Editing Matrix exponential (section)

== Evaluation by Laurent series ==
By virtue of the [[Cayley–Hamilton theorem]] the matrix exponential is expressible as a polynomial of order {{mvar|n}}−1.

If {{mvar|P}} and {{math|''Q<sub>t</sub>''}} are nonzero polynomials in one variable, such that {{math|1=''P''(''A'') = 0}}, and if the [[meromorphic function]]
<math display="block">f(z)=\frac{e^{t z}-Q_t(z)}{P(z)}</math>
is [[entire function|entire]], then
<math display="block">e^{t A} = Q_t(A).</math>
To prove this, multiply the first of the two above equalities by {{math|''P''(''z'')}} and replace {{mvar|z}} by {{mvar|A}}.

Such a polynomial {{math|''Q<sub>t</sub>''(''z'')}} can be found as follows−see [[Sylvester's formula]]. Letting {{mvar|a}} be a root of {{mvar|P}}, {{math|''Q<sub>a,t</sub>''(''z'')}} is solved from the product of {{mvar|P}} by the [[Laurent series#Principal part|principal part]] of the [[Laurent series]] of {{mvar|f}} at {{mvar|a}}: It is proportional to the relevant [[Frobenius covariant]]. Then the sum ''S<sub>t</sub>'' of the ''Q<sub>a,t</sub>'', where {{mvar|a}} runs over all the roots of {{mvar|P}}, can be taken as a particular {{math|''Q<sub>t</sub>''}}. All the other ''Q<sub>t</sub>'' will be obtained by adding a multiple of {{mvar|P}} to {{math|''S<sub>t</sub>''(''z'')}}. In particular, {{math|''S<sub>t</sub>''(''z'')}}, the [[Sylvester's formula|Lagrange-Sylvester polynomial]], is the only {{math|''Q<sub>t</sub>''}} whose degree is less than that of {{mvar|P}}.

'''Example''': Consider the case of an arbitrary {{math|2 × 2}} matrix,
<math display="block">A := \begin{bmatrix}
  a & b \\
  c & d
\end{bmatrix}.</math>

The exponential matrix {{math|e<sup>''tA''</sup>}}, by virtue of the [[Cayley–Hamilton theorem]], must be of the form
<math display="block">e^{tA} = s_0(t)\, I + s_1(t)\,A.</math>

(For any complex number {{mvar|z}} and any '''''C'''''-algebra {{mvar|B}}, we denote again by {{mvar|z}} the product of {{mvar|z}} by the unit of {{mvar|B}}.)

Let {{mvar|α}} and {{mvar|β}} be the roots of the [[characteristic polynomial]] of {{mvar|A}},
<math display="block">P(z) = z^2 - (a + d)\ z + ad - bc = (z - \alpha)(z - \beta) ~ .</math>

Then we have
<math display="block">S_t(z) = e^{\alpha t} \frac{z - \beta}{\alpha - \beta} + e^{\beta t} \frac{z - \alpha}{\beta - \alpha}~,</math>
hence
<math display="block">\begin{align}
  s_0(t) &= \frac{\alpha\,e^{\beta t} - \beta\,e^{\alpha t}}{\alpha - \beta}, &
  s_1(t) &= \frac{e^{\alpha t} - e^{\beta t}}{\alpha - \beta}
\end{align}</math>

if {{math|''α'' ≠ ''β''}}; while, if {{math|1=''α'' = ''β''}},
<math display="block">S_t(z) = e^{\alpha t} (1 + t (z - \alpha)) ~,</math>

so that
<math display="block">\begin{align}
  s_0(t) &= (1 - \alpha\,t)\,e^{\alpha t},&
  s_1(t) &= t\,e^{\alpha t}~.
\end{align}</math>

Defining
<math display="block">\begin{align}
  s &\equiv \frac{\alpha + \beta}{2} = \frac{\operatorname{tr} A}{2}~, &
  q &\equiv \frac{\alpha - \beta}{2} = \pm\sqrt{-\det\left(A - sI\right)},
\end{align}</math>

we have
<math display="block">\begin{align}
  s_0(t) &= e^{st}\left(\cosh(qt) - s\frac{\sinh(qt)}{q}\right), &
  s_1(t) &= e^{st}\frac{\sinh(qt)}{q},
\end{align}</math>

where {{math|sin(''qt'')/''q''}} is 0 if {{math|1=''t'' = 0}}, and {{mvar|t}} if {{math|1=''q'' = 0}}.

Thus,
{{Equation box 1
|indent =
|equation = <math>e^{tA}=e^{st}\left(\left(\cosh(qt) - s\frac{\sinh(qt)}{q}\right)~I~ + \frac{\sinh(qt)}{q} A\right) ~.</math>
|cellpadding= 6
|border
|border colour = #0073CF
|bgcolor=#000000
}}

Thus, as indicated above, the matrix {{mvar|A}} having decomposed into the sum of two mutually commuting pieces, the traceful piece and the traceless piece,
<math display="block">A = sI + (A-sI)~,</math>

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#Exponential of a Pauli vector|Pauli spin matrices]], that is rotations of the doublet representation of the group [[SU(2)]].

The polynomial {{math|''S<sub>t</sub>''}} can also be given the following "[[interpolation]]" characterization. Define {{math|''e<sub>t</sub>''(''z'') ≡ ''e<sup>tz</sup>''}}, and {{math|''n'' ≡ deg ''P''}}. Then {{math|''S<sub>t</sub>''(''z'')}} is the unique degree {{math|< ''n''}} polynomial which satisfies {{math|1=''S''<sub>''t''</sub><sup>(''k'')</sup>(''a'') = ''e''<sub>''t''</sub><sup>(''k'')</sup>(''a'')}} whenever {{mvar|k}} is less than the multiplicity of {{mvar|a}} as a root of {{mvar|P}}. We assume, as we obviously can, that {{mvar|P}} is the [[Minimal polynomial (linear algebra)|minimal polynomial]] of {{mvar|A}}. We further assume that {{mvar|A}} is a [[diagonalizable matrix]]. In particular, the roots of {{mvar|P}} are simple, and the "[[interpolation]]" characterization indicates that {{math|''S<sub>t</sub>''}} is given by the [[Lagrange interpolation]] formula, so it is the [[Sylvester's formula|Lagrange−Sylvester polynomial]].

At the other extreme, if {{math|1=''P'' = (''z'' - ''a'')<sup>''n''</sup>}}, then
<math display="block">S_t = e^{at}\ \sum_{k=0}^{n-1}\ \frac{t^k}{k!}\ (z - a)^k ~.</math>

The simplest case not covered by the above observations is when <math>P = (z - a)^2\,(z - b)</math> with {{math|''a'' ≠ ''b''}}, which yields
<math display="block">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}.</math>