Find $e^{At}$ of the matrix given below.

Question

Find $e^{At}$ of the matrix given below. $$A = \begin{pmatrix} 2 & 1 & -2 \\ -1 & 0 & 0 \\ 1 & 1 & -1 \end{pmatrix}$$

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If this was a diagonal matrix I could take $e$ power of every quantity. Or, else, since $e^{At}$ is the fundamental matrix such that when $t=0$ gives the identity, Is it enough to find the fundamental matrix considering this as a ODE system.

Answer 1

Just plug in:

$$e^{A t} = \sum\limits_{k=0}^\infty \frac{(A t)^k}{k!}$$

$$ = \left( \begin{array}{ccc} \sin (t)+e^t & \sin (t) & \cos (t)-e^t-\sin (t) \\ \cos (t)-e^t & \cos (t) & -\cos (t)+e^t-\sin (t) \\ \sin (t) & \sin (t) & \cos (t)-\sin (t) \\ \end{array} \right)$$

Answer 2

The eigenvalues are $\lambda_{1,2,3} = -i,i, 1$, so this matrix is diagonalizable.

The corresponding eigenvectors are

$$v_1 = \begin{pmatrix} 1\\-i\\1\end{pmatrix}, v_2 = \begin{pmatrix} 1\\i\\1\end{pmatrix}, v_3 = \begin{pmatrix} -1\\1\\0\end{pmatrix}$$

$P$ is formed using the eigenvector columns and we can write the exponential as

$$e^{At} = P e^{Dt}P^{-1} = \begin{pmatrix} 1&1&-1\\ -i & i & 1 \\ 1&1&0 \end{pmatrix}\begin{pmatrix} e^{-it}& 0 & 0\\ 0 &e^{it}&0 \\ 0&0&e^t\end{pmatrix} \begin{pmatrix} \frac{i}{2} & \frac{i}{2} & \frac{1}{2}-\frac{i}{2} \\ -\frac{i}{2} & -\frac{i}{2} & \frac{1}{2}+\frac{i}{2} \\ -1 & 0 & 1 \\ \end{pmatrix}$$

From Euler's Formula, we know $e^{i a t} = \cos at + i \sin at$.

This reduces our result to

$$e^{At} = \begin{pmatrix} e^t+\sin (t) & \sin (t) & -e^t-\sin (t)+\cos (t) \\ \cos (t)-e^t & \cos (t) & e^t-\sin (t)-\cos (t) \\ \sin (t) & \sin (t) & \cos (t)-\sin (t) \\ \end{pmatrix}$$