Matrix exponential

In mathematics, the matrix exponential is a matrix function on square matrices analogous to the ordinary exponential function. It is used to solve systems of linear differential equations. In the theory of Lie groups, the matrix exponential gives the exponential map between a matrix Lie algebra and the corresponding Lie group. Let be an real or complex matrix. The exponential of , denoted by , , or , is the matrix given by the power series e^X = \sum_{k=0}^\infty \frac{1}{k!} X^k where X^0 is defined to be the identity matrix I with the same dimensions as X , and for integer . The series always converges, so the exponential of is well-defined. Equivalently, e^X = \lim_{k \rightarrow \infty} \left(I + \frac{X}{k} \right)^k for integer-valued , where is the identity matrix. Equivalently, the matrix exponential is provided by the solution Y(t)=e^{X t} of the (matrix) differential equation \frac d {dt} Y(t) = X\, Y(t), \quad Y(0) = I, evaluated at . When is an diagonal matrix then will be an diagonal matrix with each diagonal element equal to the ordinary exponential function applied to the corresponding diagonal element of .

In mathematics, the matrix exponential is a matrix function on square matrices analogous to the ordinary exponential function. It is used to solve systems of linear differential equations. In the theory of Lie groups, the matrix exponential gives the exponential map between a matrix Lie algebra and the corresponding Lie group. Let be an real or complex matrix. The exponential of , denoted by , , or , is the matrix given by the power series e^X = \sum_{k=0}^\infty \frac{1}{k!} X^k where X^0 is defined to be the identity matrix I with the same dimensions as X , and for integer . The series always converges, so the exponential of is well-defined. Equivalently, e^X = \lim_{k \rightarrow \infty} \left(I + \frac{X}{k} \right)^k for integer-valued , where is the identity matrix. Equivalently, the matrix exponential is provided by the solution Y(t)=e^{X t} of the (matrix) differential equation \frac d {dt} Y(t) = X\, Y(t), \quad Y(0) = I, evaluated at . When is an diagonal matrix then will be an diagonal matrix with each diagonal element equal to the ordinary exponential function applied to the corresponding diagonal element of . If XY=YX then e^Xe^Y=e^{X+Y} . The proof of this last identity is the same as the standard power-series argument for the corresponding identity for the exponential of real numbers. That is to say, as long as X and Y commute, it makes no difference to the argument whether X and Y are numbers or matrices. This identity typically does not hold if X and Y do not commute (see Golden-Thompson inequality below). Special cases of this identity include: Using the above results, we can easily verify the following claims: If is symmetric then is also symmetric. If is skew-symmetric then is orthogonal. If is Hermitian then is also Hermitian. If is skew-Hermitian then is unitary. Finally, a Laplace transform of matrix exponentials amounts to the resolvent, \int_0^\infty e^{-ts}e^{tX}\,dt = (sI - X)^{-1} for all sufficiently large positive values of . \frac{d}{dt} y(t) = Ay(t), \quad y(0) = y_0, where is a constant matrix and y is a column vector, is given by y(t) = e^{At} y_0. The matrix exponential can also be used to solve the inhomogeneous equation \frac{d}{dt} y(t) = Ay(t) + z(t), \quad y(0) = y_0. See the section on applications below for examples. There is no closed-form solution for differential equations of the form \frac{d}{dt} y(t) = A(t) \, y(t), \quad y(0) = y_0, where is not constant, but the Magnus series gives the solution as an infinite sum. |indent = |equation = \det\left(e^A\right) = e^{\operatorname{tr}(A)}~. |cellpadding = 6 |border |border colour = #0073CF |bgcolor = #000000 }} In addition to providing a computational tool, this formula demonstrates