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 connection between a matrix Lie algebra and the corresponding Lie group.
Let X be an n×n real or complex matrix. The exponential of X, denoted by e^{X} or exp(X), is the n×n matrix given by the power series
where is defined to be the identity matrix with the same dimensions as .^{[1]}
The above series always converges, so the exponential of X is welldefined. If X is a 1×1 matrix the matrix exponential of X is a 1×1 matrix whose single element is the ordinary exponential of the single element of X.
Properties
Elementary properties
Let X and Y be n×n complex matrices and let a and b be arbitrary complex numbers. We denote the n×n identity matrix by I and the zero matrix by 0. The matrix exponential satisfies the following properties.^{[2]}
We begin with the properties that are immediate consequences of the definition as a power series:
 e^{0} = I
 exp(X^{T}) = (exp X)^{T}, where X^{T} denotes the transpose of X.
 exp(X^{∗}) = (exp X)^{∗}, where X^{∗} denotes the conjugate transpose of X.
 If Y is invertible then e^{YXY−1} = Ye^{X}Y^{−1}.
The next key result is this one:
 If then .
The proof of this identity is the same as the standard powerseries argument for the corresponding identity for the exponential of real numbers. That is to say, as long as and commute, it makes no difference to the argument whether and are numbers or matrices. It is important to note that this identity typically does not hold if and do not commute (see GoldenThompson inequality below).
Consequences of the preceding identity are the following:
 e^{aX}e^{bX} = e^{(a + b)X}
 e^{X}e^{−X} = I
Using the above results, we can easily verify the following claims. If X is symmetric then e^{X} is also symmetric, and if X is skewsymmetric then e^{X} is orthogonal. If X is Hermitian then e^{X} is also Hermitian, and if X is skewHermitian then e^{X} is unitary.
Finally, a Laplace transform of matrix exponentials amounts to the resolvent,
for all sufficiently large positive values of s.
Linear differential equation systems
One of the reasons for the importance of the matrix exponential is that it can be used to solve systems of linear ordinary differential equations. The solution of
where A is a constant matrix, is given by
The matrix exponential can also be used to solve the inhomogeneous equation
See the section on applications below for examples.
There is no closedform solution for differential equations of the form
where A is not constant, but the Magnus series gives the solution as an infinite sum.
The determinant of the matrix exponential
By Jacobi's formula, for any complex square matrix the following trace identity holds:^{[3]}
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 nonzero, and so det(e^{A}) ≠ 0, which implies that e^{A} must be invertible.
In the realvalued case, the formula also exhibits the map
to not be surjective, in contrast to the complex case mentioned earlier. This follows from the fact that, for realvalued matrices, the righthand side of the formula is always positive, while there exist invertible matrices with a negative determinant.
The exponential of sums
For any real numbers (scalars) x and y we know that the exponential function satisfies e^{x+y} = e^{x} e^{y}. The same is true for commuting matrices. If matrices X and Y commute (meaning that XY = YX), then,
However, for matrices that do not commute the above equality does not necessarily hold.
The Lie product formula
Even if X and Y do not commute, the exponential e^{X + Y} can be computed by the Lie product formula^{[4]}
 .
The Baker–Campbell–Hausdorff formula
In the other direction, if X and Y are sufficiently small (but not necessarily commuting) matrices, we have
where Z may be computed as a series in commutators of X and Y by means of the Baker–Campbell–Hausdorff formula:^{[5]}
 ,
where the remaining terms are all iterated commutators involving X and Y. If X and Y commute, then all the commutators are zero and we have simply Z = X + Y.
Inequalities for exponentials of Hermitian matrices
For Hermitian matrices there is a notable theorem related to the trace of matrix exponentials.
If A and B are Hermitian matrices, then
 ^{[6]}
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, tr(exp(A)exp(B)exp(C)) is not guaranteed to be real for Hermitian A, B, C. However, Lieb proved^{[7]}^{[8]} that it can be generalized to three matrices if we modify the expression as follows
The exponential map
The exponential of a matrix is always an invertible matrix. The inverse matrix of e^{X} is given by e^{−X}. This is analogous to the fact that the exponential of a complex number is always nonzero. The matrix exponential then gives us a map
from the space of all n×n matrices to the general linear group of degree n, 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^{[9]} (for this, it is essential to consider the field C of complex numbers and not R).
