Find the limit a matrix raised to $n$ when $n\to\infty$

Find $$\lim_{n \to \infty} \begin{pmatrix} 1 & \frac{\alpha}{n} \\ - \frac{\alpha}{n} & 1 \end{pmatrix}^n, \quad \text{where} ~ \alpha \in \mathbb{R}$$

Well, at first I tried using diagonalization, but there is no eigenvalue, since the equation $$(1- \lambda)^2 + \frac{\alpha^2}{n^2} = 0$$ has no solutions.

Then I had an idea to decompose the matrix into elementary matrices, but I do not know what to do next Also, I am not quite sure how to find the limit so any hints would be greatly appreciated. Thanks!

• Diagonalize over $\mathbb{C}$. Oct 22 '20 at 12:34
• Over $\mathbb{C}$, there are always eigenvalues. Oct 22 '20 at 13:09

As mentioned in comments, there is a diagonalisation of the given matrix over the complexes into $$PDP^{-1}$$ where $$P=\begin{bmatrix}i&-i\\1&1\end{bmatrix}$$ $$D=\begin{bmatrix}1-\frac\alpha ni&0\\0&1+\frac\alpha ni\end{bmatrix}$$ In the limit, $$D^n$$ tends to $$D=\begin{bmatrix}e^{-i\alpha}&0\\0&e^{i\alpha}\end{bmatrix}$$ and multiplying back and simplifying produces the final result as $$\begin{bmatrix} \cos\alpha&\sin\alpha\\ -\sin\alpha&\cos\alpha\end{bmatrix}$$

The two column vectors are orthogonal and you can rewrite the matrix as that of a similarity transform.

$$\begin{pmatrix} 1 & \dfrac{\alpha}{n} \\ - \dfrac{\alpha}{n} & 1 \end{pmatrix}=r_n\begin{pmatrix}\cos\theta_n&\sin\theta_n\\-\sin\theta_n&\cos\theta_n\end{pmatrix}$$ where $$r_n=\sqrt{1+\dfrac{\alpha^2}{n^2}}$$ and $$\tan\theta_n=\frac\alpha n.$$

After $$n$$ iterations,

$$r_n^n\begin{pmatrix}\cos n\theta_n&\sin n\theta_n\\-\sin n\theta_n&\cos n\theta_n\end{pmatrix}\to\color{green}{\begin{pmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{pmatrix}}$$ because $$n\theta_n\to\alpha$$ and $$r_n^n\to1.$$

By the isomorphism of matrices of type $$\begin{pmatrix}a&-b\\b&a\end{pmatrix}$$ and complex numbers $$a+ib$$, then $$\begin{pmatrix} 1 & \frac{\alpha}{n} \\ - \frac{\alpha}{n} & 1 \end{pmatrix}^n\leftrightarrow (1-i\alpha/n)^n\to e^{-i\alpha}\leftrightarrow\begin{pmatrix} \cos\alpha & \sin\alpha\\ -\sin\alpha & \cos\alpha \end{pmatrix}$$

• IMO this is the most straightforward answer.
– user65203
Oct 22 '20 at 16:31

$$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ $$\bbox[5px,#ffd]{\ds{\lim_{n \to \infty}}\,\,\,% \pars{\begin{array}{rr} \ds{1} & \ds{\alpha \over n} \\ \ds{-{\alpha \over n}} & \ds{1} \end{array}}^{n}}\quad$$ where $$\ds{\,\,\alpha\ \in\ \mathbb{R}}$$.

Note that $$\ds{\pars{\begin{array}{rr} \ds{1} & \ds{\alpha \over n} \\ \ds{-{\alpha \over n}} & \ds{1} \end{array}} = {\bf 1} + \ic{\alpha \over n}\sigma_{y}}$$ where $$\ds{{\bf 1}}$$ is the identity matrix and $$\ds{\sigma_{y} \equiv \pars{\begin{array}{rr} \ds{0} & \ds{-\ic} \\ \ds{\ic} & \ds{0} \end{array}}}$$ is a Pauli Matrix which satisfies $$\ds{\sigma_{y}^{2} = {\bf 1}}$$.