that a matrix exponential is always an invertible matrix. This follows from the fact that the right hand side of the above equation is always non-zero, and so , which implies that must be invertible. In the real-valued case, the formula also exhibits the map \exp \colon M_n(\R) \to \mathrm{GL}(n, \R) to not be surjective, in contrast to the complex case mentioned earlier. This follows from the fact that, for real-valued matrices, the right-hand side of the formula is always positive, while there exist invertible matrices with a negative determinant. x^Te^Sx=x^Te^{S/2}e^{S/2}x=x^T(e^{S/2})^Te^{S/2}x =(e^{S/2}x)^Te^{S/2}x=\lVert e^{S/2}x\rVert^2\geq 0. Since e^{S/2} is invertible, the equality only holds for x=0 , and we have x^Te^Sx > 0 for all non-zero x . Hence e^S is positive definite. A \,{\overline {\oplus }}\, B = A \otimes \mathrm{I} _{m}+\mathrm{I} _{n}\otimes B In this case the exponential is simply the tensor product \otimes of the exponentials of the matrices : \exp{\left(A \,{\overline {\oplus }}\, B \right)}= \exp{A}\,\otimes\,\exp{B} Here we assumed A, B to be of order n, m respectively and \mathrm{I} _{k} is the Identity matrix of order k . This follows from the commutation of the summands of the Kronecker sum and the properties discussed above. This result is tied to the Direct Product of Lie groups and its associated Lie algebra for which [A \,{\overline {\oplus }}\, B, C \,{\overline {\oplus }}\, D] = [A, C] \,{\overline {\oplus }}\, [B, D] as a representation of the Direct sum of Lie algebras. Another application of this formula is the physics of non-interacting systems. Its reverse formula \log{A}\,{\overline {\oplus }}\,\log{B} = \log{\left(A \otimes B \right)} leads to the additivity of the von Neumann entropy for independent systems if these logarithmic expressions exist. e^{X+Y} = e^Xe^Y. However, for matrices that do not commute the above equality does not necessarily hold. e^{X+Y} = \lim_{k\to\infty} \left(e^{\frac{1}{k}X}e^{\frac{1}{k}Y}\right)^k. Using a large finite to approximate the above is basis of the Suzuki-Trotter expansion, often used in numerical time evolution. e^Xe^Y = e^Z, where may be computed as a series in commutators of and by means of the Baker–Campbell–Hausdorff formula: Z = X + Y + \frac{1}{2}[X,Y] + \frac{1}{12}[X,[X,Y]] - \frac{1}{12}[Y,[X,Y]]+ \cdots, where the remaining terms are all iterated commutators involving and . If and commute, then all the commutators are zero and we have simply . \operatorname{tr}\exp(A + B) \leq \operatorname{tr}\left[\exp(A)\exp(B)\right]. There is no requirement of commutativity. There are counterexamples to show that the Golden–Thompson inequality cannot be extended to three matrices – and, in any event, is not guaranteed to be real for Hermitian , , . However, Lieb proved that it can be generalized to three matrices if we modify the expression as follows \operatorname{tr}\exp(A + B + C) \leq \int_0^\infty \mathrm{d}t\, \operatorname{tr}\left[e^A\left(e^{-B} + t\right)^{-1}e^C \left(e^{-B} + t\right)^{-1}\right]. \exp \colon M_n(\Complex) \to \mathrm{GL}(n, \Complex) from the space of all n × n matrices to the general linear group of degree , i.e. the group of all n × n invertible matrices. In fact, this map is surjective which means that every invertible matrix can be written as the exponential of some other matrix (for this, it is essential to consider the field C of complex numbers and not R). For any two matrices and , \left\| e^{X+Y} - e^X\right\| \le \|Y\| e^{\|X\|} e^{\|Y\|}, where denotes an arbitrary matrix norm. It follows that the exponential map is continuous and Lipschitz continuous on compact subsets of . The map t \mapsto e^{tX}, \qquad t \in \R defines a smooth curve in the general linear group which passes through the identity element at . In fact, this gives a one-parameter subgroup of the general linear group since e^{tX}e^{sX} = e^{(t + s)X}. The derivative of this curve (or tangent vector) at a point t is given by The derivative at is just the matrix X, which is to say that X generates this one-parameter subgroup. More generally, for a generic -dependent exponent, , Taking the above expression outside the integral sign and expanding the integrand with the help of the Hadamard lemma one can obtain the following useful expression for the derivative of the matrix exponent, e^{-X(t)}\left(\frac{d}{dt}e^{X(t)}\right) = \frac{d}{dt}X(t) - \frac{1}{2!