ACM 104. Homework Set 5 Solutions. February 21, 2001

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1 ACM 04 Homework Set 5 Solutions February, 00 Franklin Chapter 4, Problem 4, page 0 Let A be an n n non-hermitian matrix Suppose that A has distinct eigenvalues λ,, λ n Show that A has the eigenvalues λ,, λ n If u j is an eigenvector of A belonging to λ j, and if v k is an eigenvector of A belonging to λ k, show that (u j, v k ) = 0 if j k, and show that (u j, v j ) 0 This is known as the principle of biorthogonality Using theorem on page 77 we are able to say that since A has distinct eigenvalues, it has distinct eigenvectors Furthermore, there exists a matrix T such that Λ = T AT where Λ is the diagonal matrix Λ = λ 0 or A = T ΛT A = (T ΛT ) = (T ) Λ T But we know that (T ) = (T ) and Λ = Λ so A = (T ) ΛT Therefore, A is similar to the matrix Λ, and we conclude that A has n distinct eigenvalues (and n distinct eigenvectors) with Λ = λ 0 and Now consider Therefore, (Au j, v k ) = (λ j u j, v k ) = λ j (u j, v k ) (Au j, v k ) = (u j, A v k ) = (u j, λ k v k ) = λ k (u j, v k ) λ j (u j, v k ) = λ k (u j, v k ) Since the eigenvalues are distinct and non-zero, we must have that for j k, (u j, v k ) = 0 for this equality to hold Now suppose that (u j, v j ) = 0 We know that the u j s are independent and there are n of them in an n dimensional space so they form a basis Therefore, for any v k we may write v k = a i u i i=

2 with not all a i = 0 Now consider ( n ) (v j, u j ) = 0 a i u i, u j = 0 i= a i (u i, u j ) = 0 for any j Since this is true we conclude that the u i s must be dependent since a i 0 for all i This is a contradiction, therefore (u j, v k ) 0 when j = k i= Franklin Chapter 4, Problem 4, page 09 If a Hermitian matrix H is positive definite, show that H = P, where P is also positive definite The diagonalization of H is such that U HU = Λ, Λ = λ 0 Since H is positive definite we know that the diagonal elements of Λ are all positive therefore define λ 0 A = 0 λn If we now also define P = U AU then we have We can P = U AUU AU = U AAU = U ΛU = H, ie P = H Note also that P has only positive eigenvalues and so is positive definite 3 Franklin Chapter 4, Problem 5, page 09 If a Hermitian matrix H is positive definite, show that it satisfies the generalized Schwarz inequality, and the generalized triangle-inequality, (Hx, y) (Hx, x) (Hy, y) (H(x + y), x + y) (Hx, x) + (Hy, y) Since H is Hermitian, there exists a unitary matrix U such that U HU = Λ with λ 0 Λ = But H is positive definite so all λ i > 0 Therefore, we can write Λ = A where A = λ 0 0 λn

3 Since A is diagonal and real, note that A = A and we also have U HU = A = A A = AA so H = UAA U = UA(UA) Now, therefore Now note that therefore (Hx, y) = (UA(UA) x, y) = ((UA) x, (UA) y) (UA) x (UA) y = ((UA) x, (UA) x) ((UA) y, (UA) y) = (UA(UA) x, x) (UA(UA) y, y) = (Hx, x) (Hy, y) (Hx, y) (Hx, x) (Hy, y) (H(x + y), x + y) = (Hx + Hy, x + y) = (Hx, x) + (Hy, x) + (Hx, y) + (Hy, y) (Hx, x) + (Hy, x) + (Hy, y) = (Hx, x) + (Hx, x) (Hy, y) + (Hy, y) = ((Hx, x) + (Hy, y) ) (H(x + y), x + y) (Hx, x) + (Hy, y) 4 Franklin Chapter 4, Problem 4, page 5 If U AU = T is a unitary triangularization, show that a ij = t ij i= j= i j We have AA = a a n a n a nn a a n a n a nn = n k= a k n k= a nk Similarly, T T = n k= t k n k= t nk Then note that tr(aa ) = i= j= a ij, tr(t T ) = i= j= t ij = t ij i j So, in order to prove the above result we need only show that tr(aa ) = tr(t T ) Now, since A = UT U we have AA = UT U (UT U ) = UT U U T U = UT U UT U = UT T U, ie AA is similar to T T Since the trace is invariant under a change of coordinates we are done 3

