For the matrix $$P=\left( \begin{matrix} 0&1&0 \\ 0&0&1 \\ 0&1&0 \end{matrix} \right)$$ how do you find $e^{Pt}$ using the Cayley-Hamilton theorem?

I have found it by diagonalising $P$, but the question states to use the Cayley-Hamilton theorem.


If you calculate the characteristic polynomial $\chi_A(x)=x^3-x$, you know from Cayley-Hamilton theorem that $P^3-P=0$, i.e. $P^3=P$.

This implies that $P=P^3=P^5=\dots$ and $P^2=P^4=P^6=\dots$. Hence $$ \begin{align} e^{Pt} &= I+Pt + \frac{P^2t^2}2 + \frac{P^3t^3}{3!} + \dots = \\ &= I + P\left(t+\frac{t^3}{3!}+\frac{t^5}{5!}+\dots\right) + P^2\left(\frac{t^2}2+\frac{t^4}{4!}+\frac{t^6}{6!}+\dots\right) = \\ &= I+ P \frac{e^t-e^{-t}}2 + P^2 \left(\frac{e^t+e^{-t}}2-1\right) = \\ &= I+ P\sinh t + P^2(\cosh t-1) \end{align} $$

Using $P=\begin{pmatrix}0&1&0\\0&0&1\\0&1&0\end{pmatrix}$ and $P^2=\begin{pmatrix}0&0&1\\0&1&0\\0&0&1\end{pmatrix}$ I get $$e^{Pt}= \begin{pmatrix} 1&\sinh t&\cosh t-1\\ 0&\cosh t&\sinh t\\ 0&\sinh t&\cosh t \end{pmatrix}.$$

If I did not miss something, this seems to be the same result as WolframAlpha returns. (Up to some algebraic manipulation.)

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  • $\begingroup$ Thank you for your help! how did you get from the line P(t+t^3/3!+t^5/5!+.... to the following line? $\endgroup$ – user0692 Mar 24 '14 at 15:23
  • $\begingroup$ It is more or less standard trick: In one of the expressions I want to get rid of all terms in Taylor series of $e^x$ with odd exponent, in another one I want to cancel out the terms with even exponents. (Or you may simply notice that they are Taylor series of hyperbolic functions, if you remember those. They are very similar to Taylor series for $\sin x$ and $\cos x$, but there are different signs.) $\endgroup$ – Martin Sleziak Mar 24 '14 at 15:28

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