Dr. Marques Sophie Linear algebra II SpringSemester 2016 Problem Set # 11

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1 Dr. Marques Sophie Linear algebra II SpringSemester 2016 Office 519 Problem Set # 11 Justify all your answers completely (Or with a proof or with a counter example) unless mentioned differently. No step should be a mystery or bring a question. The grader cannot be expected to work his way through a sprawling mess of identities presented without a coherent narrative through line. If he can t make sense of it in finite time you could lose serious points. Coherent, readable exposition of your work is half the job in mathematics. You will loose serious points if your exposition is messy, incomplete, uses mathematical symbols not adapted... Exercise 1: Find the Jordan canonical form of the following matrices over C and deduce the corresponding canonical form for those matrices over R and give for both the base change matrices that permit to get those forms: 1. A = 2. B = 3. C = Solution: The characteristic polynomial is 4 x 1 1 det(a xi) = 2 x x = x3 + 6x 2 16x + 16 = (x 2)(x 2 4x + 8). Hence, the eigenvalues are λ 1 = 2, λ 2 = 2 2i, λ 3 = 2 + 2i. 1

2 Denote the corresponding eigenvectors v 1, v 2, v 3. Since the eigenvalues are distinct, the Jordan canonical form is simply given by the diagonal matrix [A] X = 0 2 2i i where X = {v 1, v 2, v 3 } is an ordered basis of eigenvectors. Solving for ker(a λ i I) for i = 1, 2, 3, we have We can now assign ker(a λ 1 I) = Span{(1, 1, 1)} ker(a λ 2 I) = Span{((3 i), (3 + 4i), 5)} ker(a λ 3 I) = Span{((3 + i), (3 4i), 5)}. v 1 = (1, 1, 1), v 2 = ((3 i), (3 + 4i), 5), v 3 = ((3 + i), (3 4i), 5). Since the matrix is diagonalizable, the dimension of each eigenspace is 1. Let f = x + iy E λ2 (T C ). Using the notation from class where J is the conjugate operator, it follows that J(f) E λ3 (T C ). If W = Cf CJ(f), we have shown in class that E λ1 E λ2 = W and the set Y = {x, y} is a basis for the T -invariant subspace (W ) R (T := L A ). Using polar coordinates for λ 2 = 2 2i gives λ 2 = re iθ = 8e i/4 so ( ) cos θ sin θ [T (W )R ] Y = r sin θ cos θ = ( cos( 8 4 ) sin( 4 ) sin( 4 ) cos( 4 ) where Y = {x, y}. Since v 3 = J(v 2 ), let f = v 2 so Y = {(3, 3, 5), ( 1, 4, 0)}. Notice that Z = {v 1 } Y forms an ordered basis for R 3. The corresponding canonical form for A over R is then [T ] Z = 0 8 cos( ) sin( ) sin( ) = cos( with change of basis matrix so A = S[T ] Z S 1. 4 ) S = ).

3 2. The characteristic polynomial is 2 x det(b xi) = 1 4 x x x = (2 x)(4 x)(x ) + (x ) = (x i)(x + i)(x 3) 2. The eigenvalues (including multiplicity) given by the roots are λ 1 = 3, λ 2 = 3, λ 3 = i, λ 4 = i. Solving for x in (B λ i I)x = 0 for i = 1, 3, 4 and (B λ 1 I) 2 x = 0 gives ker(b λ 1 I) = {(1, 1, 1, 1)} ker(b λ 1 I) 2 = {(1, 0, 0, 0), (0, 1, 1, 1)} ker(b λ 3 I) = {(0, 0, 1, 1 i)} ker(b λ 4 I) = {(0, 0, 1, 1 + i)}. Define vectors that are in each of the above kernels v 1 = (1, 1, 1, 1), v 2 = ( 1, 0, 0, 0), v 3 = (0, 0, 1, 1 i), v 4 = (0, 0, 1, 1 + i), respectively. Hence, X = {v 1, v 2, v 3, v 4 } is an ordered basis for the vector space that puts B in Jordan canonical form J = i i with change of basis matrix S that has ordered columns given by the vectors in X. To find the corresponding form over R, let f = v 3. If f = x + iy, then x = (0, 0, 1, 1) and y = (0, 0, 0, 1). Since λ 4 = λ 3, we have that Y = {x, y} is an ordered basis for M λ3,λ 4 = E λ3 E λ4. Writing λ 3 in polar coordinates gives us λ 3 = i = e i 2 = re iθ so r = 1 and θ =. Letting T = L 2 B, ( ) ( ) cos sin [T Eλ3,λ 4 ] Y = sin cos =

