on R. That is, L(f ) = g 0 f + g 1 f + g 2 f. Laplace s operator : C (R 2 ) C (R 2 ), f = 2 f x f y 2 (a.k.a. the Laplacian; also denoted by 2 ).

8 Basis and coordinates If {v 1,v 2,...,v n } is a basis for a vector space V, then any vector v V has a unique representation v = x 1 v 1 + x 2 v x n v n, where x i R. The coefficients x 1, x 2,...,x n are called the coordinates of v with respect to the ordered basis v 1,v 2,...,v n. The mapping vector v its coordinates (x 1, x 2,...,x n ) is a one-to-one correspondence between V and R n. This correspondence respects linear operations in V and in R n, i.e., it is a linear transformation.

9 Change of coordinates Let V be a vector space of dimension n. Let v 1,v 2,...,v n be a basis for V and g 1 : V R n be the coordinate mapping corresponding to this basis. Let u 1,u 2,...,u n be another basis for V and g 2 : V R n be the coordinate mapping corresponding to this basis. V g 1 g 2 ւ ց R n R n The composition g 2 g 1 1 is a linear transformation of R n. It has the form x Ux, where U is an n n matrix. U is called the transition matrix from v 1,v 2...,v n to u 1,u 2...,u n. Columns of U are coordinates of the vectors v 1,v 2,...,v n with respect to the basis u 1,u 2,...,u n.

10 Example. Let V be the subspace of C(R) spanned by the function f (x) = xe x + 1 and its derivatives. Then dim V = 3. One basis for V is v 1 = f, v 2 = f, v 3 = f. Another basis is u 1 (x) = xe x, u 2 (x) = e x, u 3 (x) = 1. v 1 (x) = xe x + 1 = u 1 (x) + u 3 (x), v 2 (x) = xe x + e x = u 1 (x) + u 2 (x), v 3 (x) = xe x + 2e x = u 1 (x) + 2u 2 (x). Transition matrix from v 1, v 2, v 3 to u 1, u 2, u 3 : Notice that (u 1, u 2, u 3 ) = (v 1, v 2, v 3 )

11 Range and kernel Let V,W be vector spaces and L : V W be a linear mapping. Definition. The range (or image) of L is the set of all vectors w W such that w = L(v) for some v V. The range of L is denoted L(V). The kernel of L, denoted ker(l), is the set of all vectors v V such that L(v) = 0. Theorem (i) The range of L is a subspace of W. (ii) The kernel of L is a subspace of V.

12 x x Example. L : R 3 R 3, L y = y. z z The kernel ker(l) is the nullspace of the matrix. x L y = x 1 + y 2 + z 1 z The range L(R 3 ) is the column space of the matrix.

13 x x Example. L : R 3 R 3, L y = y. z z The range of L is spanned by vectors (1, 1, 1), (0, 2, 0), and ( 1, 1, 1). It follows that L(R 3 ) is the plane spanned by (1, 1, 1) and (0, 1, 0). To find ker(l), we apply row reduction to the matrix: Hence (x, y, z) ker(l) if x z = y = 0. It follows that ker(l) is the line spanned by (1, 0, 1).

14 Example. L : C 3 (R) C(R), L(u) = u 2u + u. According to the theory of differential equations, the initial value problem u (x) 2u (x) + u (x) = g(x), x R, u(a) = b 0, u (a) = b 1, u (a) = b 2 has a unique solution for any g C(R) and any b 0, b 1, b 2 R. It follows that L(C 3 (R)) = C(R). Also, the initial data evaluation I(u) = (u(a), u (a), u (a)), which is a linear mapping I : C 3 (R) R 3, is one-to-one when restricted to ker(l). Hence dim ker(l) = 3. It is easy to check that L(xe x ) = L(e x ) = L(1) = 0. It follows that ker(l) = Span(xe x, e x, 1).

15 General linear equations Definition. A linear equation is an equation of the form L(x) = b, where L : V W is a linear mapping, b is a given vector from W, and x is an unknown vector from V. The range of L is the set of all vectors b W such that the equation L(x) = b has a solution. The kernel of L is the solution set of the homogeneous linear equation L(x) = 0. Theorem If the linear equation L(x) = b is solvable and dim ker L <, then the general solution is x 0 + t 1 v t k v k, where x 0 is a particular solution, v 1,...,v k is a basis for the kernel of L, and t 1,...,t k are arbitrary scalars.

16 { x + y + z = 4, Example. x + 2y = 3. x ( ) L : R 3 R 2, L y = x y z z ( ) 4 Linear equation: L(x) = b, where b =. 3 ( ) ( ) ( ) { { x + 2z = 5 x = 5 2z y z = 1 y = 1 + z (x, y, z) = (5 2t, 1 + t, t) = (5, 1, 0) + t( 2, 1, 1).

17 Example. u (x) 2u (x) + u (x) = e 2x. Linear operator L : C 3 (R) C(R), Lu = u 2u + u. Linear equation: Lu = b, where b(x) = e 2x. We already know that functions xe x, e x and 1 form a basis for the kernel of L. It remains to find a particular solution. L(e 2x ) = 8e 2x 2(4e 2x ) + 2e 2x = 2e 2x. Since L is a linear operator, L ( 1 2 e2x) = e 2x. Particular solution: u 0 (x) = 1 2 e2x. Thus the general solution is u(x) = 1 2 e2x + t 1 xe x + t 2 e x + t 3.

18 Matrix transformations Any m n matrix A gives rise to a transformation L : R n R m given by L(x) = Ax, where x R n and L(x) R m are regarded as column vectors. This transformation is linear. x x Example. L y = y. z z Let e 1 = (1, 0, 0), e 2 = (0, 1, 0), e 3 = (0, 0, 1) be the standard basis for R 3. We have that L(e 1 ) = (1, 3, 0), L(e 2 ) = (0, 4, 5), L(e 3 ) = (2, 7, 8). Thus L(e 1 ), L(e 2 ), L(e 3 ) are columns of the matrix.

19 Problem. Find a linear mapping L : R 3 R 2 such that L(e 1 ) = (1, 1), L(e 2 ) = (0, 2), L(e 3 ) = (3, 0), where e 1,e 2,e 3 is the standard basis for R 3. L(x, y, z) = L(xe 1 + ye 2 + ze 3 ) = xl(e 1 ) + yl(e 2 ) + zl(e 3 ) = x(1, 1) + y(0, 2) + z(3, 0) = (x + 3z, x 2y) ( ) ( ) x + 3z L(x, y, z) = = x y x 2y z Columns of the matrix are vectors L(e 1 ), L(e 2 ), L(e 3 ).

20 Theorem Suppose L : R n R m is a linear map. Then there exists an m n matrix A such that L(x) = Ax for all x R n. Columns of A are vectors L(e 1 ), L(e 2 ),...,L(e n ), where e 1,e 2,...,e n is the standard basis for R n. y = Ax y 1 y 2. y m = x 1 y 1 y 2. y m a 11 a a 1n x 1 = a 21 a a 2n x a m1 a m2... a mn x n a 11 a 21. a m1 + x 2 a 12 a 22. a m2 + + x n a 1n a 2n. a mn