A simplification regarding matrices

Question

Let $$A = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} \right )$$

What can I say about the matrix $A^k$ for arbitrary natural $k$. Is there some way to express $A^k$ in terms of $A$. I have observed that $A^4 = I$.

EDIT: I want a better form of $A^k$ in order to find a closed form expression in terms of elementary functions for $\exp(A)$

Answer 1

More generally, if $$B(\theta) = \left( \begin{array}{cc} \cos \theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{array} \right )$$

Then $$B(\theta)^n=\left( \begin{array}{cc} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \\ \end{array} \right ) = B(n\theta)$$

Your case is $A=B(\frac\pi 2)$.

In that particular case:

$$A^n = \begin{cases}I & \text{if }n\equiv 0 \pmod 4\\A&\text{if }n\equiv 1\pmod 4\\-I &\text{if }n\equiv 2\pmod 4\\-A&\text{if }n\equiv 3\pmod 4\end{cases}$$

Writing out the terms for $\exp(A)$, we get:

$$\exp(A) = (I-I^2/2!+I^4/4!-...) + A(I/1!-I/3!+I/5!...) = \\ \left( \begin{array}{cc} \cos 1 & 0 \\ 0 & \cos 1 \\ \end{array} \right )+A\left( \begin{array}{cc} \sin 1 & 0 \\ 0 & \sin 1 \\ \end{array} \right )=\\\left( \begin{array}{cc} \cos 1 & -\sin 1 \\ \sin 1 & \cos 1 \\ \end{array} \right )$$

More generally, $$\exp(B(\theta)) =\left( \begin{array}{cc} e^{\cos\theta}\cos(\sin\theta) & -e^{\cos\theta}\sin(\sin\theta) \\ e^{\cos\theta}\sin(\sin\theta) & e^{\cos\theta}\cos(\sin\theta) \\ \end{array} \right )$$

Fundamentally, what is going on here is that the matrices of the form $$\left( \begin{array}{cc} a & -b \\ b & a \end{array} \right )=aI + bA, a,b\in\mathbb R$$ form a sub-algebra of the algebra of matrices. This sub-algebra is completely isomorphic[*] to the complex numbers, with $A$ corresponding to $i$. In particular, then $\exp(A)$ corresponds to $\exp(i)=cos 1 + i\sin 1$ which corresponds to $cos 1 + (\sin 1) A$

[*] I am using this term, "completely isomorphic," loosely, but basically, I mean that not only are the algebras the same, but also they are topologically equivalent - convergence of series in one algebra implies convergence of the corresponding series in the other algebra, for example.

Answer 2

We can diagonalize $A$ by calculating the Jordan Normal Form, using it's eigenvalues, $\lambda_{1, 2} = \pm i$, where $i$ is the imaginary unit for this matrix as:

$$A = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} \right ) = S \cdot J \cdot S^{-1} = \left( \begin{array}{cc} -i & \ i \\ 1 & 1 \\ \end{array} \right ) \left( \begin{array}{cc} -i & \ 0 \\ 0 & i \\ \end{array} \right ) \left( \begin{array}{cc} \frac{i}{2} & \ \frac{1}{2} \\ \frac{-i}{2} & \frac{1}{2} \\ \end{array} \right )$$

We can now calculate the exponential of $A$ using

$$\exp(A)=S \cdot \exp(J) \cdot S^{-1}$$

Regards

Answer 3

Note that matrices of the form $xI+yA$ with real $x$ and $y$ behave exactly like complex numbers $x+yi$ under matrix addition and matrix multiplication, as well as topologically. Since additions, multiplications, and limits is all you need to define the exponential function (with the same definition in both cases), the answer is the same as $e^i$, translated back to a matrix by the correspondence $x+yi\mapsto xI+yA$.

Or in other words, $$\exp(A) = \begin{pmatrix} \cos(1) & -\sin(1) \\ \sin(1) & \cos(1)\end{pmatrix}$$