For any two matrices X and Y,
where ‖ · ‖ denotes an arbitrary matrix norm. It follows that the exponential map is continuous and Lipschitz continuous on compact subsets of M_{n}(C).
The map
defines a smooth curve in the general linear group which passes through the identity element at t = 0.
In fact, this gives a oneparameter subgroup of the general linear group since
The derivative of this curve (or tangent vector) at a point t is given by

(1)
The derivative at t = 0 is just the matrix X, which is to say that X generates this oneparameter subgroup.
More generally,^{[10]} for a generic tdependent exponent, X(t),
Taking the above expression e^{X(t)} 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,^{[11]}
The coefficients in the expression above are different from what appears in the exponential. For a closed form, see derivative of the exponential map.
Computing the matrix exponential
Finding reliable and accurate methods to compute the matrix exponential is difficult, and this is still a topic of considerable current research in mathematics and numerical analysis. Matlab, GNU Octave, and SciPy all use the Padé approximant.^{[12]}^{[13]}^{[14]} In this section, we discuss methods that are applicable in principle to any matrix, and which can be carried out explicitly for small matrices.^{[15]} Subsequent sections describe methods suitable for numerical evaluation on large matrices.
Diagonalizable case
If a matrix is diagonal:
 ,
then its exponential can be obtained by exponentiating each entry on the main diagonal:
 .
This result also allows one to exponentiate diagonalizable matrices. If
 A = UDU^{−1}
and D is diagonal, then
 e^{A} = Ue^{D}U^{−1}.
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 elementwise addition and multiplication, and hence exponentiation; in particular, the "onedimensional" exponentiation is felt elementwise for the diagonal case.)
Example : Diagonalizable
For example, the matrix
can be diagonalized as
Thus,
Nilpotent case
A matrix N is nilpotent if N^{q} = 0 for some integer q. In this case, the matrix exponential e^{N} can be computed directly from the series expansion, as the series terminates after a finite number of terms:
Since the series has a finite number of steps, it is a matrix polynomial, which can be computed efficiently.
General case
Using the Jordan–Chevalley decomposition
By the Jordan–Chevalley decomposition, any matrix X with complex entries can be expressed as
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:
Note that we need the commutativity of A and N for the last step to work.
Using the Jordan canonical form
A closely related method is, if the field is algebraically closed, to work with the Jordan form of X. Suppose that X = PJP^{ −1} where J is the Jordan form of X. Then
Also, since
Therefore, we need only know how to compute the matrix exponential of a Jordan block. But each Jordan block is of the form
where N is a special nilpotent matrix. The matrix exponential of J is then given by
Projection case
If P is a projection matrix (i.e. is idempotent: P^{2} = P), its matrix exponential is:
 e^{P} = I + (e  1)P.
Deriving this by expansion of the exponential function, each power of P reduces to P which becomes a common factor of the sum:
Rotation case
For a simple rotation in which the perpendicular unit vectors a and b specify a plane,^{[16]} the rotation matrix R can be expressed in terms of a similar exponential function involving a generator G and angle θ.^{[17]}^{[18]}
The formula for the exponential results from reducing the powers of G in the series expansion and identifying the respective series coefficients of G^{2} and G with −cos(θ) and sin(θ) respectively. The second expression here for e^{Gθ} is the same as the expression for R(θ) in the article containing the derivation of the generator, R(θ) = e^{Gθ}.
In two dimensions, if and , then , , and
reduces to the standard matrix for a plane rotation.
The matrix P = −G^{2} projects a vector onto the abplane and the rotation only affects this part of the vector. An example illustrating this is a rotation of 30° = π/6 in the plane spanned by a and b,
Let N = I  P, so N^{2} = N and its products with P and G are zero. This will allow us to evaluate powers of R.
Evaluation by Laurent series
By virtue of the Cayley–Hamilton theorem the matrix exponential is expressible as a polynomial of order n−1.
