Inmathematics, thematrix exponential is amatrix function onsquare matrices analogous to the ordinaryexponential function. It is used to solve systems of linear differential equations. In the theory of Lie groups, the matrix exponential gives theexponential map between a matrixLie algebra and the correspondingLie group.

LetX be ann ×nreal orcomplexmatrix. The exponential ofX, denoted bye^X orexp(X), is then ×n matrix given by thepower series

$${\displaystyle e^X=\sum _{k=0}^{\mathrm{\infty }}\frac{1}{k!}{X}^k}$$

where${\displaystyle X^0}$ is defined to be the identity matrix${\displaystyle I}$ with the same dimensions as${\displaystyle X}$, and⁠${\displaystyle X^k=X{X}^{k-1}}$⁠.^[1] The series always converges, so the exponential ofX is well-defined.

Equivalently,$${\displaystyle e^X=\underset{k\to \mathrm{\infty }}{lim}{\left(I+\frac{X}{k}\right)}^k}$$

for integer-valuedk, whereI is then ×nidentity matrix.

Equivalently, the matrix exponential is provided by the solution${\displaystyle Y(t)=e^{Xt}}$ of the (matrix) differential equation

$${\displaystyle \frac{d}{dt}Y(t)=XY(t),Y(0)=I.}$$

WhenX is ann ×ndiagonal matrix thenexp(X) will be ann ×n diagonal matrix with each diagonal element equal to the ordinaryexponential applied to the corresponding diagonal element ofX.

LetX andY ben ×n complex matrices and leta andb be arbitrary complex numbers. We denote then ×nidentity matrix byI and thezero 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:

• e⁰ =I
• exp(X^T) = (expX)^T, whereX^T denotes thetranspose ofX.
• exp(X^∗) = (expX)^∗, whereX^∗ denotes theconjugate transpose ofX.
• IfY isinvertible thene^YXY−1 =Ye^XY⁻¹.

The next key result is this one:

• If${\displaystyle XY=YX}$ then${\displaystyle e^X{e}^Y=e^{X+Y}}$.

The proof of this 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${\displaystyle X}$ and${\displaystyle Y}$ commute, it makes no difference to the argument whether${\displaystyle X}$ and${\displaystyle Y}$ are numbers or matrices. It is important to note that this identity typically does not hold if${\displaystyle X}$ and${\displaystyle Y}$ do not commute (seeGolden-Thompson inequality below).

Consequences of the preceding identity are the following:

• e^aXe^bX =e^{(a +b)X}
• e^Xe^−X =I

Using the above results, we can easily verify the following claims. IfX issymmetric thene^X is also symmetric, and ifX isskew-symmetric thene^X isorthogonal. IfX isHermitian thene^X is also Hermitian, and ifX isskew-Hermitian thene^X isunitary.

Finally, aLaplace transform of matrix exponentials amounts to theresolvent,$${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{-ts}{e}^{tX}dt=(sI-X{)}^{-1}}$$for all sufficiently large positive values ofs.

One of the reasons for the importance of the matrix exponential is that it can be used to solve systems of linearordinary differential equations. The solution of$${\displaystyle \frac{d}{dt}y(t)=Ay(t),y(0)=y_0,}$$whereA is a constant matrix andy is a column vector, is given by$${\displaystyle y(t)=e^{At}{y}_0.}$$

The matrix exponential can also be used to solve the inhomogeneous equation$${\displaystyle \frac{d}{dt}y(t)=Ay(t)+z(t),y(0)=y_0.}$$See the section onapplications below for examples.

There is no closed-form solution for differential equations of the form$${\displaystyle \frac{d}{dt}y(t)=A(t)y(t),y(0)=y_0,}$$whereA is not constant, but theMagnus series gives the solution as an infinite sum.

ByJacobi's formula, for any complex square matrix the followingtrace identity holds:^[3]

In addition to providing a computational tool, this formula demonstrates that a matrix exponential is always aninvertible matrix. This follows from the fact that the right hand side of the above equation is always non-zero, and sodet(e^A) ≠ 0, which implies thate^A must be invertible.

