Matrix exponential

Matrix exponential

In mathematics, the matrix exponential is a matrix function on square matrices analogous to the ordinary exponential function. Abstractly, 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 eX or exp(X), is the n×n matrix given by the power series

e^\mathbf{X} = \sum_{k=0}^\infty{1 \over k!}\mathbf{X}^k.

The above series always converges, so the exponential of X is well-defined. Note that if X is a 1×1 matrix the matrix exponential of X is a 1×1 matrix consisting of the ordinary exponential of the single element of X.



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:

Linear differential equation systems

Main article: matrix differential equation

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

 \frac{d}{dt} y(t) = Ay(t), \quad y(0) = y_0,

where A is a constant matrix, 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 A is not constant, but the Magnus series gives the solution as an infinite sum.

The exponential of sums

We know that the exponential function satisfies ex + y = exey for any real numbers (scalars) x and y. The same goes for commuting matrices: If the matrices X and Y commute (meaning that XY = YX), then

e^{X+Y} = e^Xe^Y. \,

However, if they do not commute, then the above equality does not necessarily hold, in which case we can use the Baker–Campbell–Hausdorff formula to compute eX + Y.

The converse is false: the equation eX + Y = eXeY does not necessarily imply that X and Y commute. However, the converse is true if X and Y contain only algebraic numbers and their size is at least 2×2 (Horn & Johnson 1991, pp. 435–437).

The exponential map

Note that the exponential of a matrix is always an invertible matrix. The inverse matrix of eX is given by eX. This is analogous to the fact that the exponential of a complex number is always nonzero. The matrix exponential then gives us a map

\exp \colon M_n(\mathbb C) \to \mbox{GL}(n,\mathbb C)

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 (for this, it is essential to consider the field C of complex numbers and not R).

For any two matrices X and Y, we have

 \| e^{X+Y} - e^X \| \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 Mn(C).

The map

t \mapsto e^{tX}, \qquad t \in \mathbb R

defines a smooth curve in the general linear group which passes through the identity element at t = 0. 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

\frac{d}{dt}e^{tX} = Xe^{tX} = e^{tX}X. \qquad (1)

The derivative at t = 0 is just the matrix X, which is to say that X generates this one-parameter subgroup.

More generally,(R.M. Wilcox 1966)

\frac{d}{dt}e^{X(t)} = \int_0^1 e^{\alpha X(t)} \frac{dX(t)}{dt} e^{(1-\alpha) X(t)}\,d\alpha.

Taking in above expression eX(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 matrix exponent:

\left(\frac{d}{dt}e^{X(t)}\right)e^{-X(t)} = \frac{d}{dt}X(t) + \frac{1}{2!}[X(t),\frac{d}{dt}X(t)] + \frac{1}{3!}[X(t),[X(t),\frac{d}{dt}X(t)]]+\cdots

The determinant of the matrix exponential

It can be shown that for any complex square matrix, the following identity holds:

det(eA) = etr(A).

In addition to providing a computational tool, this formula shows that a matrix exponential is always an invertible matrix. This follows from the fact the right hand side of the above equation is always non-zero, and so \det (e^A) \neq 0 which means that eA must be invertible. Another observation is the following: in the real-valued case, we see that the map

\exp \colon M_n(\mathbb R) \to \mbox{GL}(n,\mathbb R)

is not surjective (this is in contrast with the complex case mentioned earlier). This follows from the fact that (for real-valued matrices) the right hand side of the above equation is always positive while there exist invertible matrices with a negative determinant.

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. Both Matlab and GNU Octave use Padé approximant.[1][2] Several methods are listed below.

Diagonalizable case

If a matrix is diagonal:

A=\begin{bmatrix} a_1 & 0 & \ldots & 0 \\
0 & a_2 & \ldots & 0  \\ \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & a_n \end{bmatrix},

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

e^A=\begin{bmatrix} e^{a_1} & 0 & \ldots & 0 \\
0 & e^{a_2} & \ldots & 0  \\ \vdots & \vdots & \ddots & \vdots \\
0 & 0 & \ldots & e^{a_n} \end{bmatrix}.

This also allows one to exponentiate diagonalizable matrices. If A = UDU − 1 and D is diagonal, then eA = UeDU − 1. Application of Sylvester's formula yields the same result.

