Finding the matrix for a matrix exponential

Question

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?

Answer 1

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 the matrix $A$ satisfies $$(e^{tA})' = Ae^{tA}$$ and $$ e^{tA} =I $$ at $t=0.$

Answer 2

$$\frac 12 \begin{bmatrix} e^t+e^{-t} & 0 & e^t-e^{-t}\\ 0 & 2 e^t & 0 \\ e^t-e^{-t} & 0 & e^t+e^{-t}\end{bmatrix} = e^t \left( \frac 12\begin{bmatrix} 1 & 0 & 1\\ 0 & 0 & 0 \\ 1 & 0 & 1\end{bmatrix} + \begin{bmatrix} 0 & 0 & 0\\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix} \right) + e^{-t} \left( \frac{1}{2} \begin{bmatrix} 1 & 0 & -1\\ 0 & 0 & 0 \\ -1 & 0 & 1\end{bmatrix} \right)$$

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.