In the real-valued case, the formula also exhibits the map$${\displaystyle \exp :M_n(\mathbb{R})\to \mathrm{G}\mathrm{L}(n,\mathbb{R})}$$to not besurjective, 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.

The matrix exponential of a real symmetric matrix is positive definite. Let${\displaystyle S}$ be ann ×n real symmetric matrix and${\displaystyle x\in {\mathbb{R}}^n}$ a column vector. Using the elementary properties of the matrix exponential and of symmetric matrices, we have:

$${\displaystyle x^T{e}^{S}x=x^T{e}^{S/2}{e}^{S/2}x=x^T(e^{S/2}{)}^T{e}^{S/2}x=(e^{S/2}x{)}^T{e}^{S/2}x=\Vert e^{S/2}x{\Vert }^2\ge 0.}$$

Since${\displaystyle e^{S/2}}$ is invertible, the equality only holds for${\displaystyle x=0}$, and we have${\displaystyle x^T{e}^{S}x>0}$ for all non-zero${\displaystyle x}$. Hence${\displaystyle e^S}$ is positive definite.

For any real numbers (scalars)x andy we know that the exponential function satisfiese^x+y =e^xe^y. The same is true for commuting matrices. If matricesX andY commute (meaning thatXY =YX), then,$${\displaystyle e^{X+Y}=e^X{e}^Y.}$$

However, for matrices that do not commute the above equality does not necessarily hold.

Even ifX andY do not commute, the exponentiale^{X +Y} can be computed by theLie product formula^[4]$${\displaystyle e^{X+Y}=\underset{k\to \mathrm{\infty }}{lim}{\left(e^{\frac{1}{k}X}{e}^{\frac{1}{k}Y}\right)}^k.}$$

Using a large finitek to approximate the above is basis of the Suzuki-Trotter expansion, often used innumerical time evolution.

In the other direction, ifX andY are sufficiently small (but not necessarily commuting) matrices, we have$${\displaystyle e^X{e}^Y=e^Z,}$$whereZ may be computed as a series incommutators ofX andY by means of theBaker–Campbell–Hausdorff formula:^[5]$${\displaystyle Z=X+Y+\frac{1}{2}[X,Y]+\frac{1}{12}[X,[X,Y]]-\frac{1}{12}[Y,[X,Y]]+\cdots ,}$$where the remaining terms are all iterated commutators involvingX andY. IfX andY commute, then all the commutators are zero and we have simplyZ =X +Y.

ForHermitian matrices there is a notable theorem related to thetrace of matrix exponentials.

IfA andB are Hermitian matrices, then^[6]$${\displaystyle \mathrm{tr}\exp (A+B)\le \mathrm{tr}\left[\exp (A)\exp (B)\right].}$$

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,tr(exp(A)exp(B)exp(C)) is not guaranteed to be real for HermitianA,B,C. However,Lieb proved^[7][8] that it can be generalized to three matrices if we modify the expression as follows$${\displaystyle \mathrm{tr}\exp (A+B+C)\le {\int }_0^{\mathrm{\infty }}\mathrm{d}t\mathrm{tr}\left[e^A{\left(e^{-B}+t\right)}^{-1}{e}^C{\left(e^{-B}+t\right)}^{-1}\right].}$$

The exponential of a matrix is always aninvertible matrix. The inverse matrix ofe^X is given bye^−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$${\displaystyle \exp :M_n(\mathbb{C})\to \mathrm{G}\mathrm{L}(n,\mathbb{C})}$$from the space of alln ×n matrices to thegeneral linear group of degreen, i.e. thegroup of alln ×n invertible matrices. In fact, this map issurjective 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 fieldC of complex numbers and notR).

