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:

• e⁰ = 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^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 (see Golden-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. If X is symmetric then e^X is also symmetric, and if X is skew-symmetric then e^X is orthogonal. If X is Hermitian then e^X is also Hermitian, and if X is skew-Hermitian then e^X is unitary.

Finally, a Laplace transform of matrix exponentials amounts to the resolvent,

$${\displaystyle \int _{0}^{\infty }e^{-ts}e^{tX}\,dt=(sI-X)^{-1}}$$

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

$${\displaystyle {\frac {d}{dt}}y(t)=Ay(t),\quad y(0)=y_{0},}$$

where A is a constant matrix, is given by

$${\displaystyle y(t)=e^{At}y_{0}.}$$

The matrix exponential can also be used to solve the inhomogeneous equation

$${\displaystyle {\frac {d}{dt}}y(t)=Ay(t)+z(t),\quad y(0)=y_{0}.}$$

See the section on applications below for examples.

There is no closed-form solution for differential equations of the form

$${\displaystyle {\frac {d}{dt}}y(t)=A(t)\,y(t),\quad y(0)=y_{0},}$$

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 non-zero, and so det(e^A) ≠ 0, which implies that e^A must be invertible.

In the real-valued case, the formula also exhibits the map

$${\displaystyle \exp \colon M_{n}(\mathbb {R} )\to \mathrm {GL} (n,\mathbb {R} )}$$

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.

Real symmetric matrices

The matrix exponential of a real symmetric matrix is positive definite. Let ${\displaystyle S}$ be an n×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:

$${\displaystyle x^{T}e^{S}x=x^{T}e^{S/2}e^{S/2}x=x^{T}(e^{S/2})^{T}e^{S/2}x=(e^{S/2}x)^{T}e^{S/2}x=\lVert e^{S/2}x\rVert ^{2}\geq 0.}$$

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.

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]

$${\displaystyle e^{X+Y}=\lim _{k\to \infty }\left(e^{{\frac {1}{k}}X}e^{{\frac {1}{k}}Y}\right)^{k}.}$$

Using a large finite k to approximate the above is basis of the Suzuki-Trotter expansion, often used in numerical time evolution.

The Baker–Campbell–Hausdorff formula

In the other direction, if X and Y are sufficiently small (but not necessarily commuting) matrices, we have

$${\displaystyle e^{X}e^{Y}=e^{Z},}$$

where Z may be computed as a series in commutators of X and Y by means of the Baker–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 involving X and Y. If X and Y commute, then all the commutators are zero and we have simply Z = X + Y.

Directional derivatives when restricted to Hermitian matrices

Let ${\displaystyle X}$ be a ${\displaystyle n\times n}$ Hermitian matrix with distinct eigenvalues. Let ${\displaystyle X=E{\textrm {diag}}(\Lambda )E^{*}}$ be its eigen-decomposition where ${\displaystyle E}$ is a unitary matrix whose columns are the eigenvectors of ${\displaystyle X}$, ${\displaystyle E^{*}}$ is its conjugate transpose, and ${\displaystyle \Lambda =\left(\lambda _{1},\ldots ,\lambda _{n}\right)}$ the vector of corresponding eigenvalues. Then, for any ${\displaystyle n\times n}$ Hermitian matrix ${\displaystyle V}$, the directional derivative of ${\displaystyle \exp :X\to e^{X}}$ at ${\displaystyle X}$ in the direction ${\displaystyle V}$ is ^{[12] [13]}

$${\displaystyle D\exp(X)[V]\triangleq \lim _{\epsilon \to 0}{\frac {1}{\epsilon }}\left(\displaystyle e^{X+\epsilon V}-e^{X}\right)=E(G\odot {\bar {V}})E^{*}}$$

where ${\displaystyle {\bar {V}}=E^{*}VE}$, the operator ${\displaystyle \odot }$ denotes the Hadamard product, and, for all ${\displaystyle 1\leq i,j\leq n}$, the matrix ${\displaystyle G}$ is defined as

$${\displaystyle G_{i,j}=\left\{{\begin{aligned}&{\frac {e^{\lambda _{i}}-e^{\lambda _{j}}}{\lambda _{i}-\lambda _{j}}}&{\text{ if }}i\neq j,\\&e^{\lambda _{i}}&{\text{ otherwise}}.\\\end{aligned}}\right.}$$

