Linear Algebra. Week 7

Transcription

1 Linear Algebra. Week 7 Dr. Marco A Roque Sol 10 / 09 / 2018

2 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 v (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.

3 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. The composition g 2 g 1 is a transformation of R n and 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. The 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

4 Example 7.7 Find the transition matrix from the basis v 1 = (1, 2, 3), v 2 = (1, 0, 1), v 3 = (1, 2, 1) to the basis u 1 = (1, 1, 0), u 2 = (0, 1, 1), u 3 = (1, 1, 1) Solution It is convenient to make a two-step transition: first from v 1, v 2, v 3 to e 1, e 2, e 3 and then from e 1, e 2, e 3 to u 1, u 2, u 3 Let U 1 be the transition matrix from v 1, v 2, v 3 to e 1, e 2, e 3 and let U 2 be the transition matrix from u 1, u 2, u 3 to e 1, e 2, e 3

5 U 1 = U 2 = Basis v 1, v 2, v 3 coordinates Basis e 1, e 2, e 3 coordinates Basis u 1, u 2, u 3 coordinates x U 1 x U 1 2 (U 1x)

6 Thus, the transition matrix from v 1, v 2, v 3 to u 1, u 2, u 3 is U 1 2 (U 1) U = U 1 2 (U 1) = = =

7 Definition. Given vector spaces V 1 and V 2 a mapping L : V 1 V 2 is linear if 1) L(x + y) = L(x) + L(y) 2)L(rx) = rl(x) Fot any x, y V 1 and r R OBS A function f : R R given by f (x) = ax + b is a linear transformation of the vector space R if and only if b = 0

8 Basic properties of linear transformations Let L : V 1 V 2 be a linear mapping. L(r 1 v 1 + r 2 v r k vk) = r 1 L(v 1 ) + r 2 L(v 2 ) + + r k L(v k ) for all k 1 : v 1, v 2,..., v k V; r 1, r 2,..., r k R L(0 1 ) = 0 2 where 0 1 and 0 2 are zero vectors in V 1 and V 2 respectively L( v) = L(v)

9 Examples 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(sx v 0 ) = rl(x)

10 Cross 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) Multiplication by a fixed matrix L : R n R m, L(v) = Av where A M m,n L(x + y) = A(x + y) = Ax + Ay = L(x) + L(y) L(rx) = A(rx) = r(ax) = rl(x)

11 Linear mappings of functional vector spaces Evaluation at a fixed point. L : F (R) R, L(f) = f (a) where a R L(f + g) = (f + g)(a) = f (a) + g(a) = L(f ) + L(g) L(rf ) = (rf )(a) = rf (a) = rl(f ) Multiplication by a fixed function. L : F (R) F (R), L(f ) = gf where g F (R) is a fixed function. L(f + h) = (f + h)(g) = fg + hg = L(f ) + L(h) L(rf ) = (rf )g = r(fg) = rl(f )

12 Differentiation. D : C 1 (R) C(R), D(f ) = f D(f + g) = (f + h) = f + g = D(f ) + D(h) D(rf ) = (rf ) = rf = rl(f ) Integration over a finite interval. I : C(R) R, I(f ) = b a f (x)dx I(f + g) = b a (f + g)(x)dx = b a (f )(x)dx + b a (g)(x)dx = I(f ) + I(g) I(rf ) = b a (rf )(x)dx = r b a f (x)dx = ri(f )

13 Ordinary Differential Operator. ( Linear differential operator). L : C (R) C d (R), L(f ) = g 2 f df 2 + g dx 2 1 dx + fg 0 where g 0, g 1, and g 2 are smooth functions on R L(f + g) = [g 2 d 2 dx 2 + g 1 d dx + g 0](f + g)(x) = [g 2 d 2 dx 2 + g 1 d dx + g 0](f )(x)+ [g 2 d 2 dx 2 + g 1 d dx + g 0](g)(x) = L(f ) + L(g) d L(rf ) = [g 2 d 2 + g dx 2 1 dx + g 0](rf )(x) = d r[g 2 d 2 + g dx 2 1 dx + g 0](f )(x) = rl(f )

14 Antiderivative Operator.(Linear integral operator). L : C[a, b] C 1 [a, b], (Lf )(x) = x a f (t)dt L(f + g) = x a (f + g)(t)dt = x a [f (t) + g(t)]dt = x a f (t)dt + x a g(t)dt = L(f ) + L(g) L(rf ) = x a (rf )(t)dt = x a rf (t)dt = r x a f (t)dt = rl(f )

15 Hilbert-Schmidt operator.(linear integral operator). L : C[a, b] C[c, d], (Lf )(x) = b a K(x, y)f (y)dy where K C ([a, b] [c, d]) L(f + g) = b K(x, y)(f + g)(y) = b a b a K(x, y)[f (y) + g(y])dy = a K(x, y)f (y) + b a K(x, y)g(y) = L(f ) + L(g) L(rf ) = x a (rf )(t)dt = x a rf (t)dt = r x a f (t)dt = rl(f )

16 Laplace transform. (Linear integral operator). L : BC(0, ) C(0, ), (Lf )(x) = 0 e xy f (y)dy L(f +g) = 0 e xy (f +g)(y)dy = 0 e xy [f (y)+g(y)]dy = 0 e xy f (y)dy + 0 e xy g(y)dy = L(f ) + L(g) L(rf ) = 0 e xy (rf )(y)dy = 0 e xy rf (y)dy = r 0 e xy f (y)dy = rl(f )

17 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 : W X are linear mappings then the sum L 1 + L 2 is also linear.

18 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.

19 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:

20 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).

21 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)

22 Also, the initial data evaluation l(u) = u (u 0, u 0, u 0 ) which is a linear mapping l(u) : C(R) 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.

23 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.

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

25 the linear equation is L(x) = b with b = ( 4 3 Using row reduction with the augmented matrix ( ) ( ) ( ) { { 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) )

26 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

27 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

28 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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

29 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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 ).

30 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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

31 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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 +

32 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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

33 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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.

34 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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 m. 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 m.

35 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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 = ( )

36 Matrix transformations. Matrix of a linear transformation. Similar matrices. L : P 3 P 3 ; (Lp)(x) = p(x + 1) Let A L be the matrix of L P 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) 2 = 1 + 2x + x A D =

37 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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 )

38 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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

39 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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

40 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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

41 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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 }

42 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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 = ( )

43 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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

44 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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

45 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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.

46 Matrix transformations. Matrix of a linear transformation. Similar matrices. Then, N = U 1 SU ( ) ( ) S =, U = ( ) ( ) ( N = U SU = ( ) ( ) ( ) = ) =

47 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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.

48 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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

49 Matrix transformations. Matrix of a linear transformation. Similar matrices. (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.

50 Matrix transformations. Matrix of a linear transformation. Similar matrices. 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.