For any two matricesX andY,$${\displaystyle \Vert e^{X+Y}-e^X\Vert \le \Vert Y\Vert e^{\Vert X\Vert }{e}^{\Vert Y\Vert },}$$

where‖ · ‖ denotes an arbitrarymatrix norm. It follows that the exponential map iscontinuous andLipschitz continuous oncompact subsets ofM_n(C).

The map$${\displaystyle t\mapsto e^{tX},t\in \mathbb{R}}$$defines asmooth curve in the general linear group which passes through the identity element att = 0.

In fact, this gives aone-parameter subgroup of the general linear group since$${\displaystyle e^{tX}{e}^{sX}=e^{(t+s)X}.}$$

The derivative of this curve (ortangent vector) at a pointt is given by

$${\displaystyle \frac{d}{dt}{e}^{tX}=X{e}^{tX}=e^{tX}X.}$$

1

The derivative att = 0 is just the matrixX, which is to say thatX generates this one-parameter subgroup.

More generally,^[10] for a generict-dependent exponent,X(t),

Taking the above expressione^X(t) outside the integral sign and expanding the integrand with the help of theHadamard lemma one can obtain the following useful expression for the derivative of the matrix exponent,^[11]$${\displaystyle e^{-X(t)}\left(\frac{d}{dt}{e}^{X(t)}\right)=\frac{d}{dt}X(t)-\frac{1}{2!}\left[X(t),\frac{d}{dt}X(t)\right]+\frac{1}{3!}\left[X(t),\left[X(t),\frac{d}{dt}X(t)\right]\right]-\cdots }$$

The coefficients in the expression above are different from what appears in the exponential. For a closed form, seederivative of the exponential map.

Let${\displaystyle X}$ be a${\displaystyle n\times n}$ Hermitian matrix with distinct eigenvalues. Let${\displaystyle X=E\text{diag}(\mathrm{\Lambda })E^{\ast }}$ be its eigen-decomposition where${\displaystyle E}$ is a unitary matrix whose columns are the eigenvectors of${\displaystyle X}$,${\displaystyle E^{\ast }}$ is its conjugate transpose, and${\displaystyle \mathrm{\Lambda }=\left({\lambda }_1,\dots ,{\lambda }_n\right)}$ the vector of corresponding eigenvalues. Then, for any${\displaystyle n\times n}$ Hermitian matrix${\displaystyle V}$, thedirectional derivative of${\displaystyle \exp :X\to e^X}$ at${\displaystyle X}$ in the direction${\displaystyle V}$ is^[12][13]$${\displaystyle D\exp (X)[V]\triangleq \underset{ϵ\to 0}{lim}\frac{1}{ϵ}\left({\displaystyle e^{X+ϵV}-e^X}\right)=E(G\odot \overline{V})E^{\ast }}$$where${\displaystyle \overline{V}=E^{\ast }VE}$, the operator${\displaystyle \odot }$ denotes the Hadamard product, and, for all${\displaystyle 1\le i,j\le n}$, the matrix${\displaystyle G}$ is defined asIn addition, for any${\displaystyle n\times n}$ Hermitian matrix${\displaystyle U}$, the second directional derivative in directions${\displaystyle U}$ and${\displaystyle V}$ is^[13]$${\displaystyle D^2\exp (X)[U,V]\triangleq \underset{{ϵ}_u\to 0}{lim}\underset{{ϵ}_v\to 0}{lim}\frac{1}{4{ϵ}_u{ϵ}_v}\left({\displaystyle e^{X+{ϵ}_{u}U+{ϵ}_{v}V}-e^{X-{ϵ}_{u}U+{ϵ}_{v}V}-e^{X+{ϵ}_{u}U-{ϵ}_{v}V}+e^{X-{ϵ}_{u}U-{ϵ}_{v}V}}\right)=EF(U,V)E^{\ast }}$$where the matrix-valued function${\displaystyle F}$ is defined, for all${\displaystyle 1\le i,j\le n}$, as$${\displaystyle F(U,V{)}_{i,j}=\sum _{k=1}^n{\varphi }_{i,j,k}({\overline{U}}_{ik}{\overline{V}}_{jk}^{\ast }+{\overline{V}}_{ik}{\overline{U}}_{jk}^{\ast })}$$with

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,R, andSciPy all use thePadé approximant.^{[14][15][16][17]} In this section, we discuss methods that are applicable in principle to any matrix, and which can be carried out explicitly for small matrices.^[18] Subsequent sections describe methods suitable for numerical evaluation on large matrices.

