Section 7.4 Matrix Representations of Linear Operators

Transcription

1 Section 7.4 Matrix Representations of Linear Operators Definition. Φ B : V à R n defined as Φ B (c 1 v 1 +c v + + c n v n ) = [c 1 c c n ] T. Property: [u + v] B = [u] B + [v] B and [cu] B = c[u] B for all u, v V and scalar c. 1

2 Section 7.4 Matrix Representations of Linear Operators Definition. Property: [u + v] B = [u] B + [v] B and [cu] B = c[u] B for all u, v V and scalar c. Example: Let V = Span B, where B = {et cost, e t sint} is L.I. and thus a basis of V. Consider the function v = e t cos (t π /4). (Is it in V?) Then v is in V since In addition, [v] B = v = 1 p e t (cos t +sint) = 1 p e t cos t + 1 p e t sin t h 1 p 1 p i T

3 Consider a linear transformation T: V W Questions: 1) Can we define a standard matrix for T? ) If not, what kind of matrix representation of T can we formulate? In this course, we will consider only a simpler case where T is a linear operator (i.e., the domain and the codomain are the same vector space). 3

4 Let T : V V be a linear operator on an n-dimensional vector space V with a basis B. Define the linear operator Φ B T (Φ B ) -1 : R n R n, and consider its standard matrix A, called the matrix representation of T with respect to B and denoted as [T] B. With the notations, [T] B = A and T A = Φ B T (Φ B ) -1. V T V (Φ B ) -1 Φ B R n Φ B T (Φ B ) -1 R n 4

5 Let T : V V be a linear operator on an n-dimensional vector space V with a basis B. Define the linear operator Φ B T (Φ B ) -1 : R n R n, and consider its standard matrix A, called the matrix representation of T with respect to B and denoted as [T] B. With the notations, [T] B = A and T A = Φ B T (Φ B ) -1. V T V (Φ B ) -1 Φ B R n Φ B T (Φ B ) -1 Question: How to express [T] B in terms of T and b 1, b,, b n? R n 5

6 Property: If B = {v 1, v,, v n }, then [T] B = [ [T(v 1 )] B [T(v )] B [T(v n )] B ]. Proof [T] B = A Ae j = T A (e j ) = Φ B T (Φ B ) -1 (e j ) = Φ B T(v j ) = [T(v j )] B. Example: Let T: P P be defined by T(p(x)) = p(0) + 3p(1)x + p()x for all p(x) in P. Then T is linear. For B = {1, x, x }, [T] B = A = [ a 1 a a 3 ] and 6 a 1 0 = [ T(1)] = [1+ 3 x+ x ] = 3, [ T( x)] [3x x ] 3 a = = + =, 1 1 B B B B a3 = [ T( x )] [3x 4 x ] 3, so [ T] [ 1 3] B = + B = B = a a a =

7 Example: Let V = Span B, where B = {et cost, e t sint} is L.I. and thus a basis of V, and the linear operator D: V V be defined by D( f ) = f ʹ for all f V. Then D( e D( e t t cost) sin t) = = (1) e (1) e t t t cost + ( 1) e sin t [ D] t B cost + (1) e sin t 1 =

8 Theorem 7.10 Proof [T(v)] B = Φ B T(v) = Φ B T (Φ B ) -1 Φ B (v) = T A ([v] B ) = [T] B [v] B, where A = [T] B. 8

9 Example: Relative to the basis B = {1, x, x } of P, the coordinate vector of p(x) = 5-4x + 3x is [p(x)] B = [ ] T. Then [pʹ (x)] B = [D(p(x))] B = [D] B [p(x)] B,where D: P P is defined by D(p(x)) = pʹ (x), and [ D] 0 0 [ pʹ ( x)] B = B = =

10 The Matrix Representation of the Inverse of an Invertible Linear Operator Proof (a) Note that Φ B is an isomorphism with an inverse (Φ B ) -1, which is also an isomorphism. If T is invertible, then T A = Φ B T (Φ B ) -1 is a composition of isomorphisms. So T A is invertible and has an invertible standard matrix A. If A is invertible, then T A = Φ B T (Φ B ) -1 is invertible. So T = (Φ B )-1 T A Φ B is invertible. 10

