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I'm trying to find the matrix $A$ for which $$e^{tA}=\begin{pmatrix} \frac{1}{2}(e^t+e^{-t}) & 0 & \frac{1}{2}(e^t-e^{-t}) \\ 0 & e^t & 0 \\ \frac{1}{2}(e^t-e^{-t}) & 0 & \frac{1}{2}(e^t+e^{-t}) \end{pmatrix}$$

I know that $e^{tA}=\Psi(t)\cdot[\Psi(0)]^{-1}$, so $e^{tA}\cdot\Psi(0)=\Psi(t)$.

Where $\Psi(t)=(\eta^{(1)}e^{\lambda_1x},\eta^{(2)}e^{\lambda_2x},\eta^{(3)}e^{\lambda_3x})$, with $\lambda_i$ the $i$-th eigenvalue with corresponding eigenvector $\eta^{(i)}$.

However, this didn't really get me anywhere. Does anyone know how to do this?

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As you know $$e^{tA} = I + tA + (t^2/2)A^2 + ....$$

Thus if you differentiate $$e^{tA}=\begin{pmatrix} \frac{1}{2}(e^t+e^{-t}) & 0 & \frac{1}{2}(e^t-e^{-t}) \\ 0 & e^t & 0 \\ \frac{1}{2}(e^t-e^{-t}) & 0 & \frac{1}{2}(e^t+e^{-t}) \end{pmatrix}$$ and evaluate the result at $t=0$, you will get your matrix $A$

I found $$A= \begin{pmatrix} 0&0&1\\0&1&0\\1&0&0\end{pmatrix}$$

Notice that $$(e^{tA})' = Ae^{tA}$$ and $$ e^{tA} =I $$ at $t=0.$

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  • $\begingroup$ If that matrix is really $e^{tA}$ for some $A$. $\endgroup$ – user99914 Apr 11 '18 at 23:22
  • $\begingroup$ We can check it by taking the derivative and see if $(e^{tA})' =Ae^{tA}$ $\endgroup$ – Mohammad Riazi-Kermani Apr 11 '18 at 23:29
  • $\begingroup$ Yes, and probably you should mention it in the answer. $\endgroup$ – user99914 Apr 11 '18 at 23:30
  • $\begingroup$ @JohnMa Thanks, I edited my solution per your comment. $\endgroup$ – Mohammad Riazi-Kermani Apr 11 '18 at 23:36
  • $\begingroup$ I think it would improve the answer even further to actually say why you're checking the derivative (i.e. to guard against the possibility that there is no such matrix $A$). Otherwise it looks like you're just repeating the procedure from the first part of your answer. $\endgroup$ – David Z Apr 12 '18 at 4:41
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$$\frac 12 \begin{bmatrix} e^t+e^{-t} & 0 & e^t-e^{-t}\\ 0 & 2 e^t & 0 \\ e^t-e^{-t} & 0 & e^t+e^{-t}\end{bmatrix} = e^t \left( \frac 12\begin{bmatrix} 1 & 0 & 1\\ 0 & 0 & 0 \\ 1 & 0 & 1\end{bmatrix} + \begin{bmatrix} 0 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix} \right) + e^{-t} \left( \frac{1}{2} \begin{bmatrix} 1 & 0 & -1\\ 0 & 0 & 0 \\ -1 & 0 & 1\end{bmatrix} \right)$$

where each of the $3$ matrices in the right-hand side is a rank-$1$, trace-$1$ symmetric matrix. Hence, from the spectral decomposition, we obtain

$$\rm A = \frac 12\begin{bmatrix} 1 & 0 & 1\\ 0 & 0 & 0 \\ 1 & 0 & 1\end{bmatrix} + \begin{bmatrix} 0 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix} - \frac 12 \begin{bmatrix} 1 & 0 & -1\\ 0 & 0 & 0 \\ -1 & 0 & 1\end{bmatrix} = \color{blue}{\begin{bmatrix} 0 & 0 & 1\\ 0 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}}$$

which is the $3 \times 3$ reversal matrix.

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