# Compute $\lim_{N\to \infty} A_N$ where $\{A_N\}$ is a sequence of matrices [closed]

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?

## closed as off-topic by Xander Henderson, Arnaud D., Cesareo, RRL, kimchi loverOct 16 at 21:04

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• @hardmath you are correct, sir. So: must find invertible $P$ such that $A = \left( P^{-1} D P \right)^N$ with $D$ diagonal, as then we also have $A = P^{-1} D^N P$ explicit. – Will Jagy Oct 16 at 2:40
• Notation is not great. I would have called it $A_n$ instead. – Rodrigo de Azevedo Oct 16 at 9:05
• Notably, we have $A_N = (I + X/N)^N$ where $$X = \frac{\pi}{2} \pmatrix{0&1\\-1&0}.$$ In fact, $e^X = \lim_{N \to \infty}(I + X/N)^N$ is equivalent to the usual definition of a matrix exponential, so $\lim_{N \to \infty}A_N = e^X$. – Omnomnomnom Oct 16 at 9:35

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.

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 $$J$$.

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

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 $$ds^2=dx^2-dy^2$$