If a matrix isdiagonal:then its exponential can be obtained by exponentiating each entry on the main diagonal:

This result also allows one to exponentiatediagonalizable matrices. If

then

which is especially easy to compute whenD is diagonal.

Application ofSylvester'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.)

For example, the matrixcan be diagonalized as

Thus,

A matrixN isnilpotent ifN^q = 0 for some integerq. In this case, the matrix exponentiale^N can be computed directly from the series expansion, as the series terminates after a finite number of terms:

$${\displaystyle e^N=I+N+\frac{1}{2}{N}^2+\frac{1}{6}{N}^3+\cdots +\frac{1}{(q-1)!}{N}^{q-1}.}$$

Since the series has a finite number of steps, it is a matrix polynomial, which can becomputed efficiently.

By theJordan–Chevalley decomposition, any${\displaystyle n\times n}$ matrixX with complex entries can be expressed as$${\displaystyle X=A+N}$$where

• A is diagonalizable
• N is nilpotent
• Acommutes withN

This means that we can compute the exponential ofX by reducing to the previous two cases:$${\displaystyle e^X=e^{A+N}=e^A{e}^N.}$$

Note that we need the commutativity ofA andN for the last step to work.

A closely related method is, if the field isalgebraically closed, to work with theJordan form ofX. Suppose thatX =PJP⁻¹ whereJ is the Jordan form ofX. Then$${\displaystyle e^X=P{e}^J{P}^{-1}.}$$

Also, since

Therefore, we need only know how to compute the matrix exponential of aJordan block. But each Jordan block is of the form

whereN is a special nilpotent matrix. The matrix exponential ofJ is then given by$${\displaystyle e^J=e^{{\lambda }_1}{e}^{N_{a_1}}\oplus e^{{\lambda }_2}{e}^{N_{a_2}}\oplus \cdots \oplus e^{{\lambda }_n}{e}^{N_{a_n}}}$$

Another standard method by theHermite interpolation with the use ofconfluent Vandermonde matrix is presented in details in section 1.2.2, pages 4-7 of the book Higham^[19].

IfP is aprojection matrix (i.e. isidempotent:P² =P), its matrix exponential is:

Deriving this by expansion of the exponential function, each power ofP reduces toP which becomes a common factor of the sum:$${\displaystyle e^P=\sum _{k=0}^{\mathrm{\infty }}\frac{P^k}{k!}=I+\left(\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k!}\right)P=I+(e-1)P.}$$

For a simple rotation in which the perpendicular unit vectorsa andb specify a plane,^[20] therotation matrixR can be expressed in terms of a similar exponential function involving ageneratorG and angleθ.^[21][22]

The formula for the exponential results from reducing the powers ofG in the series expansion and identifying the respective series coefficients ofG² andG with−cos(θ) andsin(θ) respectively. The second expression here fore^Gθ is the same as the expression forR(θ) in the article containing the derivation of thegenerator,R(θ) =e^Gθ.

In two dimensions, if${\displaystyle a=\left[\begin{array}{c}1\\ 0\end{array}\right]}$ and${\displaystyle b=\left[\begin{array}{c}0\\ 1\end{array}\right]}$, then,, andreduces to the standard matrix for a plane rotation.

The matrixP = −G²projects a vector onto theab-plane and the rotation only affects this part of the vector. An example illustrating this is a rotation of30° = π/6 in the plane spanned bya andb,

LetN =I -P, soN² =N and its products withP andG are zero. This will allow us to evaluate powers ofR.

By virtue of theCayley–Hamilton theorem the matrix exponential is expressible as a polynomial of ordern−1.

