Matrix exponential: $\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$

Question

It is asked to calculate $e^A$, where $$A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix}$$

I begin evaluating some powers of A:

$A^0= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\; ; A=\begin{pmatrix} 0 & 1 \\ -4 & 0 \end{pmatrix} \; ; A^2 = \begin{pmatrix} -4 & 0 \\ 0 & -4\end{pmatrix} \; ; A^3 = \begin{pmatrix} 0 & -4 \\ 16 & 0\end{pmatrix}\; ; $

$ A^4=\begin{pmatrix} 16 & 0 \\ 0 & 16\end{pmatrix},\; \ldots$

I've noted that, since

$$e^A = \sum_{k=0}^\infty \frac{A^k}{k!}$$

we will have the cosine series at the principal diagonal for $\cos(2)$. But couldnt get what we will have in $(e^A)_{12}$ and $(e^A)_{21}$. Also, we know that if $B=\begin{pmatrix} 0 & \alpha \\ -\alpha & 0 \end{pmatrix}$, then $e^B = \begin{pmatrix} \cos(\alpha) & \sin(\alpha) \\ -\sin(\alpha) & \cos(\alpha) \end{pmatrix} $. Is there a general formula for $$B=\begin{pmatrix} 0& \alpha \\ \beta & 0 \end{pmatrix}$$?

Thanks!

Answer 1

Note that

$$A=P\cdot\begin{bmatrix} 2i&0\\0&-2i\end{bmatrix}\cdot P^{-1}$$

With $P=\begin{bmatrix} -1&-1\\-2i&2i\end{bmatrix}$. We have

$$e^A=P\cdot e^{D}\cdot P^{-1}$$

With $D$ the diagonal matrix above

Answer 2

By separating odd and even terms in the series, what you have found can be rewritten as: $$ e^A=I \sum_{k=0}^\infty{(-1)^k4^k\over(2k)!}+A \sum_{k=0}^\infty{(-1)^k 4^k\over(2k+1)!}=I \cos2+A{1\over2}\sin2. $$

Answer 3

For the general formula I get it to $e^A = I \cosh(\sqrt{\alpha\beta}) + A\sinh(\sqrt{\alpha\beta})/\sqrt{\alpha\beta}$ where the same square root is selected everywhere (the one under $\cosh$ doesn't matter due to symmetry).

You get this by checking the powers of $A$:

• $A^0 = I$
• $A^1 = A$
• $A^2 = (\alpha\beta)I$
• $A^3 = (\alpha\beta)A$

and so on. Expanding it results in

$e^A = I/0! + A/1! + (\alpha\beta)I/2! + (\alpha\beta)A/3! + ...$

suppose $\gamma^2 = \alpha\beta$ we get

$e^A = I\gamma^0/0! + (A/\gamma)\gamma^1/1! + I\gamma^2/2! + (A/\gamma)\gamma^3/3!+...$

then just separating the parts you get

$e^A = I\cosh(\gamma) + (A/\gamma)sinh(\gamma)$

Answer 4

here is another way to find $e^A.$ we will use the interpretation that $x = e^{At}x_0$ is the unique solution of $$\frac{dx}{dt} = Ax, x(0) = x_0.$$

for $$\frac{d}{dt}\pmatrix{x\\y} = \pmatrix{0&1\\-4&0}\pmatrix{x\\y}$$ which in component form is $$\dot x = y, \dot y = -4x \to \ddot x=-4x, \ddot y= -4y \text{ and } x(0) = x_0, y(0) = y_0.$$ the solutions are $$x = x_0\cos 2t+\frac 12y_0 \sin 2t, y = \dot x = -2x_0\sin 2t + y_0\cos 2t$$

writing the last equation in matrix form, we have $$ \pmatrix{x\\y} = \pmatrix{\cos 2t&\frac 12 \sin 2t\\-2\sin 2t&\cos 2t}\pmatrix{x_0\\y_0}.$$ therefore, $$e^{At} = \pmatrix{\cos 2t&\frac 12 \sin 2t\\-2\sin 2t&\cos 2t}.$$