In addition, for any ${\displaystyle n\times n}$ Hermitian matrix ${\displaystyle U}$, the second directional derivative in directions ${\displaystyle U}$ and ${\displaystyle V}$ is^[13]

$${\displaystyle D^{2}\exp(X)[U,V]\triangleq \lim _{\epsilon _{u}\to 0}\lim _{\epsilon _{v}\to 0}{\frac {1}{4\epsilon _{u}\epsilon _{v}}}\left(\displaystyle e^{X+\epsilon _{u}U+\epsilon _{v}V}-e^{X-\epsilon _{u}U+\epsilon _{v}V}-e^{X+\epsilon _{u}U-\epsilon _{v}V}+e^{X-\epsilon _{u}U-\epsilon _{v}V}\right)=EF(U,V)E^{*}}$$

where the matrix-valued function ${\displaystyle F}$ is defined, for all ${\displaystyle 1\leq i,j\leq n}$, as

$${\displaystyle F(U,V)_{i,j}=\sum _{k=1}^{n}\phi _{i,j,k}({\bar {U}}_{ik}{\bar {V}}_{jk}^{*}+{\bar {V}}_{ik}{\bar {U}}_{jk}^{*})}$$

with

$${\displaystyle \phi _{i,j,k}=\left\{{\begin{aligned}&{\frac {G_{ik}-G_{jk}}{\lambda _{i}-\lambda _{j}}}&{\text{ if }}i\neq j,\\&{\frac {G_{ii}-G_{ik}}{\lambda _{i}-\lambda _{k}}}&{\text{ if }}i=j{\text{ and }}k\neq i,\\&{\frac {G_{ii}}{2}}&{\text{ if }}i=j=k.\\\end{aligned}}\right.}$$

Diagonalizable case

If a matrix is diagonal:

$${\displaystyle A={\begin{bmatrix}a_{1}&0&\cdots &0\\0&a_{2}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &a_{n}\end{bmatrix}},}$$

then its exponential can be obtained by exponentiating each entry on the main diagonal:

$${\displaystyle e^{A}={\begin{bmatrix}e^{a_{1}}&0&\cdots &0\\0&e^{a_{2}}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &e^{a_{n}}\end{bmatrix}}.}$$

This result also allows one to exponentiate diagonalizable matrices. If

and D is diagonal, then

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

$${\displaystyle A={\begin{bmatrix}1&4\\1&1\\\end{bmatrix}}}$$

can be diagonalized as

$${\displaystyle {\begin{bmatrix}-2&2\\1&1\\\end{bmatrix}}{\begin{bmatrix}-1&0\\0&3\\\end{bmatrix}}{\begin{bmatrix}-2&2\\1&1\\\end{bmatrix}}^{-1}.}$$

Thus,

$${\displaystyle e^{A}={\begin{bmatrix}-2&2\\1&1\\\end{bmatrix}}e^{\begin{bmatrix}-1&0\\0&3\\\end{bmatrix}}{\begin{bmatrix}-2&2\\1&1\\\end{bmatrix}}^{-1}={\begin{bmatrix}-2&2\\1&1\\\end{bmatrix}}{\begin{bmatrix}{\frac {1}{e}}&0\\0&e^{3}\\\end{bmatrix}}{\begin{bmatrix}-2&2\\1&1\\\end{bmatrix}}^{-1}={\begin{bmatrix}{\frac {e^{4}+1}{2e}}&{\frac {e^{4}-1}{e}}\\{\frac {e^{4}-1}{4e}}&{\frac {e^{4}+1}{2e}}\\\end{bmatrix}}.}$$

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:

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

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

General case

Using the Jordan–Chevalley decomposition

By the Jordan–Chevalley decomposition, any ${\displaystyle n\times n}$ matrix X with complex entries can be expressed as

$${\displaystyle X=A+N}$$

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:

$${\displaystyle e^{X}=e^{A+N}=e^{A}e^{N}.}$$

Note that we need the commutativity of A and N for the last step to work.

