where is defined to be the identity matrix with the same dimensions as .
The above series always converges, so the exponential of X is well-defined. 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.
Let X and Y be n×n complex matrices and let a and b be arbitrary complex numbers. We denote the n×nidentity matrix by I and the zero matrix by 0. The matrix exponential satisfies the following properties.
We begin with the properties that are immediate consequences of the definition as a power series:
e0 = I
exp(XT) = (exp X)T, where XT denotes the transpose of X.
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 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 Golden-Thompson inequality below).
Consequences of the preceding identity are the following:
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(eA) ≠ 0, which implies that eA must be invertible.
In the real-valued 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 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 be an n×n real symmetric matrix and a column vector. Using the elementary properties of the matrix exponential and of symmetric matrices, we have:
Since is invertible, the equality only holds for , and we have for all non-zero . Hence is positive definite.
The exponential of sums
For any real numbers (scalars) x and y we know that the exponential function satisfies ex+y = exey. 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.
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 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 eX 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 (for this, it is essential to consider the field C of complex numbers and not R).
The derivative of this curve (or tangent vector) at a point t is given by
The derivative at t = 0 is just the matrix X, which is to say that X generates this one-parameter subgroup.
More generally, for a generic t-dependent exponent, X(t),
Taking the 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 the matrix exponent,
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. In this section, we discuss methods that are applicable in principle to any matrix, and which can be carried out explicitly for small matrices. Subsequent sections describe methods suitable for numerical evaluation on large matrices.
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
can be diagonalized as
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:
Since the series has a finite number of steps, it is a matrix polynomial, which can be computed efficiently.
Deriving this by expansion of the exponential function, each power of P reduces to P which becomes a common factor of the sum:
For a simple rotation in which the perpendicular unit vectors a and b specify a plane, the rotation matrixR can be expressed in terms of a similar exponential function involving a generatorG and angle θ.
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 eGθ is the same as the expression for R(θ) in the article containing the derivation of the generator, R(θ) = eGθ.
In two dimensions, if and , then , , and
reduces to the standard matrix for a plane rotation.
The matrix P = −G2projects 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,
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.
To prove this, multiply the first of the two above equalities by P(z) and replace z by A.
Such a polynomial Qt(z) can be found as follows−see Sylvester's formula. Letting a be a root of P, Qa,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 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(z). In particular, St(z), the Lagrange-Sylvester polynomial, is the only Qt whose degree is less than that of P.
Example: Consider the case of an arbitrary 2×2 matrix,
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 St can also be given the following "interpolation" characterization. Define et(z) ≡ etz, and n ≡ deg P. Then St(z) is the unique degree < n polynomial which satisfies St(k)(a) = et(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 St 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) = Bα exp(tα) + Bβ 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. This is illustrated here for a 4×4 example of a matrix which is not diagonalizable, and the Bs are not projection matrices.
with eigenvalues λ1 = 3/4 and λ2 = 1, each with a multiplicity of two.
Consider the exponential of each eigenvalue multiplied by t, exp(λit). Multiply each exponentiated eigenvalue by the corresponding undetermined coefficient matrix Bi. 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 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,
Substituting with the value for A yields the coefficient matrices
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 e(λI + N) = eλeN mentioned above; this yields
Therefore, the exponential of the original matrix B is
so that the general solution of the homogeneous system is
Consider now the inhomogeneous system
We again have
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 = yp(0). For more rigor, see the following generalization.
Inhomogeneous case generalization: variation of parameters
For matrix-matrix exponentials, there is a distinction between the left exponential YX and the right exponential XY, because the multiplication operator for matrix-to-matrix is not commutative. Moreover,
If X is normal and non-singular, then XY and YX have the same set of eigenvalues.
If X is normal and non-singular, Y is normal, and XY = YX, then XY = YX.
If X is normal and non-singular, and X, Y, Z commute with each other, then XY+Z = XY·XZ and Y+ZX = YX·ZX.
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Van Kortryk, T. S. (2016). "Matrix exponentials, SU(N) group elements, and real polynomial roots". Journal of Mathematical Physics. 57 (2): 021701. arXiv:1508.05859. Bibcode:2016JMP....57b1701V. doi:10.1063/1.4938418. S2CID 119647937.