The Exponential of a Matrix 5-8- The solution to the exponential growth equation dx dt kx is given by x c e kt It is natural to ask whether you can solve a constant coefficient linear system x A x in a similar way If a solution to the system is to have the same form as the growth equation solution, it should look like x e At x The first thing I need to do is to make sense of the matrix exponential e At The Taylor series for e z is e z z n It converges absolutely for all z It A is an n n matrix with real entries, define e At n t n A n n The powers A n make sense, since A is a square matrix It is possible to show that this series converges for all t and every matrix A Differentiating the series term-by-term, d dt eat n n tn A n n t n A n (n )! A n t n A n (n )! A m t m A m m! Ae At This shows that e At solves the differential equation x A x The initial condition vector x() x yields the particular solution x e At x This works, because e A I (by setting t in the power series) Another familiar property of ordinary exponentials holds for the matrix exponential: If A and B commute (that is, AB BA), then e A e B e A+B You can prove this by multiplying the power series for the exponentials on the left (e A is just e At with t ) ExampleCompute e At if A 3 Compute the successive powers of A: A 3, A 4,,A n n 9 3 n
e At n t n n 3 n n (t) n n (3t) n e t e 3t You can compute the exponential of an arbitrary diagonal matrix in the same way: λ λ A λ n, eat e λ t e λ t e λ nt Example Compute e At if A Compute the successive powers of A: A, A 4, A 3 6,,A n n Hence, e A n t n n Here s where the last equality came from: n t n n n nt n t n e t te t e t n t n et, n nt n t n t n (n )! t m t m m! tet Example Compute e At, if A 3 4 If you compute powers of A as in the last two examples, there is no evident pattern it would be difficult to compute the exponential using the power series Instead, set up the system whose coefficient matrix is A: x 3x y, y x 4y The solution is x c e t +c e t, y 5 c e t + c e t
Next, note that if B is a matrix, B first column of B and B In particular, this is true for e At Now is the solution satisfying x() x, but x Set x() (,) to get the first column of e At : x e At x c e t +c e t 5 c e t + c e t second column of B c +c 5 c + c Hence, c 5 3, c 3 So x y 5 Set x() (,) to get the second column of e At : c 3, c 3 Hence, 3 et 3 e t 3 et 3 e t c +c 5 c + c x y 3 et + 3 e t 3 et + 5 3 e t 5 e At 3 et 3 e t 3 et + 3 e t 3 et 3 e t 3 et + 5 3 e t I found e At, but I had to solve a system of differential equations in order to do it In some cases, it s possible to use linear algebra to compute the exponential of a matrix An n n matrix A is diagonalizable if it has n independent eigenvectors (This is true, for example, if A has n distinct eigenvalues) Suppose A is diagonalizable with independent eigenvectors v,, v n and corresponding eigenvalues λ,,λ n Let S be the matrix whose columns are the eigenvectors: S v v v n 3
Then As I observed above, On the other hand, since (S AS) n S A n S, Hence, e Dt λ λ S AS D λ n e λ t e Dt e λ t e λ nt ( t n (S AS) n ) S t n A n S S e At S n n e λ t e At e S λ t S e λ nt I can use this approach to compute e At in case A is diagonalizable Example Compute e At if A 3 5 The eigenvalues are λ, λ Since there are two different eigenvalues and A is a matrix, A is diagonalizable The corresponding eigenvectors are (5, ) and (, ) Thus, S 5, S 6 5 Hence, e At 5 e 4t e t ( 6 ) 5 6 5e 4t +e t 5e 4t 5e t e 4t e t e 4t +5e t Example Compute e At if 5 6 6 A 4 3 6 4 The eigenvalues are λ and λ (double) The corresponding eigenvectors are (3,,3) for λ, and (,,) and (,,) for λ Since I have 3 independent eigenvectors, the matrix is diagonalizable I have S 3, S 3 3 3 6 5 4
