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!
 A: 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
$$\begin{bmatrix}
\cos\alpha&\sin\alpha\\
-\sin\alpha&\cos\alpha\end{bmatrix}$$
A: 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
$$r_n^n\to1.$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\bbox[5px,#ffd]{\ds{\lim_{n \to \infty}}\,\,\,% 
\pars{\begin{array}{rr}
\ds{1} & \ds{\alpha \over n}
\\
\ds{-{\alpha \over n}} & \ds{1}
    \end{array}}^{n}}\quad$ 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}
A: By the isomorphism of matrices of type $\begin{pmatrix}a&-b\\b&a\end{pmatrix}$ and complex numbers $a+ib$, then $$\begin{pmatrix}    1 & \frac{\alpha}{n} \\
    - \frac{\alpha}{n} & 1
    \end{pmatrix}^n\leftrightarrow (1-i\alpha/n)^n\to e^{-i\alpha}\leftrightarrow\begin{pmatrix}    \cos\alpha & \sin\alpha\\
    -\sin\alpha & \cos\alpha
    \end{pmatrix} $$
A: ** 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, $$A^n=\begin{bmatrix} \cos a & \sin a\\ -\sin a & \cos a \end{bmatrix}$$