Nilpotent case

A matrix N is nilpotent if Nq = 0 for some integer q. In this case, the matrix exponential eN can be computed directly from the series expansion, as the series terminates after a finite number of terms:

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


When the minimal polynomial of a matrix X can be factored into a product of first degree polynomials, it can be expressed as a sum

X = A + N \,


  • A is diagonalizable
  • N is nilpotent
  • A commutes with N (i.e. AN = NA)

This is the Jordan–Chevalley decomposition.

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.

Another (closely related) method if the field is algebraically closed is to work with the Jordan form of X. Suppose that X = PJP −1 where J is the Jordan form of X. Then


Also, since

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_k}(\lambda_k) \big).

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

J_{a}(\lambda) = \lambda I + N \,

where N is a special nilpotent matrix. The matrix exponential of this block is given by

e^{\lambda I + N} = e^{\lambda}e^N. \,


If P and Qt are nonzero polynomials in one variable, such that P(A) = 0, and if the meromorphic function

f(z)=\frac{e^{t z}-Q_t(z)}{P(z)}

is entire, then

etA = Qt(A).

To prove this, multiply the first of the two above equalities by P(z) and replace z by A.

Such a polynomial Qt can be found as follows. Let a be a root of P, and Qa,t the product of P by the principal part of the Laurent series of f at a. Then the sum St of the Qa,t, where a runs over all the roots of P, can be taken as a particular Qt. All the other Qt will be obtained by adding a multiple of P to St. In particular St is the only Qt whose degree is less than that of P.

Consider the case of a 2-by-2 matrix

a & b \\
c & d \end{bmatrix}.

The exponential matrix etA is of the form e^{tA}=s_0(t)\,I+s_1(t)\,A. (For any complex number z and any \mathbb{C}-algebra 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

X^2-(a+d)\ X+ ad-bc.

Then we have

s_0(t)=\frac{\alpha\,e^{\beta t}
-\beta\,e^{\alpha t}}{\alpha-\beta},\quad 
s_1(t)=\frac{e^{\alpha t}-e^{\beta t}}{\alpha-\beta}\quad

if \alpha\not=\beta, and

s_0(t)=(1-\alpha\,t)\,e^{\alpha t},\quad 
s_1(t)=t\,e^{\alpha t}\quad

if α = β.

In either case, writing:

s = \frac{\alpha + \beta}{2}=\frac{\operatorname{tr} A}{2},


q = \frac{\alpha-\beta}{2}=\pm\sqrt{-\det\left(A-s I\right)},
s_0(t) = e^{s t}\left(\cosh (q t) - s \frac{\sinh (q t)}{q t}\right),\quad s_1(t) =e^{s t}\frac{\sinh(q t)}{q t},


\frac{\sinh 0}{0} is 0 if t = 0, and 1 if q = 0.

The polynomial St can also be given the following "interpolation" characterization. Put et(z): = etz, n: = deg P. Then St 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 also assume that A is a diagonalizable matrix. In particular, the roots of P are simple, and the "interpolation" characterization tells us that St is given by the Lagrange interpolation formula.

At the other extreme, if P = (Xa)n, then

S_t=e^{at}\ \sum_{k=0}^{n-1}\ \frac{t^k}{k!}\ (X-a)^k.

The simplest case not covered by the above observations is when P=(X-a)^2\,(X-b) with a\not=b, which gives

S_t=e^{at}\ \frac{X-b}{a-b}\ \Bigg(1+\left(t+\frac{1}{b-a}\right)(X-a)\Bigg)+e^{bt}\ \frac{(X-a)^2}{(b-a)^2}\quad.

via Laplace Transform

As above we know that the solution to the system linear differential equations given by \frac{d}{dt} y(t) = Ay(t), y(0) = y_0 is y(t) = eAty0. Using the Laplace transform, letting Y(s) = \mathcal{L}\{y\}, and applying to the differential equation we get

sY(s) - y_0 = AY(s) \Rightarrow (sI - A)Y(s) = y_0

where I is the identity matrix. Therefore y(t) = \mathcal{L}^{-1}\{(sI-A)^{-1}\}y_0. Thus it can be concluded that e^{At} = \mathcal{L}^{-1}\{(sI-A)^{-1}\}. And from this we can find eA by setting t = 1.


