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If $a \in \mathbb{R}$, evaluate $$ \lim_{n \to \infty}\left(\begin{matrix} 1&\frac{a}{n}\\\frac{-a}{n}&1\end{matrix}\right)^{n}$$

My attempt: Let $$A = \left(\begin{matrix} 0&a\\-a&0\end{matrix}\right) = -a\left(\begin{matrix} \cos(\frac{\pi}{2})&-\sin(\frac{\pi}{2})\\\sin(\frac{\pi}{2})&\cos(\frac{\pi}{2})\end{matrix}\right)$$ so that $$A^k = (-a)^k \left(\begin{matrix} \cos(\frac{k\pi}{2})&-\sin(\frac{k\pi}{2})\\\sin(\frac{k\pi}{2})&\cos(\frac{k\pi}{2})\end{matrix}\right)$$

Thus, \begin{align}\displaystyle \lim_{n \to \infty}\left(\begin{matrix} 1&\dfrac{a}{n}\\\dfrac{-a}{n}&1\end{matrix}\right)^{n} &=\displaystyle \lim_{n \to \infty} \left(I+\dfrac{A}{n}\right)^n =e^A=\displaystyle \sum_{k=0}^{\infty}\dfrac{A^k}{k!}\\&= \sum_{k=0}^{\infty} \left(\begin{matrix} \dfrac{(-a)^k\cos(\frac{k\pi}{2})}{k!}&-\dfrac{(-a)^k\sin(\frac{k\pi}{2})}{k!}\\\dfrac{(-a)^k\sin(\frac{k\pi}{2})}{k!}&\dfrac{(-a)^k\cos(\frac{k\pi}{2})}{k!}\end{matrix}\right) \end{align}

and since $\displaystyle \sum_{k=0}^{\infty}\dfrac{(-a)^k\cos(\frac{k\pi}{2})}{k!}=1+0-\dfrac{a^2}{2!}+0+\dfrac{a^4}{4!}+\cdots= \cos a$ and

$\displaystyle \sum_{k=0}^{\infty}\dfrac{(-a)^k\sin(\frac{k\pi}{2})}{k!}=0-a+0+\dfrac{a^3}{3!}+0-\dfrac{a^5}{5!}+\cdots= -\sin a$ therefore the required answer is

$\left(\begin{matrix} \cos a&\sin a\\-\sin a&\cos a\end{matrix}\right).$

However the above answer does not match the choices provided which are $I, 0$ and none of the above. So my question is: Is my answer correct?

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    $\begingroup$ your result is correct. $\endgroup$ – julian Nov 18 '17 at 14:55
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Your answer is correct as well. but there is a short way to reach it. Merely by putting ourselves in the complex plan where we identify

$$ 1 \equiv \left(\begin{matrix} 1& 0\\0&1 \end{matrix}\right)~~~\text{and}~~~~ i \equiv \left(\begin{matrix} 0& 1\\-1&0 \end{matrix}\right).$$

Thus, $$\begin{align}\displaystyle \lim_{n \to \infty}\left(\begin{matrix} 1&\dfrac{a}{n}\\\dfrac{-a}{n}&1\end{matrix}\right)^{n} &= \displaystyle \lim_{n \to \infty}\color{blue}{\left(1+\dfrac{ai}{n}\right)^n} \\&=\color{red}{e^{ai} = \cos a+i\sin a} \\&= \left(\begin{matrix} \cos a&\sin a\\-\sin a&\cos a\end{matrix}\right).\end{align}$$

Given that, for any $z\in \Bbb C$ we have, $\lim\limits_{n \to \infty}\left(1+\dfrac{z}{n}\right)^n =e^{z} $ See here also here,

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    $\begingroup$ Thanks for the elegant answer, I should of thought of using the complex plane $\endgroup$ – Aryaman Jal Nov 18 '17 at 15:15
  • $\begingroup$ Yes I had the same problem few year ago where I had to solve some linear ODE which reduce to the compact form $z'= az$ $\endgroup$ – Guy Fsone Nov 18 '17 at 15:23
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    $\begingroup$ What an answer!! An artwork! $\endgroup$ – Shashi Nov 18 '17 at 16:42
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    $\begingroup$ @Shashi you might fine mor application here: math.stackexchange.com/questions/164422/… $\endgroup$ – Guy Fsone Dec 1 '17 at 19:21
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The best approach is to diagonalize your matrix. Both $(1,i)$ and $(1,-i)$ are eigenvectors. So, let$$T=\begin{pmatrix}1&1\\i&-i\end{pmatrix}.$$Then$$T^{-1}.\begin{pmatrix}1&\frac an\\-\frac an&1\end{pmatrix}.T=\begin{pmatrix}1+\frac ani&0\\0&1-\frac 1ni\end{pmatrix}.$$Therefore$$T^{-1}.\begin{pmatrix}1&\frac an\\-\frac an&1\end{pmatrix}^n.T=\begin{pmatrix}1+\frac ani&0\\0&1-\frac 1ni\end{pmatrix}^n=\begin{pmatrix}\left(1+\frac{ai}n\right)^n&0\\0&\left(1-\frac{ai}n\right)^n\end{pmatrix}$$ and so$$\lim_{n\to\infty}T^{-1}.\begin{pmatrix}1&\frac an\\-\frac an&1\end{pmatrix}^n.T=\begin{pmatrix}e^{ai}&0\\0&e^{-ai}\end{pmatrix}.$$So$$\lim_{n\to\infty}\begin{pmatrix}1&\frac an\\-\frac an&1\end{pmatrix}^n=T.\begin{pmatrix}e^{ai}&0\\0&e^{-ai}\end{pmatrix}.T^{-1}=\begin{pmatrix}\cos a&\sin a\\-\sin a&\cos a\end{pmatrix}.$$

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