From this, it follows that e At 3et +4e t 6e t 6e t 6e t 6e t e t e t e t +3e t e t +e t 3e t +3e t 6e t 6e t 6e t 5e t Here s a quick check on the computation: If you set t in the right side, you get This checks, since e A I Note that this check isn t foolproof just because you get I by setting t doesn t mean your answer is right However, if you don t get I, your answer is surely wrong! How do you compute e At is A is not diagonalizable? I ll describe an iterative algorithm for computing e At that only requires that one know the eigenvalues of A There are various algorithms for computing the matrix exponential; this one, which is due to Williamson, seems to me to be the easiest for hand computation (Note that finding the eigenvalues of a matrix is, in general, a difficult problem: Any method for finding e At will have to deal with it) Let A be an n n matrix Let {λ,λ,,λ n } be a list of the eigenvalues, with multiple eigenvalues repeated according to their multiplicity Let a e λt, a k e λ kt a k (t) e λ k(t u) a k (u)du, k,,n, Then B I, B k (A λ k I) B k, k,,n, e At a B +a B ++a n B n To prove this, I ll show that the expression on the right satisfies the differential equation x A x To do this, I ll need two facts about the characteristic polynomial p(x) (x λ )(x λ ) (x λ n ) ±p(x) (Cayley-Hamilton Theorem) p(a) Observe that if p(x) is the characteristic polynomial, then using the first fact and the definition of the B s, p(x) ±(x λ )(x λ ) (x λ n ) p(a) ±(A λ I)(A λ I) (A λ n I) ±I(A λ I)(A λ I) (A λ n I) ±B (A λ I)(A λ I) (A λ n I) ±B (A λ I) (A λ n I) ±B n (A λ n I) 5
By the Cayley-Hamilton Theorem, ±B n (A λ n I) ( ) I will use this fact in the proof below Example I ll illustrate the Cayley-Hamilton theorem with the matrix A 3 The characteristic polynomial is ( λ)( λ) 6 λ 3λ 4 The Cayley-Hamilton theorem asserts that if you plug A into λ 3λ 4, you ll get the zero matrix First, A 3 3 9 6 7 A A 4I 9 6 7 6 9 6 3 4 4 Proof of the algorithm First, a k e λ k(t u) a k (u)du e λ kt Recall that the Fundamental Theorem of Calculus says that d dt f(u)du f(t) e λ ku a k (u)du Applying this and the Product Rule, I can differentiate a k to obtain a k λ k e λ kt e λ ku a k (u)du+e λ kt e λ kt a k (t), a k λ k a k +a k (a B +a B ++a n B n ) λ a B + λ a B +a B + λ 3 a 3 B 3 +a B 3 + λ n a n B n +a n B n Expand the a i B i terms using a i B i a i (A λ i I)B i a i AB i λ i a i B i 6
Making this substitution and telescoping the sum, I have λ a B + λ a B +a AB λ a B + λ 3 a 3 B 3 +a AB λ a B + λ n a n B n +a n AB n λ n a n B n λ n a n B n +A(a B +a B ++a n B n ) λ n a n B n Aa n B n +A(a B +a B ++a n B n ) a n (A λ n I)B n +A(a B +a B ++a n B n ) a n +A(a B +a B ++a n B n ) A(a B +a B ++a n B n ) (The result (*) proved above was used in the next-to-the-last equality) Combining the results above, I ve shown that (a B +a B ++a n B n ) A(a B +a B ++a n B n ) This shows that M a B +a B ++a n B n satisfies M AM Using the power series expansion, I have e ta A Ae ta So (e ta M) Ae ta M +e ta AM e ta AM +e ta AM (Remember that matrix multiplication is not commutative in general!) It follows that e ta M is a constant matrix Set t Since a a n, it follows that M() I In addition, e A I e ta M I, and hence M e At Example Use the matrix exponential to solve x 3 x, x() 3 4 The characteristic polynomial is (λ ) You can check that there is only one independent eigenvector, so I can t solve the system by diagonalizing I could use generalized eigenvectors to solve the system, but I will use the matrix exponential to illustrate the algorithm First, list the eigenvalues: {,} Since λ is a double root, it is listed twice First, I ll compute the a k s: a e t, Here are the B k s: a e t a (t) e (t u) e u du e t du te t B I, B (A I)B A I e At e t +te t e t +te t te t te t e t te t As a check, note that setting t produces the identity) 7