Then, \begin{align} &\bbox[5px,#ffd]{\ds{\lim_{n \to \infty}}\,\,\,% \pars{\begin{array}{rr} \ds{1} & \ds{\alpha \over n} \\ \ds{-{\alpha \over n}} & \ds{1} \end{array}}^{n}} = \lim_{n \to \infty}\pars{% {\bf 1} + \ic{\alpha \over n}\sigma_{y}}^{n} = \expo{\ic\alpha\sigma_{y}} \end{align} Note that, as a function of $$\ds{\alpha}$$, $$\ds{\expo{\ic\alpha\sigma_{y}}}$$ satisfies $$\pars{\totald[2]{}{\alpha} + \alpha^{2}} \expo{\ic\alpha\sigma_{y}} = 0,\,\,\, \left\{\begin{array}{rcl} \ds{\left.\expo{\ic\alpha\sigma_{y}} \,\right\vert_{\alpha\ =\ 0}} & \ds{=} & \ds{\bf 1} \\[2mm] \ds{\left.\partiald{\expo{\ic\alpha\sigma_{y}}}{\alpha} \,\right\vert_{\alpha\ =\ 0}} & \ds{=} & \ds{\ic\sigma_{y}} \end{array}\right.$$ such that \begin{align} &\expo{\ic\alpha\sigma_{y}} = \cos\pars{\alpha}{\bf 1} + \sin\pars{\alpha}\ic\sigma_{y} \\[5mm] = &\ \pars{\begin{array}{cc} \ds{\phantom{-}\cos\pars{\alpha}} & \ds{\sin\pars{\alpha}} \\ \ds{-\sin\pars{\alpha}} & \ds{\cos\pars{\alpha}} \end{array}} \\[5mm] &\ \mbox{Finally,} \\[2mm] &\ \bbox[5px,#ffd]{\ds{\lim_{n \to \infty}}\,\,\,% \pars{\begin{array}{rr} \ds{1} & \ds{\alpha \over n} \\ \ds{-{\alpha \over n}} & \ds{1} \end{array}}^{n}} = \pars{\begin{array}{cc} \ds{\phantom{-}\cos\pars{\alpha}} & \ds{\sin\pars{\alpha}} \\ \ds{-\sin\pars{\alpha}} & \ds{\cos\pars{\alpha}} \end{array}} \end{align}
• +1 for the reference to Pauli Matrices. Really nice Felix. Oct 22 '20 at 18:39
• @Aryadeva You're welcome. Thanks. Oct 22 '20 at 20:26

** Corrected and edited:** $$A=I+\frac{a}{n}B, B=\begin{bmatrix} 0 & 1 \\ -1 & 0\end{bmatrix} B^2=-I, B^4=I$$ $$B$$ is a periodic matrix, take $$A^n=(I+\frac{a}{n}B)^n$$, $$\implies A^n=I+{n \choose 1}\frac{a}{n}B+{n \choose 2}\frac{a^2}{n^2}B^2+{n \choose 3} \frac{a^3}{n^3} B^3+...+{n \choose n}\frac{a^b}{n^n}B^n$$ $$\implies A^n=I\sum_{k=0}^{[n/2]} (-1)^k {n \choose 2k} \left(\frac{a}{n}\right)^{2k}+B\sum_{k=0}^{[n/2]} (-1)^k{n \choose 2k+1} \left(\frac{a}{n}\right)^{2k+1}$$ $$\implies A^n=I\left(\frac{(1+ia/n)^n+(1-ia/n)^n}{2}\right)+B\left(\frac{(1+i a/n)^n-(1-i/n)^n}{2i}\right)$$ Taking limit $$n \rightarrow \infty$$, ewe get $$\lim_{n \to \infty} A^n=I (e^{ia}+e^{-ia})/2+B(e^{ia}-e^{-ia})/(2i)= I\cos a+ B \sin a$$ Finally, $$A^n=\begin{bmatrix} \cos a & \sin a\\ -\sin a & \cos a \end{bmatrix}$$

• Oh ! yes thanks I corrected it now you may see the edited version. Oct 22 '20 at 14:23
• You too are right, I have corrected it now. Oct 22 '20 at 14:30
• Thanks for revisit. Otherwise people do not come back to see correction. Oct 22 '20 at 14:32
• Today I have really gone crazy, you are right. I have corrected it. Oct 22 '20 at 14:45