Editing Matrix exponential (section)

== 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 {{math|''n'' × ''n''}} matrix {{math|exp(''tA'')}} amounts to a linear combination of the first {{mvar|n}}−1 powers of {{mvar|A}} by the [[Cayley–Hamilton theorem]]. For [[diagonalizable matrix|diagonalizable]] matrices, as illustrated above, e.g. in the {{math|2 × 2}} case, [[Sylvester's formula]] yields {{math|1=exp(''tA'') = ''B<sub>α</sub>'' exp(''tα'') + ''B<sub>β</sub>'' exp(''tβ'')}}, where the {{mvar|B}}s are the [[Frobenius covariant]]s of {{mvar|A}}.

It is easiest, however, to simply solve for these {{mvar|B}}s directly, by evaluating this expression and its first derivative at {{math|1=''t'' = 0}}, in terms of {{mvar|A}} and {{mvar|I}}, to find the same answer as above.

But this simple procedure also works for [[defective matrix|defective]] matrices, in a generalization due to Buchheim.<ref>Rinehart, R. F. (1955). "[https://www.jstor.org/stable/2306996 The equivalence of definitions of a matric function]". ''The American Mathematical Monthly'', '''62''' (6), 395-414.</ref> This is illustrated here for a {{math|4 × 4}} example of a matrix which is ''not diagonalizable'', and the {{mvar|B}}s are not projection matrices.

Consider
<math display="block">A = \begin{bmatrix}
  1 & 1 &           0 &            0 \\
  0 & 1 &           1 &            0 \\
  0 & 0 &           1 & -\frac{1}{8} \\
  0 & 0 & \frac{1}{2} &  \frac{1}{2}
\end{bmatrix} ~,</math>
with eigenvalues {{math|1=''λ''<sub>1</sub> = 3/4}} and {{math|1=''λ''<sub>2</sub> = 1}}, each with a multiplicity of two.

Consider the exponential of each eigenvalue multiplied by {{mvar|t}}, {{math|exp(''λ<sub>i</sub>t'')}}. Multiply each exponentiated eigenvalue by the corresponding undetermined coefficient matrix {{math|''B''<sub>''i''</sub>}}.  If the eigenvalues have an algebraic multiplicity greater than 1, then repeat the process, but now multiplying by an extra factor of {{mvar|t}} for each repetition, to ensure linear independence.

(If one eigenvalue had a multiplicity of three, then there would be the three terms: <math>B_{i_1} e^{\lambda_i t}, ~ B_{i_2} t e^{\lambda_i t}, ~ B_{i_3} t^2 e^{\lambda_i t} </math>. By contrast, when all eigenvalues are distinct, the {{mvar|B}}s are just the [[Frobenius covariant]]s, 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,
<math display="block">\begin{align}
  e^{At} &= B_{1_1} e^{\lambda_1 t} + B_{1_2} t e^{\lambda_1 t} + B_{2_1} e^{\lambda_2 t} + B_{2_2} t e^{\lambda_2 t} , \\
  e^{At} &= B_{1_1} e^{\frac{3}{4} t} + B_{1_2} t e^{\frac{3}{4} t} + B_{2_1} e^{1 t} + B_{2_2} t e^{1 t} ~.
\end{align}</math>

To solve for all of the unknown matrices {{mvar|B}} in terms of the first three powers of {{mvar|A}} and the identity, one needs four equations, the above one providing one such at {{mvar|t}} = 0. Further, differentiate it with respect to {{mvar|t}},
<math display="block">A e^{A t} = \frac{3}{4} B_{1_1} e^{\frac{3}{4} t} + \left( \frac{3}{4} t + 1 \right) B_{1_2} e^{\frac{3}{4} t} + 1 B_{2_1} e^{1 t} + \left(1 t + 1 \right) B_{2_2} e^{1 t} ~,</math>

and again,
<math display="block">\begin{align}
  A^2 e^{At} &= \left(\frac{3}{4}\right)^2 B_{1_1} e^{\frac{3}{4} t} + \left( \left(\frac{3}{4}\right)^2 t + \left( \frac{3}{4} + 1 \cdot \frac{3}{4}\right) \right) B_{1_2} e^{\frac{3}{4} t} + B_{2_1} e^{1 t} + \left(1^2 t + (1 + 1 \cdot 1 )\right) B_{2_2} e^{1 t} \\
             &= \left(\frac{3}{4}\right)^2 B_{1_1} e^{\frac{3}{4} t} + \left( \left(\frac{3}{4}\right)^2 t + \frac{3}{2} \right) B_{1_2} e^{\frac{3}{4} t} + B_{2_1} e^{t} + \left(t + 2\right) B_{2_2} e^{t} ~,
\end{align}</math>