If P and Q_{t} are nonzero polynomials in one variable, such that P(A) = 0, and if the meromorphic function
is entire, then
 .
To prove this, multiply the first of the two above equalities by P(z) and replace z by A.
Such a polynomial Q_{t}(z) can be found as follows−see Sylvester's formula. Letting a be a root of P, Q_{a,t}(z) is solved from the product of P by the principal part of the Laurent series of f at a: It is proportional to the relevant Frobenius covariant. Then the sum S_{t} of the Q_{a,t}, where a runs over all the roots of P, can be taken as a particular Q_{t}. All the other Q_{t} will be obtained by adding a multiple of P to S_{t}(z). In particular, S_{t}(z), the LagrangeSylvester polynomial, is the only Q_{t} whose degree is less than that of P.
Example: Consider the case of an arbitrary 2×2 matrix,
The exponential matrix e^{tA}, by virtue of the Cayley–Hamilton theorem, must be of the form
 .
(For any complex number z and any Calgebra B, we denote again by z the product of z by the unit of B.)
Let α and β be the roots of the characteristic polynomial of A,
Then we have
hence
if α ≠ β; while, if α = β,
so that
Defining
we have
where sin(qt)/q is 0 if t = 0, and t if q = 0.
Thus,
Thus, as indicated above, the matrix A having decomposed into the sum of two mutually commuting pieces, the traceful piece and the traceless piece,
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 S_{t} can also be given the following "interpolation" characterization. Define e_{t}(z) ≡ e^{tz}, and n ≡ deg P. Then S_{t}(z) is the unique degree < n polynomial which satisfies S_{t}^{(k)}(a) = e_{t}^{(k)}(a) whenever k is less than the multiplicity of a as a root of P. We assume, as we obviously can, that P is the minimal polynomial of A. We further assume that A is a diagonalizable matrix. In particular, the roots of P are simple, and the "interpolation" characterization indicates that S_{t} is given by the Lagrange interpolation formula, so it is the Lagrange−Sylvester polynomial .
At the other extreme, if P = (z  a)^{n}, then
The simplest case not covered by the above observations is when with a ≠ b, which yields
Evaluation by implementation of Sylvester's formula
A practical, expedited computation of the above reduces to the following rapid steps. Recall from above that an n×n matrix exp(tA) amounts to a linear combination of the first n−1 powers of A by the Cayley–Hamilton theorem. For diagonalizable matrices, as illustrated above, e.g. in the 2×2 case, Sylvester's formula yields exp(tA) = B_{α} exp(tα) + B_{β} exp(tβ), where the Bs are the Frobenius covariants of A.
It is easiest, however, to simply solve for these Bs directly, by evaluating this expression and its first derivative at t = 0, in terms of A and I, to find the same answer as above.
But this simple procedure also works for defective matrices, in a generalization due to Buchheim.^{[19]} This is illustrated here for a 4×4 example of a matrix which is not diagonalizable, and the Bs are not projection matrices.
Consider
with eigenvalues λ_{1} = 3/4 and λ_{2} = 1, each with a multiplicity of two.
Consider the exponential of each eigenvalue multiplied by t, exp(λ_{i}t). Multiply each exponentiated eigenvalue by the corresponding undetermined coefficient matrix B_{i}. If the eigenvalues have an algebraic multiplicity greater than 1, then repeat the process, but now multiplying by an extra factor of t for each repetition, to ensure linear independence.
(If one eigenvalue had a multiplicity of three, then there would be the three terms: . By contrast, when all eigenvalues are distinct, the Bs 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,
To solve for all of the unknown matrices B in terms of the first three powers of A and the identity, one needs four equations, the above one providing one such at t = 0. Further, differentiate it with respect to t,
and again,
and once more,
(In the general case, n−1 derivatives need be taken.)
Setting t = 0 in these four equations, the four coefficient matrices Bs may now be solved for,
to yield
Substituting with the value for A yields the coefficient matrices
so the final answer is
The procedure is much shorter than Putzer's algorithm sometimes utilized in such cases.
