MIT OpenCourseWarehttp://ocw.mit.edu 18.034 Honors Differential Equations Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � LECTURE 28. REPEATED EIGENVALUES AND THE MATRIX EXPONENTIAL Repeated eigenvalues. Again, we start with the real n × n system (28.1) �y � = A�y. We say an eigenvalue λ or A is repeated if it is a multiple root of pA(λ). That is, pA(λ) has∗ (λ − λ )2 as a factor. We suppose that λ∗ is a double root of pA(λ), so that in principle two linearly ∗independent solutions of (28.1) correspond to λ∗. If �v is the corresponding eigenvector, then ∗ �y = �v∗eλ∗t is a solution. The problem is to find the second solution of (28.1), linearly independent of �y∗. We first discuss an easy case. Example 28.1 (The complete case). Still supposing that λ is a double root of pA(λ), we say λ is a ∗ ∗ complete eigenvalue if there are two linearly independent eigenvectors corresponding to λ∗, say �v1 and �v2. Using them, we obtain two linearly independent solutions of (28.1), namely �y1(t) = �v1e λ1t and �y2(t) = �v2e λ1t Exercise. Let A be a 2 × 2 matrix. If λ is a repeated and complete eigenvalue of A, show that λ 0 A = .0 λ The converse holds true. In general, if an eigenvalue λ∗ of A is k-tuply repeated, that is, pA(λ) = |A − λI| has the power (λ−λ )k as a factor, but no higher, the eigenvalue is called complete if k linearly independent eigen-∗vectors correspond to λ . In this case, these eigenvectors produce k linearly independent solutions ∗of (28.1). An important result in linear algebra is that if a real square matrix A is symmetric, that is, A = AT , then all its eigenvalues are real and complete. However, in general, an eigenvalue of multiplicity k (> 1) has less than k eigenvectors, and we cannot construct the general solution from the techniques of eigenvalues. The matrix exponential. To motivate, the initial value problem y� = ay, y(0) = 1 has the solution y(t) = eat in the form of exponential. We want to define the expression eAt for a general n × n matrix A, n � 1, so that Y (t) = eAt is a solution of Y � = AY, Y (0) = I and moreover eAt is easy to compute. Recall that if a is a real number then (at)2 (at)n �∞ (at)n e at = 1 + at + + + + = ,2! · · · n! · · · n! n=0 1�� � � � � � � � � � � � where t ∈ R. A natural way to define the matrix exponential eAt, for an n × n matrix A, seems to use the series expression t2A2 tnAn � tnAn∞(28.2) e At = I + tA + + + + = .2! · · · n! · · · n! n=0 In order to make sense of the above expression, we first introduce the matrix norm. Definition 28.2. For an n × n matrix A, define the matrix norm as �A� = sup |A�y|, �=0 y|y|�where |�y| = |�y T �y|1/2 = (y2 + · · · + yn2 )1/2 and |A�y| = |(A�y)T (A�y)|1/2 .1 Exercise. Show that t�A + B� � �A� + �B�, �AB� = �A� �B�, �tA|| = | | �A� for any matrices A, B and for any t ∈ R. For a matrix-valued function A(t) = (aij (t))n , then A(t) A(t0) means equivalently: i,j=1→ (i) aij (t) aij (t0) as t t0 for all 1 � i, j � n.→ →(ii) ||A(t) − A(t0)|| → 0 as t → t0. Under the matrix norm, the infinite series (28.2) converges for all A and for all t, and it defines the matrix exponential. We now compute the derivative of eAt by differentiating the right side of (28.2) term by term d d t2A2 tnAn e At = I + tA + + + + dt dt 2! · · · n! · · · tn−1An =A + tA + + +· · · (n − 1)! · · · =AeAt . Moreover, since eA0 = I, by definition, the matrix exponential eAt is a solution of Y � = AY, Y (0) = I. Theorem 28.3. Let Y (t) be a fundamental matrix of A. Then Y (t) = Y (0)eAt . For several classes of A, the infinite series in (28.2) can be summed up exactly. Exercise. 1. Show that exp(diag (λ1 . . . λn)) = diag (eλ1 . . . eλn ). 0 1 At 0 t2. Show that if A = then e = .0 0 0 0 Exercise. Prove the exponential law (28.3) e(A+B)t = e At e Bt if AB = BA. We use (28.3) to compute the matrix exponentials of more matrices. 2 1Example 28.4. Let A = . We write A = B + C, where 0 2 2 0 0 1 B = , C = .0 2 0 0 Lecture 28 2 18.034 Spring 2009� � Since BC = CB, by the results from the previous exercise and by (28.3) � �� � � � At e2t 0 1 t 2t 1 t e = = e .0 e2t 0 1 0 1 Finding the fundamental matrix. In general, it doesn’t seem possible to express eAt in closed form. Nevertheless, we can find n independent vectors for which eAt can be computed exactly. The idea is to write e At�v = e(A−λI)t e λIt�v = e(A−λI)t e λt�v. If (A − λI)m�v = 0 for some integer m > 0, then (A − λI)m+l�v = 0 for all integers l � 0. Hence, e(A−λI)t�v = �v + t(A − λI)�v + · · · +(mtm− −1 1)!(A − λI)m−1�v is a finite sum, and tm−1 e At�v = e λt �v + t(A − λI)�v + · · · +(m − 1)!(A − λI)m−1�v can be computed exactly, although eAt itself cannot be computed. Example 28.5. Solve the system Y � = AY , where ⎛ ⎞ 1 1 0 A = ⎝0 1 0⎠ . 0 0 2 SOLUTION. Its characteristic polynomial is pA(λ) = (1 − λ)2(2 − λ), and it has a double root λ = 1 and a simple root λ = 2. If λ = 1, then ⎛ ⎞ 0 1 0 (A − λI)�v = ⎝0 0 0⎠ �v = 0 0 0 1 ⎛ ⎞ 1 has a solution �v1 = ⎝0⎠. Furthermore, it is the only eigenvector for λ = 1 up to a constant 0 multiple. (In fact, the well-known result from linear algebra tells us that the solution space of the above linear system of equations has dimension 1. ) One solution of the system Y � = AY is⎛ ⎞ 1 obtained and �y1(t) = et ⎝0⎠. 0 To find the second solution associated to λ = 1, we compute ⎛ ⎞ 0 0 0 (A − λI)2�v = ⎝0 0 0⎠ �v = 0 0 0 1 ⎛ ⎞ 0 has a nontrivial solution …
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