IfP andQ_t are nonzero polynomials in one variable, such thatP(A) = 0, and if themeromorphic function$${\displaystyle f(z)=\frac{e^{tz}-Q_t(z)}{P(z)}}$$isentire, then$${\displaystyle e^{tA}=Q_t(A).}$$To prove this, multiply the first of the two above equalities byP(z) and replacez byA.

Such a polynomialQ_t(z) can be found as follows−seeSylvester's formula. Lettinga be a root ofP,Q_a,t(z) is solved from the product ofP by theprincipal part of theLaurent series off ata: It is proportional to the relevantFrobenius covariant. Then the sumS_t of theQ_a,t, wherea runs over all the roots ofP, can be taken as a particularQ_t. All the otherQ_t will be obtained by adding a multiple ofP toS_t(z). In particular,S_t(z), theLagrange-Sylvester polynomial, is the onlyQ_t whose degree is less than that ofP.

Example: Consider the case of an arbitrary2 × 2 matrix,

The exponential matrixe^tA, by virtue of theCayley–Hamilton theorem, must be of the form$${\displaystyle e^{tA}=s_0(t)I+s_1(t)A.}$$

(For any complex numberz and anyC-algebraB, we denote again byz the product ofz by the unit ofB.)

Letα andβ be the roots of thecharacteristic polynomial ofA,$${\displaystyle P(z)=z^2-(a+d)z+ad-bc=(z-\alpha )(z-\beta ).}$$

Then we have$${\displaystyle S_t(z)=e^{\alpha t}\frac{z-\beta }{\alpha -\beta }+e^{\beta t}\frac{z-\alpha }{\beta -\alpha },}$$hence

ifα ≠β; while, ifα =β,$${\displaystyle S_t(z)=e^{\alpha t}(1+t(z-\alpha )),}$$

so that

Defining

we have

wheresin(qt)/q is 0 ift = 0, andt ifq = 0.

Thus,

Thus, as indicated above, the matrixA having decomposed into the sum of two mutually commuting pieces, the traceful piece and the traceless piece,$${\displaystyle A=sI+(A-sI),}$$

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 ofEuler's formula forPauli spin matrices, that is rotations of the doublet representation of the groupSU(2).

The polynomialS_t can also be given the following "interpolation" characterization. Definee_t(z) ≡e^tz, andn ≡ degP. ThenS_t(z) is the unique degree<n polynomial which satisfiesS_t^(k)(a) =e_t^(k)(a) wheneverk is less than the multiplicity ofa as a root ofP. We assume, as we obviously can, thatP is theminimal polynomial ofA. We further assume thatA is adiagonalizable matrix. In particular, the roots ofP are simple, and the "interpolation" characterization indicates thatS_t is given by theLagrange interpolation formula, so it is theLagrange−Sylvester polynomial.

At the other extreme, ifP = (z -a)ⁿ, then$${\displaystyle S_t=e^{at}\sum _{k=0}^{n-1}\frac{t^k}{k!}(z-a{)}^k.}$$

The simplest case not covered by the above observations is when${\displaystyle P=(z-a{)}^2(z-b)}$ witha ≠b, which yields$${\displaystyle S_t=e^{at}\frac{z-b}{a-b}\left(1+\left(t+\frac{1}{b-a}\right)(z-a)\right)+e^{bt}\frac{(z-a{)}^2}{(b-a{)}^2}.}$$

A practical, expedited computation of the above reduces to the following rapid steps. Recall from above that ann ×n matrixexp(tA) amounts to a linear combination of the firstn−1 powers ofA by theCayley–Hamilton theorem. Fordiagonalizable matrices, as illustrated above, e.g. in the2 × 2 case,Sylvester's formula yieldsexp(tA) =B_α exp(tα) +B_β exp(tβ), where theBs are theFrobenius covariants ofA.

It is easiest, however, to simply solve for theseBs directly, by evaluating this expression and its first derivative att = 0, in terms ofA andI, to find the same answer as above.

