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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!

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    $\begingroup$ Diagonalize over $\mathbb{C}$. $\endgroup$ Oct 22 '20 at 12:34
  • $\begingroup$ Over $\mathbb{C}$, there are always eigenvalues. $\endgroup$
    – Zhanxiong
    Oct 22 '20 at 13:09
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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 $$\begin{bmatrix} \cos\alpha&\sin\alpha\\ -\sin\alpha&\cos\alpha\end{bmatrix}$$

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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 $$r_n^n\to1.$$

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

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  • $\begingroup$ IMO this is the most straightforward answer. $\endgroup$
    – user65203
    Oct 22 '20 at 16:31
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\bbox[5px,#ffd]{\ds{\lim_{n \to \infty}}\,\,\,% \pars{\begin{array}{rr} \ds{1} & \ds{\alpha \over n} \\ \ds{-{\alpha \over n}} & \ds{1} \end{array}}^{n}}\quad$ 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}
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  • $\begingroup$ +1 for the reference to Pauli Matrices. Really nice Felix. $\endgroup$
    – MtGlasser
    Oct 22 '20 at 18:39
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    $\begingroup$ @Aryadeva You're welcome. Thanks. $\endgroup$ Oct 22 '20 at 20:26
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** 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, $$A^n=\begin{bmatrix} \cos a & \sin a\\ -\sin a & \cos a \end{bmatrix}$$

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

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