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

$${\displaystyle e^{X}=Pe^{J}P^{-1}.}$$

Also, since

$${\displaystyle {\begin{aligned}J&=J_{a_{1}}(\lambda _{1})\oplus J_{a_{2}}(\lambda _{2})\oplus \cdots \oplus J_{a_{n}}(\lambda _{n}),\\e^{J}&=\exp {\big (}J_{a_{1}}(\lambda _{1})\oplus J_{a_{2}}(\lambda _{2})\oplus \cdots \oplus J_{a_{n}}(\lambda _{n}){\big )}\\&=\exp {\big (}J_{a_{1}}(\lambda _{1}){\big )}\oplus \exp {\big (}J_{a_{2}}(\lambda _{2}){\big )}\oplus \cdots \oplus \exp {\big (}J_{a_{n}}(\lambda _{n}){\big )}.\end{aligned}}}$$

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

$${\displaystyle {\begin{aligned}&&J_{a}(\lambda )&=\lambda I+N\\&\Rightarrow &e^{J_{a}(\lambda )}&=e^{\lambda I+N}=e^{\lambda }e^{N}.\end{aligned}}}$$

where N is a special nilpotent matrix. The matrix exponential of J is then given by

$${\displaystyle e^{J}=e^{\lambda _{1}}e^{N_{a_{1}}}\oplus e^{\lambda _{2}}e^{N_{a_{2}}}\oplus \cdots \oplus e^{\lambda _{n}}e^{N_{a_{n}}}}$$

Projection case

If P is a projection matrix (i.e. is idempotent: P² = P), its matrix exponential is:

Deriving this by expansion of the exponential function, each power of P reduces to P which becomes a common factor of the sum:

$${\displaystyle e^{P}=\sum _{k=0}^{\infty }{\frac {P^{k}}{k!}}=I+\left(\sum _{k=1}^{\infty }{\frac {1}{k!}}\right)P=I+(e-1)P~.}$$

Rotation case

For a simple rotation in which the perpendicular unit vectors a and b specify a plane,^[19] the rotation matrix R can be expressed in terms of a similar exponential function involving a generator G and angle θ.^[20][21]

$${\displaystyle {\begin{aligned}G&=\mathbf {ba} ^{\mathsf {T}}-\mathbf {ab} ^{\mathsf {T}}&P&=-G^{2}=\mathbf {aa} ^{\mathsf {T}}+\mathbf {bb} ^{\mathsf {T}}\\P^{2}&=P&PG&=G=GP~,\end{aligned}}}$$

$${\displaystyle {\begin{aligned}R\left(\theta \right)=e^{G\theta }&=I+G\sin(\theta )+G^{2}(1-\cos(\theta ))\\&=I-P+P\cos(\theta )+G\sin(\theta )~.\\\end{aligned}}}$$

The formula for the exponential results from reducing the powers of G in the series expansion and identifying the respective series coefficients of G² 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 ${\displaystyle a=\left[{\begin{smallmatrix}1\\0\end{smallmatrix}}\right]}$ and ${\displaystyle b=\left[{\begin{smallmatrix}0\\1\end{smallmatrix}}\right]}$, then ${\displaystyle G=\left[{\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}}\right]}$, ${\displaystyle G^{2}=\left[{\begin{smallmatrix}-1&0\\0&-1\end{smallmatrix}}\right]}$, and

$${\displaystyle R(\theta )={\begin{bmatrix}\cos(\theta )&-\sin(\theta )\\\sin(\theta )&\cos(\theta )\end{bmatrix}}=I\cos(\theta )+G\sin(\theta )}$$

reduces to the standard matrix for a plane rotation.

The matrix P = −G² projects a vector onto the ab-plane 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,

$${\displaystyle {\begin{aligned}\mathbf {a} &={\begin{bmatrix}1\\0\\0\\\end{bmatrix}}&\mathbf {b} &={\frac {1}{\sqrt {5}}}{\begin{bmatrix}0\\1\\2\\\end{bmatrix}}\end{aligned}}}$$

$${\displaystyle {\begin{aligned}G={\frac {1}{\sqrt {5}}}&{\begin{bmatrix}0&-1&-2\\1&0&0\\2&0&0\\\end{bmatrix}}&P=-G^{2}&={\frac {1}{5}}{\begin{bmatrix}5&0&0\\0&1&2\\0&2&4\\\end{bmatrix}}\\P{\begin{bmatrix}1\\2\\3\\\end{bmatrix}}={\frac {1}{5}}&{\begin{bmatrix}5\\8\\16\\\end{bmatrix}}=\mathbf {a} +{\frac {8}{\sqrt {5}}}\mathbf {b} &R\left({\frac {\pi }{6}}\right)&={\frac {1}{10}}{\begin{bmatrix}5{\sqrt {3}}&-{\sqrt {5}}&-2{\sqrt {5}}\\{\sqrt {5}}&8+{\sqrt {3}}&-4+2{\sqrt {3}}\\2{\sqrt {5}}&-4+2{\sqrt {3}}&2+4{\sqrt {3}}\\\end{bmatrix}}\\\end{aligned}}}$$

