0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ It's better to use \tag ;-) $\endgroup$ – Rodrigo de Azevedo Oct 20 '19 at 19:45
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.