Using the Cayley–Hamilton theorem, I got the following matrix exponential (for $3 \times 3$ matrix $A$):

$$e^{At}=\left(e^t - te^t+\dfrac{1}{2}\, t^2e^t\right) E+\left(te^t-t^2e^t\right)A+\dfrac{1}{2}\, e^t A^2 \tag{1}$$

where $e^{At}$ is a matrix exponential, $A$ a coefficient matrix and $E$ a unit matrix. Is it possible to obtain (by simplifying $\texttt{(1)}$):

$$e^{At} = e^t \:e^{(A-E)\,t} = e^t \left[E+\left(A-E\right)t\right]+\left(A-E\right)^2 \,\dfrac{t^2}{2} \tag{2}$$

Given that

$$A= \begin{bmatrix}2&1&1\\1&2&1\\-2&-2&-1\end{bmatrix}$$

everything nicely simplifies from $\texttt{(2)}$ while $\left(A-E\right)^2 =0$. I just can’t simplify the $\texttt{(1)}$ to get $\texttt{(2)}$. If anyone have a time to check this, I’d really appreciate it. Thanks.

  • $\begingroup$ It's better to use \tag ;-) $\endgroup$ – Rodrigo de Azevedo Oct 20 '19 at 19:45
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    $\begingroup$ Double-check the last term of (1) and your bracketing in (2). $\endgroup$ – amd Oct 20 '19 at 20:02
  • $\begingroup$ Thank you! That was the problem! $\endgroup$ – Josh E. Oct 21 '19 at 8:38

As @amd points out, you can see that your (1) is wrong, since, at t =0 , it does not give you an identity, nor for the first and second derivatives w.r.t. t, as it should!

The correct expression, instead, is $$e^{At}=e^t\left(\left (1 - t+\dfrac{t^2}{2} \right) E+\left(t-t^2\right)A+\dfrac{t^2}{2} A^2\right ) \tag{1}$$ amounting to $$ = e^t \left( E+(A-E)t +\left(A-E\right)^2 \dfrac{t^2}{2} \right )\\ =e^t e^{t(A-E)}, \tag{2}$$ alright, by virtue of the C-H relation, $(A-E)^3=0$.

You may check (1) does satisfy these three conditions.

  • $\begingroup$ Thanks and I was looking at the equation for too long :). I wasn't paying attention to the first steps and concentrating too much for the results to match. $\endgroup$ – Josh E. Oct 21 '19 at 8:34

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