Solving exponential matrix using Cayley–Hamilton theorem, I got stuck.

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.

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

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, $$(A-E)^3=0$$.