10.4 Matrix Exponential
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1 104 Matrix Exponential Matrix Exponential The prolem x t = Axt, x0 = x 0 has a unique solution, according to the Picard-Lindelöf theorem Solve the prolem n times, when x 0 equals a column of the identity matrix, and write w 1 t,, w n t for the n solutions so otained Define the matrix exponential y packaging these n solutions into a matrix: e At augw 1 t,,w n t By construction, any possile solution of x = Ax can e uniquely expressed in terms of the matrix exponential e At y the formula xt = e At x0 Matrix Exponential Identities Announced here and proved elow are various formulae and identities for the matrix exponential e At : e At = Ae At e 0 = I Be At = e At B e At e Bt = e A+Bt e At e As = e At+s e At 1 = e At Columns satisfy x = Ax Where 0 is the zero matrix If AB = BA If AB = BA At and As commute Equivalently, e At e At = I e At = r 1 tp r n tp n e At = e λ 1t I + eλ 1t e λ 2t A λ 1 I e At = e λ 1t I + te λ 1t A λ 1 I Putzer s spectral formula See page 508 A is 2 2, λ 1 λ 2 real A is 2 2, λ 1 = λ 2 real e At = e at cos t I + eat sin t A ai A is 2 2, λ 1 = λ 2 = a+i, > 0 e At = n=0 A ntn n! Picard series See page 510 e At = P 1 e Jt P Jordan form J = PAP 1
2 506 Putzer s Spectral Formula The spectral formula of Putzer applies to a system x = Ax to find the general solution, using matrices P 1,,P n constructed from A and the eigenvalues λ 1,,λ n of A, matrix multiplication, and the solution rt of the first order n n initial value prolem r t = λ λ λ λ n rt, r0 = The system is solved y first order scalar methods and ack-sustitution We will derive the formula separately for the 2 2 case the one used most often and the n n case Putzer s Spectral Formula for a 2 2 matrix A The general solution of x = Ax is given y the formula xt = r 1 tp 1 + r 2 tp 2 x0, where r 1, r 2, P 1, P 2 are defined as follows The eigenvalues r = λ 1,λ 2 are the two roots of the quadratic equation deta ri = 0 Define 2 2 matrices P 1, P 2 y the formulae P 1 = I, P 2 = A λ 1 I The functions r 1 t, r 2 t are defined y the differential system r 1 = λ 1 r 1, r 1 0 = 1, r 2 = λ 2 r 2 + r 1, r 2 0 = 0 Proof: The Cayley-Hamilton formula A λ 1 IA λ 2 I = 0 is valid for any 2 2 matrix A and the two roots r = λ 1, λ 2 of the determinant equality deta ri = 0 The Cayley-Hamilton formula is the same as A λ 2 P 2 = 0, which implies the identity AP 2 = λ 2 P 2 Compute as follows x t = r 1tP 1 + r 2tP 2 x0 = λ 1 r 1 tp 1 + r 1 tp 2 + λ 2 r 2 tp 2 x0 = r 1 ta + λ 2 r 2 tp 2 x0 = r 1 ta + r 2 tap 2 x0 = Ar 1 ti + r 2 tp 2 x0 = Axt This proves that xt is a solution Because Φt r 1 tp 1 + r 2 tp 2 satisfies Φ0 = I, then any possile solution of x = Ax can e represented y the given formula The proof is complete
3 104 Matrix Exponential 507 Real Distinct Eigenvalues Suppose A is 2 2 having real distinct eigenvalues λ 1, λ 2 and x0 is real Then and r 1 = e λ 1t, r 2 = eλ 1t e λ 2T xt = e λ1t I + eλ1t e λ 2t A λ 1 I x0 The matrix exponential formula for real distinct eigenvalues: e At = e λ 1t I + eλ 1t e λ 2t A λ 1 I Real Equal Eigenvalues Suppose A is 2 2 having real equal eigenvalues λ 1 = λ 2 and x0 is real Then r 1 = e λ 1t, r 2 = te λ 1t and xt = e λ1t I + te λ1t A λ 1 I x0 The matrix exponential formula for real equal eigenvalues: e At = e λ 1t I + te λ 1t A λ 1 I Complex Eigenvalues Suppose A is 2 2 having complex eigenvalues λ 1 = a + i with > 0 and λ 2 = a i If x0 is real, then a real solution is otained y taking the real part of the spectral formula This formula is formally identical to the case of real distinct eigenvalues Then Rext = Rer 1 ti + Rer 2 ta λ 1 Ix0 = Ree a+it at sin t I + Ree A a + ii x0 = e at cos t I + e atsin t A ai x0 The matrix exponential formula for complex conjugate eigenvalues: e At = e at cos t I + sin t A ai How to Rememer Putzer s Formula for a 2 2 Matrix A The expressions 1 e At = r 1 ti + r 2 ta λ 1 I, r 1 t = e λ1t, r 2 t = eλ1t e λ 2t
