Linear Algebra. Session 8

Transcription

1 Linear Algebra. Session 8 Dr. Marco A Roque Sol 08/01/2017

2 Abstract Linear Algebra 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 by L(V The kernel of L, denoted by ker(l is the set of all vectors v V such that L(v = 0.

3 Abstract Linear Algebra Example 8.1 Consider the linear tranformation, L : R 3 R 3, given by x x L y = = y = AX z z Find ker(l and range of L Solution The kernel, ker(l, is the nullspace of the matrix A. To find ker(l, we apply row reduction to the matrix A:

4 Abstract Linear Algebra Hence (x, y, z belongs to ker(l if x z = y = 0. It follows that ker(l is the line spanned by (1, 0, 1 Since L is given by L x y z = x y z then, 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, (0, 2, 0.

5 Abstract Linear Algebra Example 8.2 Consider the linear tranformation, L : C (R 3 C (R, given by L(u = u 2u + u. Find range and ker(l of L Solution According to the theory of differential equations, the initial value problem u(x 2u(x +u(x = g(x; u(x 0 = u 0 ; u (x 0 = u 0; u (x 0 = u 0 has a unique solution for any g C (R and any u 0, u 0, u 0. It Follows that L(C(R 3 = C(R

6 Abstract Linear Algebra Also, the initial data evaluation l(u = u (u 0, u 0, u 0 which is a linear mapping l(u : C(R 3 R 3, becomes invertible when restricted to ker(l. Hence dim (ker(l = 3 Now, the functions xe x, e x, 1 satisfy L(xe x = L(e x = L(1 = 0 and W (xe x, e x, 1 0. Therefore, ker(l = Span(xe x, e x, 1 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.

7 Abstract Linear Algebra 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 ker(l, and t 1,, t k are arbitrary scalars.

8 Abstract Linear Algebra Example 8.3 Consider the linear equation. { x + y + z = 4 x + 2y = 3 Find its solution. Solution L : R 3 R 3 ; L x y z = ( x y z

9 Abstract Linear Algebra the linear equation is L(x = b with b = ( 4 3 Using row reduction with the augmented matrix ( L : R 3 R ; ( ( { { x + 2z = 5 x = 2z + 5 y z = 1 y = z 1 (x, y, z = (5 2t, 1 + t, t = (5, 1, 0 + t( 2, 1, 1

10 Abstract Linear Algebra Example 8.4 Consider the linear equation. u (x 2u (x + u (x = e 2x Find its solution. Solution Linear operator L : C (R 3 C (R, given by L(u = u 2u + u Linear equation L(u = b where b(x = e 2x

11 Abstract Linear Algebra We already know that functions xe x, e x, 1 form a basis for the kernel of L. It remains to find a particular solution. But, since L is a linear operator, L( 1 2 e2x = e 2x. Thus, the particular solution is u 0 = 1 2 e2x and the general solution is u(x = c 1 xe x + c 2 e x + c e2x

12 Matrix transformations Any m n matrix A gives rise to a transformation L : R n R m, given by Ax = b. where x R n and L(x R m are regarded as column vectors. This transformation is linear. Thus, for instance L x y z = Let 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 A. x y z

13 Example 8.5 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 = (1, 0, 0, e 2 = (0, 1, 0, e 3 = (0, 0, 1 is the standard basis for R 3 Solution L(x, y, z = L(xe 1 + ye 2 + ze 3 = xl(e 1 + yl(e 2 + zl(e 3 = ( x x(1, 1+y(0, 2+z(3, 0 = (x+3z, x 2y y z Columns of the matrix are vectors L(e 1, L(e 2, L(e 3.

14 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 = a 11 a 12 a 1n a 21 a 22 a 2n a m1 a m2 a mn. x 1 x 2. x n

15 y 1 y 2. y m = x 1 a 11 a 21. a mn + x 2 a 12 a 22. a m2 + + x n a 1n a 2n. a mn +

16 Abstract Linear Algebra Change of coordinates Let V be a vector space. 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. R n g 1 V g 2 g 1 1 g 2 R n

17 The composition g 2 g 1 1 is a linear mapping of R n to itself. Hence it s represented as 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. Matrix of a linear transformation Let V and W be vector spaces and f : V W be a linear map. 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.

18 Abstract Linear Algebra Let w 1, w 2,..., w m be another basis for V and g 2 : V R m be the coordinate mapping corresponding to this basis. V f W g 1 R n g 2 f g 1 1 g 2 R m The composition g 2 f g 1 is a linear mapping of R n R m. Hence it s represented as x Ax, where A is an m n matrix, called the matrix of f with respect to bases v 1, v 2,..., v n and w 1, w 2,..., w n. Columns of A are coordinates of vectors f (v 1, f (v 2,..., f (v n with respect to the basis w 1, w 2,..., w n.