} \left[X(t), \frac{d}{dt}X(t)\right] + \frac{1}{3!} \left[X(t), \left[X(t), \frac{d}{dt}X(t)\right]\right] - \cdots The coefficients in the expression above are different from what appears in the exponential. For a closed form, see derivative of the exponential map. Let X be a n \times n Hermitian matrix with distinct eigenvalues. Let X = E \textrm{diag}(\Lambda) E^* be its eigen-decomposition where E is a unitary matrix whose columns are the eigenvectors of X , E^* is its conjugate transpose, and \Lambda = \left(\lambda_1, \ldots, \lambda_n\right) the vector of corresponding eigenvalues. Then, for any n \times n Hermitian matrix V , the directional derivative of \exp: X \to e^X at X in the direction V is D \exp (X) [V] \triangleq \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left(\displaystyle e^{X + \epsilon V} - e^{X} \right) G_{i, j} = \left\{\begin{align} & \frac{e^{\lambda_i} - e^{\lambda_j}}{\lambda_i - \lambda_j} & \text{ if } i \neq j,\\ & e^{\lambda_i} & \text{ otherwise}.\\ \end{align}\right. In addition, for any n \times n Hermitian matrix U , the second directional derivative in directions U and V is D^2 \exp (X) [U, V] \triangleq \lim_{\epsilon_u \to 0} \lim_{\epsilon_v \to 0} \frac{1}{4 \epsilon_u \epsilon_v} \left(\displaystyle e^{X + \epsilon_u U + \epsilon_v V} - e^{X - \epsilon_u U + \epsilon_v V} - e^{X + \epsilon_u U - \epsilon_v V} + e^{X - \epsilon_u U - \epsilon_v V} \right) = E F(U, V) E^* where the matrix-valued function F is defined, for all 1 \leq i, j \leq n , as F(U, V)_{i,j} = \sum_{k=1}^n \phi_{i,j,k}(\bar{U}_{ik}\bar{V}_{jk}^* + \bar{V}_{ik}\bar{U}_{jk}^*) with \phi_{i,j,k} = \left\{\begin{align} & \frac{G_{ik} - G_{jk}}{\lambda_i - \lambda_j} & \text{ if } i \ne j,\\ & \frac{G_{ii} - G_{ik}}{\lambda_i - \lambda_k} & \text{ if } i = j \text{ and } k \ne i,\\ & \frac{G_{ii}}{2} & \text{ if } i = j = k.\\ \end{align}\right. A = \begin{bmatrix} a_1 & 0 & \cdots & 0 \\ 0 & a_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_n \end{bmatrix} , then its exponential can be obtained by exponentiating each entry on the main diagonal: e^A = \begin{bmatrix} e^{a_1} & 0 & \cdots & 0 \\ 0 & e^{a_2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & e^{a_n} \end{bmatrix} . This result also allows one to exponentiate diagonalizable matrices. If then which is especially easy to compute when is diagonal. Application of Sylvester's formula yields the same result. (To see this, note that addition and multiplication, hence also exponentiation, of diagonal matrices is equivalent to element-wise addition and multiplication, and hence exponentiation; in particular, the "one-dimensional" exponentiation is felt element-wise for the diagonal case.)

Example : Diagonalizable

For example, the matrix A = \begin{bmatrix} 1 & 4\\ 1 & 1\\ \end{bmatrix} can be diagonalized as \begin{bmatrix} -2 & 2\\ 1 & 1\\ \end{bmatrix}\begin{bmatrix} -1 & 0\\ 0 & 3\\ \end{bmatrix}\begin{bmatrix} -2 & 2\\ 1 & 1\\ \end{bmatrix}^{-1}. Thus, e^A = \begin{bmatrix} -2 & 2\\ 1 & 1\\ \end{bmatrix}e^\begin{bmatrix} -1 & 0\\ 0 & 3\\ \end{bmatrix}\begin{bmatrix} -2 & 2\\ 1 & 1\\ \end{bmatrix}^{-1}=\begin{bmatrix} -2 & 2\\ 1 & 1\\ \end{bmatrix}\begin{bmatrix} \frac{1}{e} & 0\\ 0 & e^3\\ \end{bmatrix}\begin{bmatrix} -2 & 2\\ 1 & 1\\ \end{bmatrix}^{-1} = \begin{bmatrix} \frac{e^4+1}{2e} & \frac{e^4-1}{e}\\ \frac{e^4-1}{4e} & \frac{e^4+1}{2e}\\ \end{bmatrix}. e^N = I + N + \frac{1}{2}N^2 + \frac{1}{6}N^3 + \cdots + \frac{1}{(q - 1)!