Illustrations
Suppose that we want to compute the exponential of
Its Jordan form is
where the matrix P is given by
Let us first calculate exp(J). We have
The exponential of a 1×1 matrix is just the exponential of the one entry of the matrix, so exp(J_{1}(4)) = [e^{4}]. The exponential of J_{2}(16) can be calculated by the formula e^{(λI + N)} = e^{λ} e^{N} mentioned above; this yields^{[20]}
Therefore, the exponential of the original matrix B is
Applications
Linear differential equations
The matrix exponential has applications to systems of linear differential equations. (See also matrix differential equation.) Recall from earlier in this article that a homogeneous differential equation of the form
has solution e^{At} y(0).
If we consider the vector
we can express a system of inhomogeneous coupled linear differential equations as
Making an ansatz to use an integrating factor of e^{−At} and multiplying throughout, yields
The second step is possible due to the fact that, if AB = BA, then e^{At}B = Be^{At}. So, calculating e^{At} leads to the solution to the system, by simply integrating the third step with respect to t.
Example (homogeneous)
Consider the system
The associated defective matrix is
The matrix exponential is
so that the general solution of the homogeneous system is
amounting to
Example (inhomogeneous)
Consider now the inhomogeneous system
We again have
and
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,
which could be further simplified to get the requisite particular solution determined through variation of parameters. Note c = y_{p}(0). For more rigor, see the following generalization.
Inhomogeneous case generalization: variation of parameters
For the inhomogeneous case, we can use integrating factors (a method akin to variation of parameters). We seek a particular solution of the form y_{p}(t) = exp(tA) z(t),
For y_{p} to be a solution,
Thus,
where c is determined by the initial conditions of the problem.
More precisely, consider the equation
with the initial condition Y(t_{0}) = Y_{0}, where
 A is an n by n complex matrix,
 F is a continuous function from some open interval I to ℂ^{n},
 is a point of I, and
 is a vector of ℂ^{n}.
Leftmultiplying the above displayed equality by e^{−tA} yields
We claim that the solution to the equation
with the initial conditions for 0 ≤ k < n is
where the notation is as follows:
 is a monic polynomial of degree n > 0,
 f is a continuous complex valued function defined on some open interval I,
 is a point of I,
 is a complex number, and
s_{k}(t) is the coefficient of in the polynomial denoted by in Subsection Evaluation by Laurent series above.
To justify this claim, we transform our order n scalar equation into an order one vector equation by the usual reduction to a first order system. Our vector equation takes the form
where A is the transpose companion matrix of P. 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 n = 2 we get the following statement. The solution to
is
where the functions s_{0} and s_{1} are as in Subsection Evaluation by Laurent series above.
Matrixmatrix exponentials
The matrix exponential of another matrix (matrixmatrix exponential),^{[21]} is defined as
for any normal and nonsingular n×n matrix X, and any complex n×n matrix Y.
For matrixmatrix exponentials, there is a distinction between the left exponential ^{Y}X and the right exponential X^{Y}, because the multiplication operator for matrixtomatrix is not commutative. Moreover,
 If X is normal and nonsingular, then X^{Y} and ^{Y}X have the same set of eigenvalues.
 If X is normal and nonsingular, Y is normal, and XY = YX, then X^{Y} = ^{Y}X.
 If X is normal and nonsingular, and X, Y, Z commute with each other, then X^{Y+Z} = X^{Y}·X^{Z} and ^{Y+Z}X = ^{Y}X·^{Z}X.
See also
 Matrix function
 Matrix logarithm
 C_{0}semigroup
 Exponential function
 Exponential map (Lie theory)
 Magnus expansion
 Derivative of the exponential map
 Vector flow
 Golden–Thompson inequality
 Phasetype distribution
 Lie product formula
 Baker–Campbell–Hausdorff formula
 Frobenius covariant
 Sylvester's formula
 Trigonometric functions of matrices
References
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 ^ Hall 2015 Proposition 2.3
 ^ Hall 2015 Theorem 2.12
 ^ Hall 2015 Theorem 2.11
 ^ Hall 2015 Chapter 5
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