The solution to the given initial value problem is e x t +te t te t te t e t te t 3 4 You can get the general solution by replacing (3,4) with (c,c ) Example Find e At if A The eigenvalues are obviously λ (double) and λ First, I ll compute the a k s I have a e t, and a a 3 Next, I ll compute the B k s B I, and e t u e u du e t du te t, e (t u) ue u du te t e t +e t B A I, B 3 (A I)B e t e At te t e t te t +e t e t e t e t e t Example Use the matrix exponential to solve x 5 x 4 This example will demonstrate how the algorithm for e At works when the eigenvalues are complex The characteristic polynomial is λ + λ + The eigenvalues are λ ± i I will list them as { +i, i} First, I ll compute the a k s a e ( +i)t, and a e ( +i)(t u) e ( i)u du e ( +i)t e ( i)u e ( i)u du e ( +i)t e iu du e ( +i)ti ( e it ) i ( e ( i)t e ( +i)t) 8
Next, I ll compute the B k s B I, and B A ( +i)i 3 i 5 3 i e At e ( +i)t + i ( e ( i)t e ( +i)t) 3 i I want a real solution, so I ll use DeMoivre s Formula to simplify: e ( +i)t e t (cost+isint) 5 3 i e ( i)t e ( +i)t e t (cost isint) e t (cost+isint) ie t sint i ( e ( i)t e ( +i)t) e t sint Plugging these into the expression for e At above, I have e At e t (cost+isint) +e t 3 i 5 sint 3 i e t cost+3sint 5sint sint cost 3sint Notice that all the i s have dropped out! This reflects the obvious fact that the exponential of a real matrix must be a real matrix Finally, the general solution to the original system is x y e t cost+3sint 5sint sint cost 3sint c c Example I ll compare the matrix exponential and the eigenvector solution methods by solving the following system both ways: x x The characteristic polynomial is λ 4λ+5 The eigenvalues are λ ±i Consider λ +i: i A (+i)i i As this is an eigenvector matrix, it must be singular, and hence the rows must be multiples So ignore the second row I want a vector (a,b) such that ( i)a+( )b To get such a vector, switch the i and and negate one of them: a, b i Thus, (, i) is an eigenvector The corresponding solution is e (+i)t e t cost+isint i sint icost Take the real and imaginary parts: ree (+i)t i ime (+i)t i e t cost, sint e t sint cost 9
The solution is ( ) x e t cost sint c +c sint cost Now I ll solve the equation using the exponential The eigenvalues are {+i, i} Compute the a k s a e (+i)t, and a e ( i)t e (+i)t e ( i)(t u) e (+i)u du e ( i)t e iu du e ( i)t i eiu t i e( i)t( e it) i et( e it e it) e t sint (Here and below, I m cheating a little in the comparison by not showing all the algebra involved in the simplification You need to use DeMoivre s Formula to eliminate the complex exponentials) Next, compute the B k s B I, and i B A (+i)i i The solution is e At e (+i)t +e t i sint e t cost sint i sint cost x e t cost sint sint cost Taking into account some of the algebra I didn t show for the matrix exponential, I think the eigenvector approach is easier c c Example Solve the system x 5 8 x 3 For comparison, I ll do this first using the generalized eigenvector method, then using the matrix exponential The characteristic polynomial is λ λ+ The eigenvalue is λ (double) A I 4 8 4 Ignore the first row, and divide the second row by, obtaining the vector (, ) I want (a,b) such that ()a+( )b Swap and and negate the : I get (a,b) (,) This is an eigenvector for λ Since I only have one eigenvector, I need a generalized eigenvector This means I need (a,b ) such that 4 8 a 4 b Row reduce: 4 8 4 This means that a b + Setting b yields a The generalized eigenvector is (, )
The solution is ( ) x c e t +c te t +e t Next, I ll solve the system using the matrix exponential The eigenvalues are {, } First, I ll compute the a k s a e t, and a e t e t e t u e u du e t du te t Next, compute the B k s B I, and The solution is B A I 4 8 4 e At e t +te t 4 8 e t +4te t 8te t 4 te t e t 4te t e x t +4te t 8te t te t e t 4te t c In this case, finding the solution using the matrix exponential may be a little bit easier c Richard Williamson, Introduction to differential equations Englewood Cliffs, NJ: Prentice-Hall, 986 c by Bruce Ikenaga