}N^{q-1} ~. Since the series has a finite number of steps, it is a matrix polynomial, which can be computed efficiently. X = A + N where A is diagonalizable N is nilpotent A commutes with N This means that we can compute the exponential of X by reducing to the previous two cases: e^X = e^{A+N} = e^A e^N. Note that we need the commutativity of A and N for the last step to work. e^{X} = Pe^{J}P^{-1}. Also, since \begin{align} J &= J_{a_1}(\lambda_1) \oplus J_{a_2}(\lambda_2) \oplus \cdots \oplus J_{a_n}(\lambda_n), \\ e^J &= \exp \big( J_{a_1}(\lambda_1) \oplus J_{a_2}(\lambda_2) \oplus \cdots \oplus J_{a_n}(\lambda_n) \big) \\ &= \exp \big( J_{a_1}(\lambda_1) \big) \oplus \exp \big( J_{a_2}(\lambda_2) \big) \oplus \cdots \oplus \exp \big( J_{a_n}(\lambda_n) \big). \end{align} Therefore, we need only know how to compute the matrix exponential of a Jordan block. But each Jordan block is of the form \begin{align} & & J_a(\lambda) &= \lambda I + N \\ &\Rightarrow & e^{J_a(\lambda)} &= e^{\lambda I + N} = e^\lambda e^N. \end{align} where is a special nilpotent matrix. The matrix exponential of is then given by e^J = e^{\lambda_1} e^{N_{a_1}} \oplus e^{\lambda_2} e^{N_{a_2}} \oplus \cdots \oplus e^{\lambda_n} e^{N_{a_n}} e^P = \sum_{k=0}^{\infty} \frac{P^k}{k!} = I + \left(\sum_{k=1}^{\infty} \frac{1}{k!}\right)P = I + (e - 1)P ~. \begin{align} G &= \mathbf{ba}^\mathsf{T} - \mathbf{ab}^\mathsf{T} & P &= -G^2 = \mathbf{aa}^\mathsf{T} + \mathbf{bb}^\mathsf{T} \\ P^2 &= P & PG &= G = GP ~, \end{align} \begin{align} R\left( \theta \right) = e^{G\theta} &= I + G\sin (\theta) + G^2(1 - \cos(\theta)) \\ &= I - P + P\cos (\theta) + G\sin (\theta ) ~.\\ \end{align} The formula for the exponential results from reducing the powers of in the series expansion and identifying the respective series coefficients of and with and respectively. The second expression here for is the same as the expression for in the article containing the derivation of the generator, . In two dimensions, if a = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right] and b = \left[ \begin{smallmatrix} 0 \\ 1 \end{smallmatrix} \right] , then G = \left[ \begin{smallmatrix} 0 & -1 \\ 1 & 0\end{smallmatrix} \right] , G^2 = \left[ \begin{smallmatrix}-1 & 0 \\ 0 & -1\end{smallmatrix} \right] , and R(\theta) = \begin{bmatrix}\cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{bmatrix} = I \cos(\theta) + G \sin(\theta) reduces to the standard matrix for a plane rotation. The matrix projects a vector onto the -plane and the rotation only affects this part of the vector. An example illustrating this is a rotation of in the plane spanned by and , \begin{align} \mathbf{a} &= \begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix} & \mathbf{b} &= \frac{1}{\sqrt{5}}\begin{bmatrix} 0 \\ 1 \\ 2 \\ \end{bmatrix} \end{align} \begin{align} G = \frac{1}{\sqrt{5}}&\begin{bmatrix} 0 & -1 & -2 \\ 1 & 0 & 0 \\ 2 & 0 & 0 \\ \end{bmatrix} & P = -G^2 &= \frac{1}{5}\begin{bmatrix} 5 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 2 & 4 \\ \end{bmatrix} \\ P\begin{bmatrix} 1 \\ 2 \\ 3 \\ \end{bmatrix} = \frac{1}{5}&\begin{bmatrix} 5 \\ 8 \\ 16 \\ \end{bmatrix} = \mathbf{a} + \frac{8}{\sqrt{5}}\mathbf{b} & R\left(\frac{\pi}{6}\right) &= \frac{1}{10}\begin{bmatrix} 5\sqrt{3} & -\sqrt{5} & -2\sqrt{5} \\ \sqrt{5} & 8 + \sqrt{3} & -4 + 2\sqrt{3} \\ 2\sqrt{5} & -4 + 2\sqrt{3} & 2 + 4\sqrt{3} \\ \end{bmatrix} \\ \end{align} Let , so and its products with and are zero. This will allow us to evaluate powers of . \begin{align} R\left( \frac{\pi}{6} \right) &= N + P\frac{\sqrt{3}}{2} + G\frac{1}{2} \\ R\left( \frac{\pi}{6} \right)^2 &= N + P\frac{1}{2} + G\frac{\sqrt{3}}{2} \\ R\left( \frac{\pi}{6} \right)^3 &= N + G \\ R\left( \frac{\pi}{6} \right)^6 &= N - P \\ R\left( \frac{\pi}{6} \right)^{12} &= N + P = I \\ \end{align} f(z)=\frac{e^{t z}-Q_t(z)}{P(z)} is entire, then e^{t A} = Q_t(A). To prove this, multiply the first of the two above equalities by and replace by . Such a polynomial can be found as follows−see Sylvester's formula. Letting be a root of , is solved from the product of by the principal part of the Laurent series of at : It is proportional to the relevant Frobenius covariant. Then the sum S of the Q , where runs over all the roots of , can be taken as a particular . All the other Q will be obtained by adding a multiple of to . In particular, , the Lagrange-Sylvester polynomial, is the only whose degree is less than that of . Example: Consider the case of an arbitrary matrix, A := \begin{bmatrix} a & b \\ c & d \end{bmatrix}. The exponential matrix , by virtue of the Cayley–Hamilton theorem, must be of the form e^{tA} = s_0(t)\, I + s_1(t)\,A. (For any complex number and any C-algebra , we denote again by the product of by the unit of .) Let and be the roots of the characteristic polynomial of , P(z) = z^2 - (a + d)\ z + ad - bc = (z - \alpha)(z - \beta) ~ . Then we have S_t(z) = e^{\alpha t} \frac{z - \beta}{\alpha - \beta} + e^{\beta t} \frac{z - \alpha}{\beta - \alpha}~, hence \begin{align} s_0(t) &= \frac{\alpha\,e^{\beta t} - \beta\,e^{\alpha t}}{\alpha - \beta}, & s_1(t) &= \frac{e^{\alpha t} - e^{\beta t}}{\alpha - \beta} \end{align} if ; while, if , S_t(z) = e^{\alpha t} (1 + t (z - \alpha)) ~, so that \begin{align} s_0(t) &= (1 - \alpha\,t)\,e^{\alpha t},& s_1(t) &= t\,e^{\alpha t}~. \end{align} Defining \begin{align} s &\equiv \frac{\alpha + \beta}{2} = \frac{\operatorname{tr} A}{2}~, & q &\equiv \frac{\alpha - \beta}{2} = \pm\sqrt{-\det\left(A - sI\right)}, \end{align} we have \begin{align} s_0(t) &= e^{st}\left(\cosh(qt) - s\frac{\sinh(qt)}{q}\right), & s_1(t) &= e^{st}\frac{\sinh(qt)}{q}, \end{align} where is 0 if , and if . Thus, Thus, as indicated above, the matrix having decomposed into the sum of two mutually commuting pieces, the traceful piece and the traceless piece, A = sI + (A-sI)~, the matrix exponential reduces to a plain product of the exponentials of the two respective pieces. This is a formula often used in physics, as it amounts to the analog of Euler's formula for Pauli spin matrices, that is rotations of the doublet representation of the group SU(2). The polynomial can also be given the following "interpolation" characterization. Define , and . Then is the unique degree polynomial which satisfies whenever is less than the multiplicity of as a root of . We assume, as we obviously can, that is the minimal polynomial of . We further assume that is a diagonalizable matrix. In particular, the roots of are simple, and the "interpolation" characterization indicates that is given by the Lagrange interpolation formula, so it is the Lagrange−Sylvester polynomial. At the other extreme, if , then S_t = e^{at}\ \sum_{k=0}^{n-1}\ \frac{t^k}{k!}\ (z - a)^k ~. The simplest case not covered by the above observations is when P = (z - a)^2\,(z - b) with , which yields S_t = e^{at}\ \frac{z - b}{a - b}\ \left(1 + \left(t + \frac{1}{b - a}\right)(z - a)\right) + e^{bt}\ \frac{(z - a)^2}{(b - a)^2}. A = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & -\frac{1}{8} \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{bmatrix} ~, with eigenvalues and , each with a multiplicity of two. Consider the exponential of each eigenvalue multiplied by , . Multiply each exponentiated eigenvalue by the corresponding undetermined coefficient matrix . If the eigenvalues have an algebraic multiplicity greater than 1, then repeat the process, but now multiplying by an extra factor of for each repetition, to ensure linear independence. (If one eigenvalue had a multiplicity of three, then there would be the three terms: B_{i_1} e^{\lambda_i t}, ~ B_{i_2} t e^{\lambda_i t}, ~ B_{i_3} t^2 e^{\lambda_i t} . By contrast, when all eigenvalues are distinct, the s are just the Frobenius covariants, and solving for them as below just amounts to the inversion of the Vandermonde matrix of these 4 eigenvalues.) Sum all such terms, here four such, \begin{align} e^{At} &= B_{1_1} e^{\lambda_1 t} + B_{1_2} t e^{\lambda_1 t} + B_{2_1} e^{\lambda_2 t} + B_{2_2} t e^{\lambda_2 t} , \\ e^{At} &= B_{1_1} e^{\frac{3}{4} t} + B_{1_2} t e^{\frac{3}{4} t} + B_{2_1} e^{1 t} + B_{2_2} t e^{1 t} ~. \end{align} To solve for all of the unknown matrices in terms of the first three powers of and the identity, one needs four equations, the above one providing one such at = 0. Further, differentiate it with respect to , A e^{A t} = \frac{3}{4} B_{1_1} e^{\frac{3}{4} t} + \left( \frac{3}{4} t + 1 \right) B_{1_2} e^{\frac{3}{4} t} + 1 B_{2_1} e^{1 t} + \left(1 t + 1 \right) B_{2_2} e^{1 t} ~, and again, \begin{align} A^2 e^{At} &= \left(\frac{3}{4}\right)^2 B_{1_1} e^{\frac{3}{4} t} + \left( \left(\frac{3}{4}\right)^2 t + \left( \frac{3}{4} + 1 \cdot \frac{3}{4}\right) \right) B_{1_2} e^{\frac{3}{4} t} + B_{2_1} e^{1 t} + \left(1^2 t + (1 + 1 \cdot 1 )\right) B_{2_2} e^{1 t} \\ &= \left(\frac{3}{4}\right)^2 B_{1_1} e^{\frac{3}{4} t} + \left( \left(\frac{3}{4}\right)^2 t + \frac{3}{2} \right) B_{1_2} e^{\frac{3}{4} t} + B_{2_1} e^{t} + \left(t + 2\right) B_{2_2} e^{t} ~, \end{align} and once more, \begin{align} A^3 e^{At} &= \left(\frac{3}{4}\right)^3 B_{1_1} e^{\frac{3}{4} t} + \left( \left(\frac{3}{4}\right)^3 t + \left( \left(\frac{3}{4}\right)^2 + \left(\frac{3}{2}\right) \cdot \frac{3}{4}\right) \right) B_{1_2} e^{\frac{3}{4} t} + B_{2_1} e^{1 t} + \left(1^3 t + (1 + 2) \cdot 1 \right) B_{2_2} e^{1 t} \\ &= \left(\frac{3}{4}\right)^3 B_{1_1} e^{\frac{3}{4} t}\! + \left( \left(\frac{3}{4}\right)^3 t\! + \frac{27}{16} \right) B_{1_2} e^{\frac{3}{4} t}\! + B_{2_1} e^{t}\! + \left(t + 3\cdot 1\right) B_{2_2} e^{t} ~. \end{align} (In the general case, −1 derivatives need be taken.) Setting = 0 in these four equations, the four coefficient matrices s may now be solved for, \begin{align} I &= B_{1_1} + B_{2_1} \\ A &= \frac{3}{4} B_{1_1} + B_{1_2} + B_{2_1} + B_{2_2} \\ A^2 &= \left(\frac{3}{4}\right)^2 B_{1_1} + \frac{3}{2} B_{1_2} + B_{2_1} + 2 B_{2_2} \\ A^3 &= \left(\frac{3}{4}\right)^3 B_{1_1} + \frac{27}{16} B_{1_2} + B_{2_1} + 3 B_{2_2} ~, \end{align} to yield \begin{align} B_{1_1} &= 128 A^3 - 366 A^2 + 288 A - 80 I \\ B_{1_2} &= 16 A^3 - 44 A^2 + 40 A - 12 I \\ B_{2_1} &= -128 A^3 + 366 A^2 - 288 A + 80 I \\ B_{2_2} &= 16 A^3 - 40 A^2 + 33 A - 9 I ~. \end{align} Substituting with the value for yields the coefficient matrices \begin{align} B_{1_1} &= \begin{bmatrix}0 & 0 & 48 & -16\\ 0 & 0 & -8 & 2\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}\\ B_{1_2} &= \begin{bmatrix}0 & 0 & 4 & -2\\ 0 & 0 & -1 & \frac{1}{2}\\ 0 & 0 & \frac{1}{4} & -\frac{1}{8}\\ 0 & 0 & \frac{1}{2} & -\frac{1}{4} \end{bmatrix}\\ B_{2_1} &= \begin{bmatrix}1 & 0 & -48 & 16\\ 0 & 1 & 8 & -2\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{bmatrix}\\ B_{2_2} &= \begin{bmatrix}0 & 1 & 8 & -2\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{bmatrix} \end{align} so the final answer is e^{tA} = \begin{bmatrix} e^t & te^t & \left(8t - 48\right) e^t\! + \left(4t + 48\right)e^{\frac{3}{4}t} & \left(16 - 2\,t\right)e^t\! + \left(-2t - 16\right)e^{\frac{3}{4}t}\\ 0 & e^t & 8e^t\! + \left(-t - 8\right) e^{\frac{3}{4}t} & -2e^t + \frac{t + 4}{2}e^{\frac{3}{4}t}\\ 0 & 0 & \frac{t + 4}{4}e^{\frac{3}{4}t} & -\frac{t}{8}e^{\frac{3}{4}t}\\ 0 & 0 & \frac{t}{2}e^{\frac{3}{4}t} & -\frac{t - 4}{4}e^{\frac{3}{4}t} ~. \end{bmatrix} The procedure is much shorter than Putzer's algorithm sometimes utilized in such cases. B = \begin{bmatrix} 21 & 17 & 6 \\ -5 & -1 & -6 \\ 4 & 4 & 16 \end{bmatrix}. Its