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.

  • $\begingroup$ You don't need to use the Cayley-Hamilton theorem, you just need some annihilating polynomial to of the matrix to rewrite the power series. Since you are going to need the monomials $I,P,P^2$ anyway, computing the minimal polynomial directly can actually save you the (admittedly not very hard) work of computing the characteristic polynomial. $\endgroup$ May 8, 2021 at 7:13

1 Answer 1


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.)

  • $\begingroup$ 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, 2014 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.) $\endgroup$ Mar 24, 2014 at 15:28
  • $\begingroup$ Nice answer!!$\;$ $\endgroup$
    – DatBoi
    May 7, 2021 at 17:14

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