and once more,
<math display="block">\begin{align}
  A^3 e^{At} &= \left(\frac{3}{4}\right)^3 B_{1_1} e^{\frac{3}{4} t} + \left( \left(\frac{3}{4}\right)^3 t + \left( \left(\frac{3}{4}\right)^2 + \left(\frac{3}{2}\right) \cdot \frac{3}{4}\right) \right) B_{1_2} e^{\frac{3}{4} t} + B_{2_1} e^{1 t} + \left(1^3 t + (1 + 2) \cdot 1 \right) B_{2_2} e^{1 t} \\
             &= \left(\frac{3}{4}\right)^3 B_{1_1} e^{\frac{3}{4} t}\! + \left( \left(\frac{3}{4}\right)^3 t\! + \frac{27}{16} \right) B_{1_2} e^{\frac{3}{4} t}\! + B_{2_1} e^{t}\! + \left(t + 3\cdot 1\right) B_{2_2} e^{t} ~.
\end{align}</math>

(In the general case, {{mvar|n}}−1 derivatives need be taken.)

Setting {{mvar|t}} = 0 in these four equations, the four coefficient matrices {{mvar|B}}s may now be solved for,
<math display="block">\begin{align}
I &= B_{1_1} + B_{2_1} \\
A &= \frac{3}{4} B_{1_1} + B_{1_2} + B_{2_1} + B_{2_2} \\
A^2 &= \left(\frac{3}{4}\right)^2 B_{1_1} + \frac{3}{2} B_{1_2} + B_{2_1} + 2 B_{2_2} \\
A^3 &= \left(\frac{3}{4}\right)^3 B_{1_1} + \frac{27}{16} B_{1_2} + B_{2_1} + 3 B_{2_2} ~,
\end{align} </math>

to yield
<math display="block">\begin{align}
B_{1_1} &=  128 A^3 - 366 A^2 + 288 A - 80 I \\
B_{1_2} &=   16 A^3 -  44 A^2 +  40 A - 12 I \\
B_{2_1} &= -128 A^3 + 366 A^2 - 288 A + 80 I \\
B_{2_2} &=   16 A^3 -  40 A^2 +  33 A -  9 I ~.
\end{align}</math>

Substituting with the value for {{mvar|A}} yields the coefficient matrices
<math display="block">\begin{align}
B_{1_1} &= \begin{bmatrix}0 & 0 &  48 & -16\\ 0 & 0 & -8 & 2\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}\\
B_{1_2} &= \begin{bmatrix}0 & 0 &   4 &  -2\\ 0 & 0 & -1 & \frac{1}{2}\\ 0 & 0 & \frac{1}{4} & -\frac{1}{8}\\ 0 & 0 & \frac{1}{2} & -\frac{1}{4} \end{bmatrix}\\
B_{2_1} &= \begin{bmatrix}1 & 0 & -48 &  16\\ 0 & 1 & 8 & -2\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{bmatrix}\\
B_{2_2} &= \begin{bmatrix}0 & 1 &   8 &  -2\\ 0 & 0 & 0 &  0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\end{bmatrix}
\end{align}</math>

so the final answer is
<math display="block">e^{tA} = \begin{bmatrix}
    e^t & te^t & \left(8t - 48\right) e^t\! + \left(4t + 48\right)e^{\frac{3}{4}t} & \left(16 - 2\,t\right)e^t\! + \left(-2t - 16\right)e^{\frac{3}{4}t}\\
    0 & e^t & 8e^t\! + \left(-t - 8\right) e^{\frac{3}{4}t} & -2e^t + \frac{t + 4}{2}e^{\frac{3}{4}t}\\
    0 & 0 & \frac{t + 4}{4}e^{\frac{3}{4}t} & -\frac{t}{8}e^{\frac{3}{4}t}\\
    0 & 0 & \frac{t}{2}e^{\frac{3}{4}t} & -\frac{t - 4}{4}e^{\frac{3}{4}t} ~.
  \end{bmatrix}
</math>

The procedure is much shorter than [[Matrix differential equation#Putzer Algorithm for computing eAt|Putzer's algorithm]] sometimes utilized in such cases.

{{See also|Derivative of the exponential map}}