MATH 304Linear AlgebraLecture 35:Matrix exponentials.• Initial value problem for a linear ODE:dydt= 2y , y(0) = 3.Solution: y(t) = 3e2t.• Initial value problem for a system of linear ODEs:(dxdt= 2x + 3y ,dydt= x + 4y,x(0) = 2, y (0) = 1.The system can be rewritten in vector formddtxy= Axy, where A =2 31 4.Solution:x(t)y(t)= etA21.What is etA?Exponential functionExponential function: f ( x) = exp x = ex, x ∈ R.Principal property: ex+y= ex· ey.Definition 1. ex= limn→∞1 +xnn.In particular, e = limn→∞1 +1nn≈ 2.7182818.Definition 2. ex= 1 + x +x22!+ · · · +xnn!+ · · ·Definition 3. f (x) = exis the unique solution ofthe initial value problem f′= f , f (0) = 1.Matrix exponentialsDefinition. For any square matrix A letexp A = eA= I + A +12!A2+ · · · +1n!An+ · · ·Matrix exponential is a limit of matrix polynomials .Remark. Let A(1), A(2), . . . be a sequence of n×nmatrices, A(n)= (a(n)ij). The sequence converges toan n×n matrix B = (bij) if a(n)ij→ bijas n → ∞,i.e., if each entry conv erges.Theorem The matrix exp A is well defined, i.e.,the series converges.Properties of matrix exponentialsTheorem 1 If AB = BA then eAeB= eBeA= eA+B.Corollary (a) etAesA= esAetA= e(t+s)A, t, s ∈ R;(b) eO= I ; (c) (eA)−1= e−A.Theorem 2ddtetA= AetA= etAA.Indeed, etA= I + tA +t22!A2+ · · · +tnn!An+ · · · ,and the series can be di fferentiated term by term.ddttnn!An=ddttnn!An=tn−1(n − 1)!An.Lemma Let A be an n×n matrix and x ∈ Rn.Then the vector function v(t) = etAx satisfiesv′= Av.Proof:dvdt=ddtetAx =AetAx = AetAx= Av.Theorem For any t0∈ R and x0∈ Rnthe initialvalue problemdvdt= Av, v(t0) = x0has a unique soluti on v(t) = e(t−t0)Ax0.Indeed, v(t) = e(t−t0)Ax0= etAe−t0Ax0= etAx,where x = e−t0Ax0is a cons tant vector.Evaluation of matrix exponentialsExample. A = diag(a1, a2, . . . , ak).An= diag(an1, an2, . . . , ank), n = 1, 2, 3, . . .eA= I + A +12!A2+ · · · +1n!An+ · · ·= diag(b1, b2, . . . , bk),where bi= 1 + ai+12!a2i+13!a3i+ · · · = eai.Theorem 1 If A = diag(a1, a2, . . . , ak) thenetA= diag(ea1t, ea2t, . . . , eakt).Theorem 2 If A = UBU−1, then etA= UetBU−1.Example. A =2 31 4.The eigenvalues of A: λ1= 1, λ2= 5.Eigenvectors: v1= (3 , −1 ), v2= (1 , 1 ).Therefore A = UBU−1, whereB =1 00 5, U =3 1−1 1.ThenetA= UetBU−1=3 1−1 1et00 e5t3 1−1 1−1=3ete5t−ete5t141 −11 3=143et+ e5t−3et+ 3e5t−et+ e5tet+ 3e5t.Problem. Solve a system of differential equations(dxdt= 2x + 3y,dydt= x + 4ysubject to initial conditions x(0) = 2, y (0) = 1.The unique solution:x(t)y(t)= etAx0, where A =2 31 4, x0=21.etA=143et+ e5t−3et+ 3e5t−et+ e5tet+ 3e5t=⇒(x(t) =34et+54e5t,y(t) = −14et+54e5t.Example. A =0 1 00 0 10 0 0, a Jordan block.A2=0 0 10 0 00 0 0, A3= O, An= O for n ≥ 3.eA= I + A +12A2=1 1120 1 10 0 1,etA= I + tA +t22A2=1 t12t20 1 t0 0 1.Example. A =1 10 1, another Jordan block.A2=1 20 1, A3=1 30 1, An=1 n0 1.etA= I +tA+t22!A2+ · · · +tnn!An+ · · · =a(t) b(t)0 a(t),where a(t) = 1 + t +t22!+t33!+ · · · = et,b(t) = t + 2t22!+ 3t33!+ · · · = tet.Thus etA=ettet0 et.Example. A =λ 10 λ, a general Jordan block.We have that A = λI + B, where B =0 10 0.Since (λI )B = B(λI ) , it foll ows that eA= eλIeB.Similarly, etA= etλIetB.etλI=eλt00 eλt= eλtI ,B2= O =⇒ etB= I + tB =1 t0 1.Thus etA= etλIetB=eλtteλt0 eλt.Problem. Solve a system of differential equations(dxdt= 2x + y,dydt= 2ysubject to initial conditions x(0) = y (0) = 1.The unique solution:x(t)y(t)= etAx0, where A =2 10 2, x0=11.etA=e2tte2t0 e2t= e2t1 t0 1=⇒(x(t) = e2t(1 + t),y(t) =
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