11 The Matrix Representation of the Inverse of an Invertible Linear Operator Proof (b) By (a) and the invertibility of T, T C = Φ B T -1 (Φ B ) -1, where C = [T -1 ] B. Also by (a), T A -1 = (T A ) -1 = Φ B T -1 (Φ B ) -1. T C = T A -1 C = A

12 Example: In the vector space V with a basis B = {et cost, e t sint} and a linear operator D: V V defined by D( f ) = f ʹ f V, it is known that [D] B = apple So an anti-derivative of e t sint is D -1 (e t sint). Since [D -1 ] B = ([D] B )-1 and [e t sint] B = [ 0 1]T, [D 1 (e t sin t)] B = apple apple 0 1 = apple 1 1 i.e., D -1 (e t sint) = -(e t cost)/ + (e t sint)/. 1

13 Definition. Example: For linear operator D: C C defined by D( f ) = f ʹ with C = {f f :R R, f has derivatives of all order}, it has an eigenvector e λt C corresponding to the eigenvalue λ, since D( f )(t) = (e λt )ʹ = λe λt = λ f (t). Any scalar λ is an eigenvalue of D. D has infinitely many eigenvalues. 13

14 Example: f is a solution of yʹ ʹ + 4y = 0 f C, since f must be twice-differentiable and f ʹ ʹ = -4f, which imply that the fourth derivative of f exists (f ʹ ʹ ʹ ʹ = -4f ʹ ʹ ), and so on. f 0 is an eigenvector of D : C C corresponding to the eigenvalue -4 f eigenspace of D corresponding to the eigenvalue -4 Clearly, every vector in the eigenspace of D corresponding to the eigenvalue -4 is a solution of yʹ ʹ + 4y = 0. 14

15 Example: U: R n n R n n defined by U(A) = A T is an isomorphism. The eigenspace of U corresponding to the eigenvalue 1 is the set of A Rn n such that U(A) = A T = A, i.e., the set of symmetric matrices. The eigenspace of U corresponding to the eigenvalue -1 is the set of A Rn n such that U(A) = A T = -A, i.e., the set of skew-symmetric matrices. U only has eigenvalues 1 and -1, since U(A) = A T = λa implies A = (A T ) T = (λa) T = λa T = λ(λa) = λ A. 15

16 Eigenvalues and Eigenvectors of a Matrix Representation of a Linear Operator Proof only if ( ) Suppose v 0 satisfies T(v) = λv. [v] B 0 and A[v] B = [T] B [v] B = [T(v)] B = [λv] B = λ[v] B if ( ) Suppose w 0 satisfies Aw = [T] B w = λw. Let v = (Φ B ) -1 (w) 0. Φ B T(v) = [T(v)] B = [T] B [v] B = λ[v] B = Φ B (λv) T(v) = λv since Φ B is an isomorphism. 16

17 Example: Let T: P P be defined as T(p(x)) = p(0) + 3p(1)x + p()x 17 for all p(x) in P. Then T is linear, and [T] B = A = where B = {1, x, x } is a basis of P The characteristic polynomial of A is -(t - 1) (t - 6). Span{[ 0-3 ] T } is the eigenspace of A corresponding to the eigenvalue 1 ap(x) with a 0 and p(x) = -3x + x is an eigenvector of T corresponding to the eigenvalue 1. Span{[ ] T } is the eigenspace of A corresponding to the eigenvalue 6 bq(x) with b 0 and q(x) = x + x is an eigenvector of T corresponding to the eigenvalue

18 Example: U: R R defined by U(A) = A T is an isomorphism and only has eigenvalues 1 and -1. Let a basis of R be B = apple , apple , apple , apple [U] B = The characteristic polynomial of [U] B is (t - 1) 3 (t + 1), for which indeed the only roots are 1 and

19 19 Homework Set for Section 7.4 Section 7.4: Problems 1, 5, 9, 11, 19, 1, 3, 8, 3, 36, 39, 40, 43, 44, 46.