Jordan form is J = P^{-1}BP = \begin{bmatrix} 4 & 0 & 0 \\ 0 & 16 & 1 \\ 0 & 0 & 16 \end{bmatrix}, where the matrix P is given by P = \begin{bmatrix} -\frac14 & 2 & \frac54 \\ \frac14 & -2 & -\frac14 \\ 0 & 4 & 0 \end{bmatrix}. Let us first calculate exp(J). We have J = J_1(4) \oplus J_2(16) The exponential of a matrix is just the exponential of the one entry of the matrix, so . The exponential of J (16) can be calculated by the formula mentioned above; this yields \begin{align} &\exp \left( \begin{bmatrix} 16 & 1 \\ 0 & 16 \end{bmatrix} \right) {}={} &e^{16} \left(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + {1 \over 2!}\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} + \cdots {} \right) = \begin{bmatrix} e^{16} & e^{16} \\ 0 & e^{16} \end{bmatrix}. \end{align} Therefore, the exponential of the original matrix is \begin{align} \exp(B) &= P \exp(J) P^{-1} = P \begin{bmatrix} e^4 & 0 & 0 \\ 0 & e^{16} & e^{16} \\ 0 & 0 & e^{16} \end{bmatrix} P^{-1} \\[6pt] &= {1 \over 4} \begin{bmatrix} 13e^{16} - e^4 & 13e^{16} - 5e^4 & 2e^{16} - 2e^4 \\ -9e^{16} + e^4 & -9e^{16} + 5e^4 & -2e^{16} + 2e^4 \\ 16e^{16} & 16e^{16} & 4e^{16} \end{bmatrix}. \end{align} \mathbf{y}' = A\mathbf{y} has solution . If we consider the vector \mathbf{y}(t) = \begin{bmatrix} y_1(t) \\ \vdots \\y_n(t) \end{bmatrix} ~, we can express a system of inhomogeneous coupled linear differential equations as \mathbf{y}'(t) = A\mathbf{y}(t)+\mathbf{b}(t). Making an ansatz to use an integrating factor of and multiplying throughout, yields \begin{align} & & e^{-At}\mathbf{y}'-e^{-At}A\mathbf{y} &= e^{-At}\mathbf{b} \\ &\Rightarrow & e^{-At}\mathbf{y}'-Ae^{-At}\mathbf{y} &= e^{-At}\mathbf{b} \\ &\Rightarrow & \frac{d}{dt} \left(e^{-At}\mathbf{y}\right) &= e^{-At}\mathbf{b}~. \end{align} The second step is possible due to the fact that, if , then . So, calculating leads to the solution to the system, by simply integrating the third step with respect to . A solution to this can be obtained by integrating and multiplying by e^{\textbf{A}t} to eliminate the exponent in the LHS. Notice that while e^{\textbf{A}t} is a matrix, given that it is a matrix exponential, we can say that e^{\textbf{A}t} e^{-\textbf{A}t} = I . In other words, \exp{\textbf{A}t} = \exp{{(-\textbf{A}t)}^{-1}} . \begin{matrix} x' &=& 2x & -y & +z \\ y' &=& & 3y & -1z \\ z' &=& 2x & +y & +3z \end{matrix}~. The associated defective matrix is A = \begin{bmatrix} 2 & -1 & 1 \\ 0 & 3 & -1 \\ 2 & 1 & 3 \end{bmatrix}~. The matrix exponential is e^{tA} = \frac{1}{2}\begin{bmatrix} e^{2t}\left( 1 + e^{2t} - 2t\right) & -2te^{2t} & e^{2t}\left(-1 + e^{2t}\right) \\ -e^{2t}\left(-1 + e^{2t} - 2t\right) & 2(t + 1)e^{2t} & -e^{2t}\left(-1 + e^{2t}\right) \\ e^{2t}\left(-1 + e^{2t} + 2t\right) & 2te^{2t} & e^{2t}\left( 1 + e^{2t}\right) \end{bmatrix}~, so that the general solution of the homogeneous system is \begin{bmatrix}x \\y \\ z\end{bmatrix} = \frac{x(0)}{2}\begin{bmatrix}e^{2t}\left(1 + e^{2t} - 2t\right) \\ -e^{2t}\left(-1 + e^{2t} - 2t\right) \\ e^{2t}\left(-1 + e^{2t} + 2t\right)\end{bmatrix} + \frac{y(0)}{2}\begin{bmatrix}-2te^{2t} \\ 2(t + 1)e^{2t} \\ 2te^{2t}\end{bmatrix} + \frac{z(0)}{2}\begin{bmatrix}e^{2t}\left(-1 + e^{2t}\right) \\ -e^{2t}\left(-1 + e^{2t}\right) \\ e^{2t}\left(1 + e^{2t}\right)\end{bmatrix} ~, amounting to \begin{align} 2x &= x(0)e^{2t}\left(1 + e^{2t} - 2t\right) + y(0)\left(-2te^{2t}\right) + z(0)e^{2t}\left(-1 + e^{2t}\right) \\[2pt] 2y &= x(0)\left(-e^{2t}\right)\left(-1 + e^{2t} - 2t\right) + y(0)2(t + 1)e^{2t} + z(0)\left(-e^{2t}\right)\left(-1 + e^{2t}\right) \\[2pt] 2z &= x(0)e^{2t}\left(-1 + e^{2t} + 2t\right) + y(0)2te^{2t} + z(0)e^{2t}\left(1 + e^{2t}\right) ~. \end{align} \begin{matrix} x' &=& 2x & - & y & + & z & + & e^{2t} \\ y' &=& & & 3y& - & z & \\ z' &=& 2x & + & y & + & 3z & + & e^{2t} \end{matrix} ~. We again have A = \left[\begin{array}{rrr} 