Suppose that we want to compute the exponential of

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

-\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 1×1 matrix is just the exponential of the one entry of the matrix, so exp(J1(4)) = [e4]. The exponential of J2(16) can be calculated by the formula exp(λI + N) = eλ exp(N) mentioned above; this yields[3]

\exp \left( \begin{bmatrix} 16 & 1 \\ 0 & 16 \end{bmatrix} \right) 
& = e^{16} \exp \left( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \right) \\[6pt]
& = 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}.

Therefore, the exponential of the original matrix B is

& = 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} 


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 differential equation of the form

 \mathbf{y}' = C\mathbf{y}

has solution eCty(0). If we consider the vector

 \mathbf{y}(t) = \begin{pmatrix} y_1(t) \\ \vdots \\y_n(t) \end{pmatrix}

we can express a system of coupled linear differential equations as

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

If we make an ansatz and use an integrating factor of eAt and multiply throughout, we obtain

e^{-At}\mathbf{y}'-e^{-At}A\mathbf{y} = e^{-At}\mathbf{b}
e^{-At}\mathbf{y}'-Ae^{-At}\mathbf{y} = e^{-At}\mathbf{b}
 \frac{d}{dt} (e^{-At}\mathbf{y}) = e^{-At}\mathbf{b}.

The second step is possible due to the fact that if AB=BA then eAtB = BeAt. If we can calculate eAt, then we can obtain the solution to the system.

Example (homogeneous)

Say we have the system

x' &=& 2x&-y&+z \\
y' &=&   &3y&-1z \\
z' &=& 2x&+y&+3z \end{matrix}

We have the associated matrix

2 & -1 &  1 \\
0 &  3 & -1 \\
2 &  1 &  3 \end{bmatrix}

The matrix exponential

    e^{2t}(1+e^{2t}-2t)  & -2te^{2t}    &  e^{2t}(-1+e^{2t}) \\
   -e^{2t}(-1+e^{2t}-2t) & 2(t+1)e^{2t} & -e^{2t}(-1+e^{2t}) \\
    e^{2t}(-1+e^{2t}+2t) & 2te^{2t}     &  e^{2t}(1+e^{2t})  \end{bmatrix}

so the general solution of the system is

\begin{bmatrix}x \\y \\ z\end{bmatrix}=
C_1\begin{bmatrix}e^{2t}(1+e^{2t}-2t) \\-e^{2t}(-1+e^{2t}-2t)\\e^{2t}(-1+e^{2t}+2t)\end{bmatrix}

that is,

x & = C_1(e^{2t}(1+e^{2t}-2t)) + C_2(-2te^{2t}) + C_3(e^{2t}(-1+e^{2t})) \\
y & = C_1(-e^{2t}(-1+e^{2t}-2t)) + C_2(2(t+1)e^{2t}) + C_3(-e^{2t}(-1+e^{2t})) \\
z & = C_1(e^{2t}(-1+e^{2t}+2t)) + C_2(2te^{2t}) + C_3(e^{2t}(1+e^{2t}))

Inhomogeneous case – 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 yp(t) = exp(tA) z (t) :

\mathbf{y}_p'(t) & = (e^{tA})'\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).

For yp to be a solution:

e^{tA}\mathbf{z}'(t) & = \mathbf{b}(t) \\[6pt]
\mathbf{z}'(t) & = (e^{tA})^{-1}\mathbf{b}(t) \\[6pt]
\mathbf{z}(t) & = \int_0^t e^{-uA}\mathbf{b}(u)\,du+\mathbf{c}.


\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}

where c is determined by the initial conditions of the problem.

More precisely, consider the equation

Y'-A\ Y=F(t)

with the initial condition Y(t0) = Y0, where

A is an n by n complex matrix,

F is a continuous function from some open interval I to \mathbb{C}^n,

t0 is a point of I, and

Y0 is a vector of \mathbb{C}^n.

