Matrix exponential

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, R, 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.

Diagonalizable case

If a matrix is diagonal: $$A={\begin{bmatrix}a_{1}&0&\cdots &0\\0&a_{2}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &a_{n}\end{bmatrix}},$$ then its exponential can be obtained by exponentiating each entry on the main diagonal: $$e^{A}={\begin{bmatrix}e^{a_{1}}&0&\cdots &0\\0&e^{a_{2}}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &e^{a_{n}}\end{bmatrix}}.$$

This result also allows one to exponentiate diagonalizable matrices. If

A = UDU⁻¹

and D is diagonal, then

e^A = Ue^DU⁻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 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 $$A={\begin{bmatrix}1&4\\1&1\\\end{bmatrix}}$$ can be diagonalized as $${\begin{bmatrix}-2&2\\1&1\\\end{bmatrix}}{\begin{bmatrix}-1&0\\0&3\\\end{bmatrix}}{\begin{bmatrix}-2&2\\1&1\\\end{bmatrix}}^{-1}.$$

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

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:

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

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 $n\times n$ matrix X with complex entries can be expressed as $$X=A+N$$ 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: $$e^{X}=e^{A+N}=e^{A}e^{N}.$$

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

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

Therefore, we need only know how to compute the matrix exponential of a Jordan block. But each Jordan block is of the form $${\begin{aligned}&&J_{a}(\lambda )&=\lambda I+N\\&\Rightarrow &e^{J_{a}(\lambda )}&=e^{\lambda I+N}=e^{\lambda }e^{N}.\end{aligned}}$$

where N is a special nilpotent matrix. The matrix exponential of J is then given by $$e^{J}=e^{\lambda _{1}}e^{N_{a_{1}}}\oplus e^{\lambda _{2}}e^{N_{a_{2}}}\oplus \cdots \oplus e^{\lambda _{n}}e^{N_{a_{n}}}$$

Projection case

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

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

Rotation case

For a simple rotation in which the perpendicular unit vectors a and b specify a plane, the rotation matrix R can be expressed in terms of a similar exponential function involving a generator G and angle θ. $${\begin{aligned}G&=\mathbf {ba} ^{\mathsf {T}}-\mathbf {ab} ^{\mathsf {T}}&P&=-G^{2}=\mathbf {aa} ^{\mathsf {T}}+\mathbf {bb} ^{\mathsf {T}}\\P^{2}&=P&PG&=G=GP~,\end{aligned}}$$ $${\begin{aligned}R\left(\theta \right)=e^{G\theta }&=I+G\sin(\theta )+G^{2}(1-\cos(\theta ))\\&=I-P+P\cos(\theta )+G\sin(\theta )~.\\\end{aligned}}$$

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 $a=\left[{\begin{smallmatrix}1\\0\end{smallmatrix}}\right]$ and $b=\left[{\begin{smallmatrix}0\\1\end{smallmatrix}}\right]$, then $G=\left[{\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}}\right]$, $G^{2}=\left[{\begin{smallmatrix}-1&0\\0&-1\end{smallmatrix}}\right]$, and $$R(\theta )={\begin{bmatrix}\cos(\theta )&-\sin(\theta )\\\sin(\theta )&\cos(\theta )\end{bmatrix}}=I\cos(\theta )+G\sin(\theta )$$ 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,

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

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.

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

Further information: Rodrigues' rotation formula and Axis–angle representation § Exponential map from so(3) to SO(3)