4 Let Z = {v 1, v 2, x, y} be an ordered basis for the vector space over R. The matrix with columns equal to the ordered vectors in Z gives us the desired change of basis matrix to get the corresponding canonical form over R. Denote this matrix P = The corresponding canonical form is then [T ] Z = = P 1 BP The characteristic polynomial is given by 3 x det(c xi) = 1 x x x = (3 x)( x 3 3x 2 3 3x) 2x x 2 + 4x x 2 + 2x + 6 = x 4 + 2x The roots of the polynomial gives us the eigenvalues (including multiplicity) λ 1 = i, λ 2 = i, λ 3 = i, λ 4 = i. Examining the kernels ker(c λ i I), we have generalized eigenvectors v 1 = ( 1 + i 2, 0, 1 + i 2, 1) ker(c λ 1I) v 2 = ( 2 i 2, 1 + i 2, 3 2i, 0) 2 ker(c λ 1 I) 2 ker(c λ 1 I) v 3 = ( 1 i 2, 0, 1 i 2, 1) ker(c λ 3I) v 4 = ( 2 + i 2, 1 i 2, 3 + 2i, 0) 2 ker(c λ 3 I) 2 ker(c λ 3 I) with (C λ 1 )(v 2 ) = v 1 and (C λ 3 )(v 4 ) = v 3. Hence, the Jordan canonical form is given by taking the basis X = {v 1, v 2, v 3, v 4 } for V to get i [T ] X = 0 i i i 4

5 where T = L C. For the form over R, first notice that λ 1 = λ 2 = λ 3 = λ 4. Putting this into polar coordinates gives λ 1 = i = re iθ = e i 2 so r = 1 and θ =. Let f 2 1 = x + iy = v 1 = Jv 3 and f 2 = u + iw = v 2 = Jv 4. Then Y = {x, y, u, w} gives us a basis for M λ1,λ 3. But V = M λ1 M λ3 so Y is a basis for the vector space. Hence, the corresponding canonical form is cos θ sin θ sin θ cos θ [T ] Y = 0 0 cos θ sin θ 0 0 sin θ cos θ = with base change matrix S = Exercise 2: Prove the following important property of the exponential map det(e A ) = e T r(a) for all A M(n, F) holds when F = C. Then explain how complexification can be used to show this result remains true when F = R. Solution: If F = R an R-basis {e i } for V gives a C-basis {ẽ j = e j + i0} in V C and it is trivial to check that 1. (e A ) C = e (A C) 2. det(a C ) = det(a). 3. T r(a C ) = T r(a). Thus it suffices to consider the case in which F = C, A = A C, V = V C. Then A can be put into Jordan canonical form. If sp(a) = {λ i }, V = i M λi. The generalized eigenspace decomposition, and p i = projection onto M λi. The generalized eigenspace decomposition, and p i =projection onto M λi. The operator D = r i=1 λ it i is diagonalized by any basis X that runs through M λ1, M λ2, in succession and if this basis is suitably chosen A D = N is upper triangular, with [D, N] = 0. Now T r(a) = T r(d + N) = T r 5 J J r = r m i λ i i=1