But this simple procedure also works fordefective matrices, in a generalization due to Buchheim.^[23] This is illustrated here for a4 × 4 example of a matrix which isnot diagonalizable, and theBs are not projection matrices.

Considerwith eigenvaluesλ₁ = 3/4 andλ₂ = 1, each with a multiplicity of two.

Consider the exponential of each eigenvalue multiplied byt,exp(λ_it). Multiply each exponentiated eigenvalue by the corresponding undetermined coefficient matrixB_i. If the eigenvalues have an algebraic multiplicity greater than 1, then repeat the process, but now multiplying by an extra factor oft for each repetition, to ensure linear independence.

(If one eigenvalue had a multiplicity of three, then there would be the three terms:${\displaystyle B_{i_1}{e}^{{\lambda }_{i}t},B_{i_2}t{e}^{{\lambda }_{i}t},B_{i_3}{t}^2{e}^{{\lambda }_{i}t}}$. By contrast, when all eigenvalues are distinct, theBs are just theFrobenius covariants, and solving for them as below just amounts to the inversion of theVandermonde matrix of these 4 eigenvalues.)

Sum all such terms, here four such,

To solve for all of the unknown matricesB in terms of the first three powers ofA and the identity, one needs four equations, the above one providing one such att = 0. Further, differentiate it with respect tot,$${\displaystyle A{e}^{At}=\frac{3}{4}{B}_{1_1}{e}^{\frac{3}{4}t}+\left(\frac{3}{4}t+1\right)B_{1_2}{e}^{\frac{3}{4}t}+1B_{2_1}{e}^{1t}+\left(1t+1\right)B_{2_2}{e}^{1t},}$$

and again,

and once more,

(In the general case,n−1 derivatives need be taken.)

Settingt = 0 in these four equations, the four coefficient matricesBs may now be solved for,

to yield

Substituting with the value forA yields the coefficient matrices

so the final answer is

The procedure is much shorter thanPutzer's algorithm sometimes utilized in such cases.

Suppose that we want to compute the exponential of

ItsJordan form iswhere the matrixP is given by

Let us first calculate exp(J). We have$${\displaystyle J=J_1(4)\oplus J_2(16)}$$

The exponential of a1 × 1 matrix is just the exponential of the one entry of the matrix, soexp(J₁(4)) = [e⁴]. The exponential ofJ₂(16) can be calculated by the formulae^{(λI +N)} =e^λe^N mentioned above; this yields^[24]

Therefore, the exponential of the original matrixB is

The matrix exponential has applications to systems oflinear differential equations. (See alsomatrix differential equation.) Recall from earlier in this article that ahomogeneous differential equation of the form$${\displaystyle {\mathbf{y}}^{\prime }=A\mathbf{y}}$$has solutione^Aty(0).

If we consider the vector$${\displaystyle \mathbf{y}(t)=\left[\begin{array}{c}{y}_1(t)\\ ⋮\\ y_n(t)\end{array}\right],}$$we can express a system ofinhomogeneous coupled linear differential equations as$${\displaystyle {\mathbf{y}}^{\prime }(t)=A\mathbf{y}(t)+\mathbf{b}(t).}$$Making anansatz to use an integrating factor ofe^−At and multiplying throughout, yields

The second step is possible due to the fact that, ifAB =BA, thene^AtB =Be^At. So, calculatinge^At leads to the solution to the system, by simply integrating the third step with respect tot.

A solution to this can be obtained by integrating and multiplying by${\displaystyle e^{\mathbf{\text{A}}t}}$ to eliminate the exponent in the LHS. Notice that while${\displaystyle e^{\mathbf{\text{A}}t}}$ is a matrix, given that it is a matrix exponential, we can say that${\displaystyle e^{\mathbf{\text{A}}t}{e}^{-\mathbf{\text{A}}t}=I}$. In other words,${\displaystyle \exp \mathbf{\text{A}}t=\exp {(-\mathbf{\text{A}}t)}^{-1}}$.