Given matrix, $$A_N = \begin{bmatrix} 1 & \frac{\pi}{2N}\\ \frac{-\pi}{2N} & 1\end{bmatrix}^N$$ compute $$\lim_{N \rightarrow \infty} A_N$$
I took the logarithm of both sides but was not able to figure out the limit. Any suggestions on how to approach this problem?
On
I would use the isomorphism between the complex numbers of the form $z= a + bi$ and the $2 \times 2$ matrices of the form $aI +bJ$ with $I$ the identity matrix and $$J=\begin{bmatrix}0 & 1\\-1 & 0\end{bmatrix}$$ so then this becomes $\lim_{N \rightarrow \infty}(1+\frac{\pi}{2N}i)^N$. To take powers of complex numbers it's easier to express them in polar form which is going to be the same as diagonalizing. It's not hard to see that as $N$ approaches infinity that the radius is $1$ and the angle $\theta = \frac{\pi}{2N}$ and so using that $r$ is arbitrarily close to $1$ we can see that $(e^{i\pi/2N})^N=e^{i\pi/2}=i$ and so we conclude that the limit is the matrix $J$.
On
Take out eigenvalues and eigenvectors
$$A^N = PD^NP^{-1}$$
Where P has eigenvectors and D is a diagonal matrix containing eigenvalues Now take limit for D Then multiply all the matrices
On
Define $$B= \begin{bmatrix} 1 & \frac{\pi}{2N} \\ \frac{-\pi}{2N} & 1\\ \end{bmatrix}$$therefore$$B^2=2B-1+{\pi^2\over 4N^2}$$By assuming $$B^k=a_kB+b_kI$$where $a_0=0,b_0=1,a_1=1,b_1=0$ we obtain $$B^{k+1}=BB^{k}=B(a_kB+b_k)=a_kB^2+b_kB=(2a_k+b_k)B+a_k\left(-1+{\pi^2\over 4N^2}\right)$$ from which we obtain $$a_{k+1}=2a_k+b_k\\b_{k+1}=a_k\left(-1+{\pi^2\over 4N^2}\right)$$ which leads to $$a_{k+2}=2a_{k+1}+a_k\left(-1+{\pi^2\over 4N^2}\right)$$ and by doing some calculations we immediately obtain $$a_k=C\left(1+{\pi\over 2N}\right)^k+D\left(1-{\pi\over 2N}\right)^k\\b_k=E\left(1+{\pi\over 2N}\right)^k+F\left(1-{\pi\over 2N}\right)^k$$for some constants $C,D,E,F$.
By applying the initial condition we finally have$$C=-D={N\over \pi} \\E={\pi-2N\over 2\pi}\\F={\pi+2N\over 2\pi}$$by substituting $k=N$ we obtain $$A{=(C\cdot B+E\cdot I)\left(1+{\pi\over 2N}\right)^N+(D\cdot B+F\cdot I)\left(1-{\pi\over 2N}\right)^N\\=\begin{bmatrix}{1\over 2}&{1\over 2}\\{1\over 2}&{1\over 2}\end{bmatrix}\left(1+{\pi\over 2N}\right)^N+\begin{bmatrix}{1\over 2}&-{1\over 2}\\-{1\over 2}&{1\over 2}\end{bmatrix}\left(1-{\pi\over 2N}\right)^N}$$hence $$\lim_{N\to \infty}A=\begin{bmatrix}\cosh {\pi\over 2}&\sinh {\pi\over 2}\\\sinh {\pi\over 2}&\cosh {\pi\over 2}\end{bmatrix}$$ Additional Remark An interesting property is that $$\det\left(\lim_{N\to \infty}A\right)=1$$ and $\lim_{N\to \infty}A$ show a rotation in a Lorentzian space with a metric $$ds^2=dx^2-dy^2$$
The characteristic polynomial of $A$ is given by: $$ \det \begin{bmatrix} x-1 & \frac{-\pi}{2N} \\ \frac{\pi}{2N} & x-1\end{bmatrix} =(x-1)^2+\frac{\pi^2}{4N^2}$$ To find the eigenvalues of $A$, set the characteristic polynomial equal to zero, and solve for $x$. This gives us the complex conjugate eigenvalues $x=\frac{i\pi}{2N}+1$ and $x=\frac{-i\pi}{2N}+1$.
Now, $2$ linearly independent eigenvectors of $A$ would then be: $x=\begin{bmatrix} i \\ -1 \end{bmatrix}$ and $x=\begin{bmatrix} i \\ 1 \end{bmatrix}$, corresponding to the eigenvalues $x=\frac{i\pi}{2N}+1$ and $x=\frac{-i\pi}{2N}+1$ respectively ( This part should be relatively easy to obtain ).
Hence, $A$ is diagonalisable, and we have $A=PDP^{-1}$, where the diagonal matrix $D$ is such that $D = \begin{bmatrix} 1+\frac{i\pi}{2N} & 0 \\ 0 & 1-\frac{i\pi}{2N} \end{bmatrix} $ and $P= \begin{bmatrix} i & i \\ -1 & 1 \end{bmatrix}$. In addition, we may easily evaluate $P^{-1}$ to be: $P^{-1}= \frac{1}{2i} \begin{bmatrix} 1 & -i \\ 1 & i \end{bmatrix}$.
Since $A^{N}$=$(PDP^{-1})^N$=$PD^NP^{-1}$, and $D^N \rightarrow \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}$ as $N \rightarrow \infty$ , we have that $A^N=P\begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix}P^{-1}$ as $N \rightarrow \infty$ . Furthermore, this is simply equal to: $\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$, giving us the desired limit of the matrix.