4 508 are enough to generate all three formulae The fraction r 2 t is a difference quotient with limit te λ 1t as λ 2 λ 1, therefore the formula includes the case λ 1 = λ 2 y limiting If λ 1 = λ 2 = a + i with > 0, then the fraction r 2 is already real, ecause it has for z = e λ 1t and w = λ 1 the form r 2 t = z z w w sin t = Taking real parts of expression 1 then gives the complex case formula for e At Putzer s Spectral Formula for an n n Matrix A The general solution of x = Ax is given y the formula xt = r 1 tp 1 + r 2 tp r n tp n x0, where r 1, r 2,, r n, P 1, P 2,, P n are defined as follows The eigenvalues r = λ 1,,λ n are the roots of the polynomial equation deta ri = 0 Define n n matrices P 1,, P n y the formulae P 1 = I, P k = A λ k 1 IP k 1, k = 2,,n More succinctly, P k = Π k 1 j=1 A λ ji The functions r 1 t,, r n t are defined y the differential system r 1 = λ 1 r 1, r 1 0 = 1, r 2 = λ 2 r 2 + r 1, r 2 0 = 0, r n = λ n r n + r n 1, r n 0 = 0 Proof: The Cayley-Hamilton formula A λ 1 I A λ n I = 0 is valid for any n n matrix A and the n roots r = λ 1,, λ n of the determinant equality deta ri = 0 Two facts will e used: 1 The Cayley-Hamilton formula implies AP n = λ n P n ; 2 The definition of P k implies λ k P k + P k+1 = AP k for 1 k n 1 Compute as follows 1 x t = r 1 tp r n tp nx0 n n 2 = λ k r k tp k + r k 1 P k x0 3 = 4 = n 1 n 1 k=2 n 1 λ k r k tp k + r n tλ n P n + r k P k+1 x0 r k tλ k P k + P k+1 + r n tλ n P n x0
5 104 Matrix Exponential = n 1 r k tap k + r n tap n x0 n 6 = A r k tp k x0 7 = Axt Details: 1 Differentiate the formula for xt 2 Use the differential equations for r 1,,r n 3 Split off the last term from the first sum, then re-index the last sum 4 Comine the two sums 5 Use the recursion for P k and the Cayley-Hamilton formula A λ n IP n = 0 6 Factor out A on the left 7 Apply the definition of xt This proves that xt is a solution Because Φt n r ktp k satisfies Φ0 = I, then any possile solution of x = Ax can e so represented The proof is complete Proofs of Matrix Exponential Properties Verify e At = Ae At Let x 0 denote a column of the identity matrix Define xt = e At x 0 Then e At x0 = x t = Axt = Ae At x 0 Because this identity holds for all columns of the identity matrix, then e At and Ae At have identical columns, hence we have proved the identity e At = Ae At Verify AB = BA implies Be At = e At B Define w 1 t = e At Bw 0 and w 2 t = Be At w 0 Calculate w 1t = Aw 1 t and w 2t = BAe At w 0 = ABe At w 0 = Aw 2 t, due to BA = AB Because w 1 0 = w 2 0 = w 0, then the uniqueness assertion of the Picard-Lindelöf theorem implies that w 1 t = w 2 t Because w 0 is any vector, then e At B = Be At The proof is complete Verify e At e Bt = e A+Bt Let x 0 e a column of the identity matrix Define xt = e At e Bt x 0 and yt = e A+Bt x 0 We must show that xt = yt for all t Define ut = e Bt x 0 We will apply the result e At B = Be At, valid for BA = AB The details: x t = e At ut = Ae At ut + e At u t = Axt + e At But = Axt + Be At ut = A + Bxt We also know that y t = A + Byt and since x0 = y0 = x 0, then the Picard-Lindelöf theorem implies that xt = yt for all t This completes the proof
6 510 Verify e At e As = e At+s Let t e a variale and consider s fixed Define xt = e At e As x 0 and yt = e At+s x 0 Then x0 = y0 and oth satisfy the differential equation u t = Aut By the uniqueness in the Picard-Lindelöf theorem, xt = yt, which implies e At e As = e At+s The proof is complete Verify e At = A ntn The idea of the proof is to apply Picard iteration n! n=0 By definition, the columns of e At are vector solutions w 1 t,, w n t whose values at t = 0 are the corresponding columns of the n n identity matrix According to the theory of Picard iterates, a particular iterate is defined y y n+1 t = y 0 + t 0 Ay n rdr, n 0 The vector y 0 equals some column of the identity matrix The Picard iterates can e found explicitly, as follows y 1 t = y 0 + t 0 Ay 0dr = I + Aty 0, y 2 t = y 0 + t 0 Ay 1rdr = y 0 + t 0 AI + Aty 0dr y n t = = I + At + A 2 t 2 /2 y 0, I + At + A 2 t2 2 tn + + An n! y 0 The Picard-Lindelöf theorem implies that for y 0 = column k of the identity matrix, lim n y nt = w k t This eing valid for each index k, then the columns of the matrix sum N A m tm m! m=0 converge as N to w 1 t,, w n t This implies the matrix identity The proof is complete e At = n=0 A n tn n! Theorem 12 Special Formulas for e At e diagλ 1,,λ nt = diag e λ1t,,e λnt a a e t = e at cos t sin t sint cos t Real or complex constants λ 1,, λ n Real a,
7 104 Matrix Exponential 511 Theorem 13 Computing e Jt for J Triangular If J is an upper triangular matrix, then a column ut of e Jt can e computed y solving the system u t = Jut, u0 = v, where v is the corresponding column of the identity matrix This prolem can always e solved y first-order scalar methods of growth-decay theory and the integrating factor method Theorem 14 Block Diagonal Matrix If A = diagb 1,,B k and each of B 1,, B k is a square matrix, then e At = diag e B1t,,e B kt
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