Left multiplying the above displayed equality by e tA, we get

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)(t0) = yk for 0\le k<n 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 n > 0,

f is a continuous complex valued function defined on some open interval I,

t0 is a point of I,

yk is a complex number, and

sk(t) is the coefficient of Xk in the polynomial denoted by S_t\in\mathbb{C}[X] in Subsection Alternative 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

\frac{dY}{dt}-A\ Y=F(t),\quad Y(t_0)=Y_0,

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 Alternative above.

In the case n = 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\ .


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 s0 and s1 are as in Subsection Alternative above.

Example (inhomogeneous)

Say we have the system

x' &=& 2x & - & y & + & z & + & e^{2t} \\
y' &=&    &   & 3y& - & z & \\
z' &=& 2x & + & y & + & 3z & + & e^{2t}. \end{matrix}

So we then have

M= \left[ \begin{array}{rrr}
2 & -1 &  1 \\
0 &  3 & -1 \\
2 &  1 &  3 \end{array} \right]


\mathbf{b}=e^{2t}\begin{bmatrix}1 \\0\\1\end{bmatrix}.

From before, we 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, now we only need to find the particular solution (via variation of parameters).

We have, above:

\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}
\mathbf{y}_p = e^{tA}\int_0^t
     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}
\mathbf{y}_p = e^{tA}\int_0^t
e^{2u}( 2e^u - 2ue^{2u}) \\  \\
  e^{2u}(-2e^u + 2(1 + u)e^{2u}) \\  \\
  2e^{3u} + 2ue^{4u}\end{bmatrix}+e^{tA}\mathbf{c}
\mathbf{y}_p = e^{tA}\begin{bmatrix}
-{1 \over 24}e^{3t}(3e^t(4t-1)-16) \\  \\
{1 \over 24}e^{3t}(3e^t(4t+4)-16) \\  \\
{1 \over 24}e^{3t}(3e^t(4t-1)-16)\end{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}

which can be further simplified to get the requisite particular solution determined through variation of parameters.

See also


  1. ^
  2. ^
  3. ^ This can be generalized; in general, the exponential of Jn(a) is an upper triangular matrix with ea/0! on the main diagonal, ea/1! on the one above, ea/2! on the next one, and so on.

External links

Wikimedia Foundation. 2010.

Игры ⚽ Нужен реферат?

Look at other dictionaries:

  • Exponential — may refer to any of several mathematical topics related to exponentiation, including: *Exponential function, also: **Matrix exponential, the matrix analogue to the above *Exponential decay *Exponential growth *Exponential map, in differential… …   Wikipedia

  • Exponential map — In differential geometry, the exponential map is a generalization of the ordinary exponential function of mathematical analysis to all differentiable manifolds with an affine connection. Two important special cases of this are the exponential map …   Wikipedia

  • Exponential function — The natural exponential function y = ex In mathematics, the exponential function is the function ex, where e is the number (approximately 2.718281828) such that the function ex is its own derivative …   Wikipedia

  • Matrix (mathematics) — Specific elements of a matrix are often denoted by a variable with two subscripts. For instance, a2,1 represents the element at the second row and first column of a matrix A. In mathematics, a matrix (plural matrices, or less commonly matrixes)… …   Wikipedia

  • Matrix function — In mathematics, a matrix function is a function which maps a matrix to another matrix. Contents 1 Extending scalar functions to matrix functions 1.1 Power series 1.2 Jordan decomposition …   Wikipedia

  • Matrix of ones — In mathematics, a matrix of ones is a matrix where every element is equal to one. Examples of standard notation are given below: In special contexts, the term unit matrix is used as a synonym for matrix of ones [1] This is done whenever it is… …   Wikipedia

  • Matrix normal distribution — parameters: mean row covariance column covariance. Parameters are matrices (all of them). support: is a matrix …   Wikipedia

  • Matrix chain multiplication — is an optimization problem that can be solved using dynamic programming. Given a sequence of matrices, we want to find the most efficient way to multiply these matrices together. The problem is not actually to perform the multiplications, but… …   Wikipedia

  • Matrix mechanics — Quantum mechanics Uncertainty principle …   Wikipedia

  • Exponential distribution — Not to be confused with the exponential families of probability distributions. Exponential Probability density function Cumulative distribution function para …   Wikipedia

Share the article and excerpts

Direct link
Do a right-click on the link above
and select “Copy Link”