4 5 Franklin Chapter 4, Problem 5, page 5 Consider a system of differential equations dx/dt = Ax, where A is not similar to a diagonal matrix Let U AU = T be a unitary triangularization of A Show how the change of variable x = Uy allows the system to be resolved recursively for y n, then y n,, and finally y We have dx dt = Ax replacing x with Uy we get now multiply both sides by U so that d Uy = AUy dt U d dt Uy = U AUy d dt y = T y that is, or y y n = λ y y n y n = λ n y n y n = λ n y n + t n,n y n So the first equation will allow us to solve for y n Substituting this into the second equation will allow us to solve for y n Continuing in this manner, we have y solved 6 Franklin Chapter 4, Problem 4, page 0 Let N N = NN Show that there is positive-semidefinite Hermitian matrix, P, and a unitary matrix, V, such that N = P V = V P (Use the factorization, N = UΛU If NN = UΛΛ U = P, how can P be defined?) Note that so define ΛΛ = P = U λ 0 0 λ n λ 0 0 λ n U P is clearly Hermitian, and since the eigenvalues of P are non-negative, P is semi-definite So now we want to find V unitary such that N = P V If N is nonsingular, all eigenvalues will be nonzero so we can write V = P N V is unitary since V V = P NN P = P UΛΛ U P = P P P = I Now suppose that N is singular We want to find V such that V V = I and P V = UΛU Assume V = U DU where D is diagonal So λ 0 P V = U U UDU 0 λ n 4

5 so we want λ 0 0 λ n D = Λ Looking at it term by term, we want λ i µ i = λ i so µ i = λi λ i If λ i = 0 let µ i = Let D = µ 0 0 µ where µ i = { λi λ i, if λ i 0, if λ i = 0 so V = U DU Note that V V = UDD U = I so U is unitary and by our definition V P = U 7 A skew Hermitian matrix is a matrix that satisfies K = K Show that K = UΛU, with Λ diagonal and for a unitary U Show that the eigenvalues are imaginary and the eigenvectors orthogonal 3 Show that K + I is invertible 4 Show that (I K)(I + K) is a unitary matrix 5 Show that e Kt is a unitary matrix Consider the matrix ik Then so ik is hermitian (ik) = (i) K = i( K) = ik Since ik is hermitian, by theorem on page 00 there exist a unitary matrix U and a diagonal matrix Λ, so that ik = UΛU K = U( i)λu () The matrix K has the same eigenvectors as ik and its eigenvalues are the eigenvalues of ik, multiplied by i Since ik is hermitian, K has orthogonal eigenvectors and its eigenvalues are imaginary 3 To find the eigenvalues of K + I we just add to the eigenvalues of K If v is an eigenvector of K and λ is the corresponding eigenvalue: Kv = λv (K + I)v = (λ + )v Since the eigenvalues of K are imaginary, the eigenvalues of K + I are all greater than in absolute value, therefore nonzero and K + I is invertible 5

6 4 Consider ((I K)(I + K) ) (I K)(I + K) = ((I + K) ) (I K) (I K)(I + K) Note that (I + K) and (I K) commute for any matrix K, so = ((I + K) ) (I + K)(I K)(I + K) ((I K)(I + K) ) (I K)(I + K) = ((I + K) ) (I K)(I + K)(I + K) So (I K)(I + K) is unitary = ((I + K) ) (I K) = ((I + K) ) (I + K) = ((I + K)(I + K) ) = I = I 5 Let λ,, λ n denote the imaginary eigenvalues of K: e Kt = K j t j j= (UΛU ) j t j = j= = U Λ j t j U = U = U j= j= λ j tj e λt e λnt j= U λ j n tj U Since the eigenvalues λ i are imaginary, e λi is a complex number with magnitude Using the last expression it is obvious that (e Kt ) e Kt = I so e Kt is unitary 6