Let N = I - P, so N² = N and its products with P and G are zero. This will allow us to evaluate powers of R.

$${\displaystyle {\begin{aligned}R\left({\frac {\pi }{6}}\right)&=N+P{\frac {\sqrt {3}}{2}}+G{\frac {1}{2}}\\R\left({\frac {\pi }{6}}\right)^{2}&=N+P{\frac {1}{2}}+G{\frac {\sqrt {3}}{2}}\\R\left({\frac {\pi }{6}}\right)^{3}&=N+G\\R\left({\frac {\pi }{6}}\right)^{6}&=N-P\\R\left({\frac {\pi }{6}}\right)^{12}&=N+P=I\\\end{aligned}}}$$

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

$${\displaystyle \mathbf {y} '=A\mathbf {y} }$$

has solution e^At y(0).

If we consider the vector

$${\displaystyle \mathbf {y} (t)={\begin{bmatrix}y_{1}(t)\\\vdots \\y_{n}(t)\end{bmatrix}}~,}$$

we can express a system of inhomogeneous coupled linear differential equations as

$${\displaystyle \mathbf {y} '(t)=A\mathbf {y} (t)+\mathbf {b} (t).}$$

Making an ansatz to use an integrating factor of e^−At and multiplying throughout, yields

$${\displaystyle {\begin{aligned}&&e^{-At}\mathbf {y} '-e^{-At}A\mathbf {y} &=e^{-At}\mathbf {b} \\&\Rightarrow &e^{-At}\mathbf {y} '-Ae^{-At}\mathbf {y} &=e^{-At}\mathbf {b} \\&\Rightarrow &{\frac {d}{dt}}\left(e^{-At}\mathbf {y} \right)&=e^{-At}\mathbf {b} ~.\end{aligned}}}$$

The second step is possible due to the fact that, if AB = BA, then e^AtB = Be^At. So, calculating e^At leads to the solution to the system, by simply integrating the third step with respect to t.

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

Example (homogeneous)

Consider the system

$${\displaystyle {\begin{matrix}x'&=&2x&-y&+z\\y'&=&&3y&-1z\\z'&=&2x&+y&+3z\end{matrix}}~.}$$

The associated defective matrix is

$${\displaystyle A={\begin{bmatrix}2&-1&1\\0&3&-1\\2&1&3\end{bmatrix}}~.}$$

The matrix exponential is

$${\displaystyle e^{tA}={\frac {1}{2}}{\begin{bmatrix}e^{2t}\left(1+e^{2t}-2t\right)&-2te^{2t}&e^{2t}\left(-1+e^{2t}\right)\\-e^{2t}\left(-1+e^{2t}-2t\right)&2(t+1)e^{2t}&-e^{2t}\left(-1+e^{2t}\right)\\e^{2t}\left(-1+e^{2t}+2t\right)&2te^{2t}&e^{2t}\left(1+e^{2t}\right)\end{bmatrix}}~,}$$

so that the general solution of the homogeneous system is

$${\displaystyle {\begin{bmatrix}x\\y\\z\end{bmatrix}}={\frac {x(0)}{2}}{\begin{bmatrix}e^{2t}\left(1+e^{2t}-2t\right)\\-e^{2t}\left(-1+e^{2t}-2t\right)\\e^{2t}\left(-1+e^{2t}+2t\right)\end{bmatrix}}+{\frac {y(0)}{2}}{\begin{bmatrix}-2te^{2t}\\2(t+1)e^{2t}\\2te^{2t}\end{bmatrix}}+{\frac {z(0)}{2}}{\begin{bmatrix}e^{2t}\left(-1+e^{2t}\right)\\-e^{2t}\left(-1+e^{2t}\right)\\e^{2t}\left(1+e^{2t}\right)\end{bmatrix}}~,}$$

amounting to

$${\displaystyle {\begin{aligned}2x&=x(0)e^{2t}\left(1+e^{2t}-2t\right)+y(0)\left(-2te^{2t}\right)+z(0)e^{2t}\left(-1+e^{2t}\right)\\[2pt]2y&=x(0)\left(-e^{2t}\right)\left(-1+e^{2t}-2t\right)+y(0)2(t+1)e^{2t}+z(0)\left(-e^{2t}\right)\left(-1+e^{2t}\right)\\[2pt]2z&=x(0)e^{2t}\left(-1+e^{2t}+2t\right)+y(0)2te^{2t}+z(0)e^{2t}\left(1+e^{2t}\right)~.\end{aligned}}}$$