19 D : P 3 P 2 ; (Dp(x = p (x Let A D be the matrix of D with respect to the bases {1, x, x 2 } and {1, x}. Columns of A D are coordinates of polynomials {D1, Dx, Dx 2 } w.r.t. the basis {1, x}. D1 = 0, Dx = 1, Dx 2 = 2x A D = (

20 L : P 3 P 3 ; (Lp(x = p(x + 1 Let A L be the matrix of D with respect to the basis {1, x, x 2 }. Columns of A L are coordinates of polynomials {L(1, L(x, L(x 2 } w.r.t. the basis {1, x}. L1 = 1, Lx = 1 + x, Lx 2 = (x = 1 + 2x + x A D =

21 Example 8.6 Consider a linear operator L on the vector space of 2 2 matrices given by ( x y L z w = ( ( x y z w Find the matrix of L with respect to the basis E 11 = ( ( 0 1, E 12 = 0 0 ( 0 0, E 21 = 1 0 ( 0 0, E 22 = 0 1

22 Solution Let M L be the matrix of L with respect to the basis {E 11, E 12, E 21, E 22 }. Columns of M L are coordinates of {L(E 11, L(E 12, L(E 21, L(E 22, } w.r.t. the basis {E 11, E 12, E 21, E 22 }. L(E 11 = ( ( = ( = 1E 11 +0E 12 +3E 21 +0E 22

23 L(E 12 = ( ( = ( = 0E 11 +1E 12 +0E 21 +3E 22 L(E 21 = ( ( = ( = 2E 11 +0E 12 +4E 21 +0E 22 L(E 22 = ( ( = ( = 0E 11 +2E 12 +0E 21 +4E 22

24 therefore M L = Thus, the relation ( x1 y 1 z 1 w 1 is equivalent to the relation x 1 y 1 z 1 w 1 = = ( ( x y z w x y z w

25 Example 8.7 Consider a linear operator L : R 2 R 2 given by ( ( ( x 1 1 x L = y 0 1 y Find the matrix of L with respect to the basis {v 1 = (3, 1, v 2 = (2, 1}. Solution Let N L be the matrix of L with respect to the basis {v 1, v 2 }. Columns of N L are coordinates of L(v 1, L(v 2 w.r.t. the basis {v 1, v 2 }

26 L(v 1 = ( ( 3 1 = ( 4 1 { ( α = 2 3 β = 1 ; L(v 2 = 1 = αv 1 +βv 2 = 1v 1 + 0v 2 { 3α + 2β = 4 α + β = 1 Thus N L = (

27 Change of basis for a linear operator Let L : V V be a linear operator on a vector space V. Let A be the matrix of L with respect to the basis a 1, a 2,..., a n for V. Let B be the matrix of L with respect to the basis b 1, b 2,..., b n for V. Let U be the transition matrix from the basis a 1, a 2,..., a n to b 1, b 2,..., b n

28 Abstract Linear Algebra a coordinate of v A a coordinate of L(v U b coordinate of v B U b coordinate of L(v It follows that UAx = BUx for all x R n UA = BU. Then, the diagram commutes and A = U 1 BU and B = U 1 AU

29 Example 8.8 Consider a linear operator L : R 2 R 2 given by ( ( ( x 1 1 x L = y 0 1 y Find the matrix of L with respect to the basis v 1 = (3, 1, v 2 = (2, 1 Solution Let S be the matrix of L with respect to the standard basis, let N be the matrix of L with respect to the basis v 1, v 2, and U be the transition matrix from v 1, v 2 to e 1, e 2.

30 Then, N = U 1 SU ( ( S =, U = ( ( ( N = U SU = ( ( ( = =

31 Similarity of matrices Definition. An n n matrix B is said to be similar to an n n matrix A if B = S 1 AS for some nonsingular n n matrix S. Remark. Two n n matrices are similar if and only if they represent the same linear operator on R n with respect to different bases.

32 Theorem Similarity is an equivalence relation, which means that (i Any square matrix A is similar to itself; (ii If B is similar to A, then A is similar to B; (iii If A is similar to B and B is similar to C, then A is similar to C proof (i A = I 1 AI

33 (ii If B = S 1 AS then A = SBS 1 = (S 1 1 BS 1 = (S 1 1 BS 1 where S 1 = S 1 (iii If A = S 1 BS and B = T 1 CT then A = S 1 BS = S 1 T 1 CTS = (TS 1 CTS = (S 2 1 CS 2 where S 2 = TS Corollary The set of n n matrices is partitioned into disjoint subsets (called similarity classes such that all matrices in the same subset are similar to each other while matrices from different subsets are never similar.

34 Theorem If A and B are similar matrices then they have the same (i determinant, (ii trace = the sum of diagonal entries, (iii rank, and (iv nullity.