Computing $e^{tA}$, where $A$ is a matrix

Question

Given $$A = \begin{bmatrix} 0&-1\\1&0\end{bmatrix}$$ find $e^{tA}$, where $t \in \mathbb{R}$.

I have read about calculating the matrix exponential here. I know that when $A$ is written in its diagonal form it's kind of easy. The same with nilpotent matrices. Unfortunately, I don't know how to deal with $t$. Especially when the eigenvalues are complex (in the given example they are).

Answer 1

Your matrix is diagonalizable. Indeed, if you define$$P=\begin{bmatrix}1&1\\i&-i\end{bmatrix}$$(note that the columns of $P$ ere eigenvectors of $A$), then$$P^{-1}.A.P=\begin{bmatrix}-i&0\\0&i\end{bmatrix}$$and therefore$$P^{-1}.(tA).P=\begin{bmatrix}-ti&0\\0&ti\end{bmatrix}.$$So$$P^{-1}.e^{tA}.P=\begin{bmatrix}e^{-ti}&0\\0&e^{ti}\end{bmatrix}$$and so$$e^{tA}=P.\begin{bmatrix}e^{-ti}&0\\0&e^{ti}\end{bmatrix}.P^{-1}=\begin{bmatrix}\cos t&-\sin t\\\sin t&\cos t\end{bmatrix}.$$

Answer 2

If you can diagonalize the matrix, then it's still easy.

If you have a diagonal form $$ P^{-1}AP = \begin{pmatrix}d_1 \\ & \ddots \\ &&d_n \end{pmatrix} $$ then $$P^{-1}(tA)P = t P^{-1}AP = \begin{pmatrix}td_1 \\ & \ddots \\ && td_n \end{pmatrix}$$ and you can exponentiate this as usual and eventually find $$ e^{tA} = P\begin{pmatrix}e^{td_1} \\ & \ddots \\ && e^{td_n} \end{pmatrix} P^{-1} $$ It makes no difference here that the eigenvalues are complex; the imaginary parts of the answer ought to cancel out in the end.

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In your particular case, after you plug in Euler's formulas and let the dust settle, you should end up with $$ e^{tA} = \begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{pmatrix} $$

Answer 3

We have $A^2 = -I$, $A^3 = -A$ and $A^4 = I$ so

\begin{align} e^{tA} &= \sum_{n=0}^\infty \frac{t^nA^n}{n!} \\ &= I\cdot\sum_{k=0}^\infty \frac{t^{4k}}{(4k)!} + A \cdot \sum_{k=0}^\infty \frac{t^{4k+1}}{(4k+1)!} - I\cdot\sum_{k=0}^\infty \frac{t^{4k+2}}{(4k+2)!} - A \cdot \sum_{k=0}^\infty \frac{t^{4k+3}}{(4k+3)!}\\ &= I\cdot\sum_{k=0}^\infty (-1)^k\frac{t^{2k}}{(2k)!} + A \cdot \sum_{k=0}^\infty (-1)^k\frac{t^{2k+1}}{(2k+1)!}\\ &= I \cdot \cos t + A \cdot \sin t\\ &= \pmatrix{\cos t & -\sin t \\ \sin t & \cos t} \end{align}

Answer 4

For this special case, one can start out from the observation that $A$ acts on the standard basis of $\Bbb R^2$ exactly as does multiplying by $i$ on the basis $1,i$ of $\Bbb C$.
In particular, $A^2= -\mathrm{Id}, \ A^3=-A$ and $A^4=\mathrm{Id}$.

Thus, calculating $e^{tA}$ is formally exactly the same as calculating $e^{ti}$, only that we write $\mathrm{Id}$ in place of $1$ and $A$ in place of $i$, yielding $$e^{tA} =\cos t\cdot\mathrm{Id} \ +\sin t\cdot A$$

Answer 5

$$(B(t))''=(e^{At})''=A^2e^{At}=-Ie^{At}=-B(t)$$ implies that $B$ is a solution of the ODE $$B''(t)=-B(t),$$ i.e. is of the form

$$e^{At}=B_c\cos t+B_s\sin t.$$

Now

$$\left.e^{At}\right|_{t=0}=I=B_c$$

and

$$\left.(e^{At})'\right|_{t=0}=A=B_s.$$