2 & -1 & 1 \\ 0 & 3 & -1 \\ 2 & 1 & 3 \end{array}\right] ~, and \mathbf{b} = e^{2t}\begin{bmatrix}1 \\0\\1\end{bmatrix}. From before, we already have the general solution to the homogeneous equation. Since the sum of the homogeneous and particular solutions give the general solution to the inhomogeneous problem, we now only need find the particular solution. We have, by above, \begin{align} \mathbf{y}_p &= e^{tA}\int_0^t e^{(-u)A}\begin{bmatrix}e^{2u} \\0\\e^{2u}\end{bmatrix}\,du+e^{tA}\mathbf{c} \\[6pt] &= e^{tA}\int_0^t \begin{bmatrix} 2e^u - 2ue^{2u} & -2ue^{2u} & 0 \\ -2e^u + 2(u+1)e^{2u} & 2(u+1)e^{2u} & 0 \\ 2ue^{2u} & 2ue^{2u} & 2e^u \end{bmatrix}\begin{bmatrix}e^{2u} \\0 \\e^{2u}\end{bmatrix}\,du + e^{tA}\mathbf{c} \\[6pt] &= e^{tA}\int_0^t \begin{bmatrix} e^{2u}\left( 2e^u - 2ue^{2u}\right) \\ e^{2u}\left(-2e^u + 2(1 + u)e^{2u}\right) \\ 2e^{3u} + 2ue^{4u} \end{bmatrix}\,du + e^{tA}\mathbf{c} \\[6pt] &= e^{tA}\begin{bmatrix} -{1 \over 24}e^{3t}\left(3e^t(4t - 1) - 16\right) \\ {1 \over 24}e^{3t}\left(3e^t(4t + 4) - 16\right) \\ {1 \over 24}e^{3t}\left(3e^t(4t - 1) - 16\right) \end{bmatrix} + \begin{bmatrix} 2e^t - 2te^{2t} & -2te^{2t} & 0 \\ -2e^t + 2(t + 1)e^{2t} & 2(t + 1)e^{2t} & 0 \\ 2te^{2t} & 2te^{2t} & 2e^t \end{bmatrix}\begin{bmatrix}c_1 \\c_2 \\c_3\end{bmatrix} ~, \end{align} which could be further simplified to get the requisite particular solution determined through variation of parameters. Note c = y (0). For more rigor, see the following generalization. \begin{align} \mathbf{y}_p'(t) & = \left(e^{tA}\right)'\mathbf{z}(t) + e^{tA}\mathbf{z}'(t) \\[6pt] & = Ae^{tA}\mathbf{z}(t) + e^{tA}\mathbf{z}'(t) \\[6pt] & = A\mathbf{y}_p(t) + e^{tA}\mathbf{z}'(t)~. \end{align} For to be a solution, \begin{align} e^{tA}\mathbf{z}'(t) &= \mathbf{b}(t) \\[6pt] \mathbf{z}'(t) &= \left(e^{tA}\right)^{-1}\mathbf{b}(t) \\[6pt] \mathbf{z}(t) &= \int_0^t e^{-uA}\mathbf{b}(u)\,du + \mathbf{c} ~. \end{align} Thus, \begin{align} \mathbf{y}_p(t) & = e^{tA}\int_0^t e^{-uA}\mathbf{b}(u)\,du + e^{tA}\mathbf{c} \\ & = \int_0^t e^{(t - u)A}\mathbf{b}(u)\,du + e^{tA}\mathbf{c}~, \end{align} where is determined by the initial conditions of the problem. More precisely, consider the equation Y' - A\ Y = F(t) with the initial condition , where is an by complex matrix, is a continuous function from some open interval to , t_0 is a point of , and Y_0 is a vector of . Left-multiplying the above displayed equality by yields Y(t) = e^{(t - t_0)A}\ Y_0 + \int_{t_0}^t e^{(t - x)A}\ F(x)\ dx ~. We claim that the solution to the equation P(d/dt)\ y = f(t) with the initial conditions y^{(k)}(t_0) = y_k for is y(t) = \sum_{k=0}^{n-1}\ y_k\ s_k(t - t_0) + \int_{t_0}^t s_{n-1}(t - x)\ f(x)\ dx ~, where the notation is as follows: P\in\mathbb{C}[X] is a monic polynomial of degree , is a continuous complex valued function defined on some open interval , t_0 is a point of , y_k is a complex number, and is the coefficient of X^k in the polynomial denoted by S_t\in\mathbb{C}[X] in Subsection Evaluation by Laurent series above. To justify this claim, we transform our order scalar equation into an order one vector equation by the usual reduction to a first order system. Our vector equation takes the form \frac{dY}{dt} - A\ Y = F(t),\quad Y(t_0) = Y_0, where is the transpose companion matrix of . We solve this equation as explained above, computing the matrix exponentials by the observation made in Subsection Evaluation by implementation of Sylvester's formula above. In the case = 2 we get the following statement. The solution to y - (\alpha + \beta)\ y' + \alpha\,\beta\ y = f(t),\quad y(t_0) = y_0,\quad y'(t_0) = y_1 is y(t) = y_0\ s_0(t - t_0) + y_1\ s_1(t - t_0) + \int_{t_0}^t s_1(t - x)\,f(x)\ dx, where the functions and are as in Subsection Evaluation by Laurent series above.

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