6 where J i = λ i 0 λ i with (m i = dim(m λi ) the algebraic multiplicity of λ i in p T (x)). Hence, E 1 0 e A = 0 E r where So that det(e A ) = E i = 0 e λ i e λ i r (e λ i ) m i = i=1 r e m iλ i = e i m iλ i = e T r(a) i=1 Exercise 3: Find A M(2, C) such that L A (x) = A x is diagonalizable, but not orthogonally diagonalizable with respect to the standard inner product on C 2. Solution: Let {e 1, e 2 } the standard ON basis in C 2 ; f 1 = e 1, f 2 = e 1 + e 2 and define linear operator T : C 2 C 2 such that T (f 1 ) = f 1, T (f 2 ) = 2f 2. Obviously, T is diagonalized ( ) over the bases {f 1, f 2 }. Its matrix with the standard basis X = {e 1, e 2 } is 1 1 A = since T (e ) = e 1, T (e 2 ) = e 1 + 2e 2. The eigenspaces E λ (T ), λ = 1 and λ = 2, are uniquely determined and are not orthogonal. Therefore no ON basis can diagonalize T. Exercise 4: Let V be an inner product space T : V V diagonalizable. Prove that T is orthogonally diagonalizable if and only if the eigenspace E λ (T ) in V = λ Sp(T ) E λ (T ) are orthogonal. Solution: The E λ (T ) are defined without reference to any inner product on V. If the E λ (T ) are orthogonal, we may take ON basis in each E λ (T ); combining then we get an ON basis for V consisting of T -eigenvectors (because T (e i ) = λe i, for any vector e i E λ (T ).) That proves ( ). On the other hand, if there is an ON basis {e i } such that T (e i ) = λe i, 1 i n, we 6

7 may lump together the e i associated with a single eigenvalue λ. If sp(t ) = {µ j : 1 j r} and we let I j = {i {1, n} : λ i = µ j }. Then [1, n] = r j=1i j (disjoint union) and V j = ( i Ij Ce i ) is included the eigenspace E µj (T ). Obviously, these eigenspaces are orthogonal (since e i e i, if i i in [1, n]) and V = r j=1e µj (T ) since T is diagonalizable by hypotheses. But V = r j=1v j (orthogonal direct sum ) and V j E µj (T ) then V j = E µj (T ) if V is finite dimensional. Since V i V j if j j, we must then have E µj (T ) E µ j (T ) for j = j and the eigenspace E µj (T ) are orthogonal, as claimed. Exercise 5: Decide whether A, B are unitarily equivalent. ( ) ( ) /2 1. A = and B = 1 0 1/ A = and B = Solution: 1. If B = UAU then in particular B is similar to A (conjugate by some 2 2 matrix with det(s) 0). But sp(a) = roots of λ 2 1 = { 1, 1}, sp(b) = roots of λ 2 1/4 = { 1/2, 1/2}. But sp(sas 1 ) = sp(a) for any matrix. 2. Now sp(a) = sp(b) = {1, 2, 3} (the diagonal entries of an upper triangular matrix are its eigenvalues). Solving (A λi)x = 0 for λ = 1, 2, 3, e compute the eigenspaces for A. E λ=1 = C = Ce 3, E λ=2 = C(e 1 + e 2 ), E λ=3 = C (e 1 + 2e 2 + e 3 ) Obviously, B is orthogonally diagonalizable, diagonalized by X = {e 1, e 2, e 3 } (ON basis). If B is unitarily equivalent to A, with A = UBU for some U U(2), then U AU = B then A is diagonalized by the ON basis Y = U(X ), vectors in Y are eigenvectors, but the eigenvectors are obviously not orthogonal. Hence B is not unitarily equivalent to A. Exercise 6: If T is a unitary operator on a finite dimensional inner product space. Prove that it has a unitary square root U (such that U 2 = T ). Solution: T being normal is orthogonally diagonalizable say with ON bases of eigenvectors. T (f i ) = λ i f i with each λ i sp(t ) (reapeats allowed). These eigenvalues must 7

8 have λ = 1. In fact if Uv = λv, (v 0) then v 2 = (v, v) = (Uv, Uv) = (λv, λv) = λ λ(v, v) = λ 2 v 2 hence λ 2 = 1, it has polar form λ = e iθ = cos(θ) + i(sin(θ)), for some real λ; then µ = e iθ/2 satisfies µ 2 = λ and µ = 1. Define U, so e iθ 1/2 0 [U] Y = 0 e iθn/2 when Y is the ON basis of eigenvectors for T. Obviously U U = I since e iθ 1/2 0 U = 0 e iθn/2 Exercise 7: Let U : V V be a linear operator on a finite dimensional inner product space. If U(x) = x, for all x V (U is an isometry), is U necessarily a unitary operator? Solution: If U is a bijection, the polarization identities imply that (Ux, Uy) V = (x, y) V, making U unitary. Since dim(v ) <, U is surjective if and only if it is one-to-one if and only if ker(u) = {x V : Ux = 0} is trivial. But Ux = 0 if and only Ux 2 = x 2 if and only if x = 0 in V. The answer is yes. 8