# Nice matrix limit))

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• May 1st 2010, 02:23 AM
DeMath
Nice matrix limit))
Find $\displaystyle \lim\limits_{n\to\infty}\!\left(\!\begin{array}{*{ 20}c}1&\dfrac{x}{n}\\-\dfrac{x}{n}&1\\\end{array}\!\right)^n$

P.S. I've got not very elegant solution http://s50.radikal.ru/i130/1005/60/5ac5fdcf1585.gif
• May 1st 2010, 03:33 AM
tonio
Quote:

Originally Posted by DeMath
Find $\displaystyle \lim\limits_{n\to\infty}\!\left(\!\begin{array}{*{ 20}c}1&\dfrac{x}{n}\\-\dfrac{x}{n}&1\\\end{array}\!\right)^n$

P.S. I've got not very elegant solution http://s50.radikal.ru/i130/1005/60/5ac5fdcf1585.gif

Let's see what can be done. I assume $\displaystyle x\in\mathbb{R}\,,\,\,A=\begin{pmatrix}1&\frac{x}{n }\\\!\!\!-\frac{x}{n}&1\end{pmatrix}$:

Spoiler:
The char. polynomial of this matrix is $\displaystyle t^2-2t+1+\frac{x^2}{n^2}$ . Assuming $\displaystyle x\neq 0$ (if this is the case the solution is trivial), we get that this quadratic's

roots are $\displaystyle 1\pm \frac{xi}{n}$ , so there's an invertible matrix $\displaystyle P\,\,\,s.t.\,\,\,P^{-1}AP=D=\begin{pmatrix}1+\frac{xi}{n}&0\\0&1-\frac{xi}{n}\end{pmatrix}$ $\displaystyle \Longrightarrow A=PDP^{-1}\Longrightarrow A^n=PD^nP^{-1}$

After getting the eigenvectors we get that $\displaystyle P=\begin{pmatrix}1&\!\!\!-1\\i&i\end{pmatrix}$ , so

$\displaystyle A^n=\begin{pmatrix}1&\!\!\!-1\\i&i\end{pmatrix}\begin{pmatrix}1+\frac{xi}{n}&0 \\0&1-\frac{xi}{n}\end{pmatrix}^n\begin{pmatrix}1/2&\!\!\!-i/2\\\!\!\!-1/2&\!\!\!-i/2\end{pmatrix}$ $\displaystyle =\begin{pmatrix}1&\!\!\!-1\\i&i\end{pmatrix}\begin{pmatrix}\left(1+\frac{xi }{n}\right)^n&0\\0&\left(1-\frac{xi}{n}\right)^n\end{pmatrix}\begin{pmatrix}1/2&\!\!\!-i/2\\\!\!\!-1/2&\!\!\!-i/2\end{pmatrix}$ $\displaystyle \xrightarrow [n\to\infty]{}\begin{pmatrix}1&\!\!\!-1\\i&i\end{pmatrix}\begin{pmatrix}e^{xi}&0\\0&e^{-xi}\end{pmatrix}\begin{pmatrix}1/2&\!\!\!-i/2\\\!\!\!-1/2&\!\!\!-i/2\end{pmatrix}=$

$\displaystyle =\begin{pmatrix}\cos x&\sin x\\\!\!\!-\sin x&\!\!\!-\cos x\end{pmatrix}$

Tonio
• May 1st 2010, 04:11 AM
DeMath
Quote:

Originally Posted by tonio
Let's see what can be done. I assume $\displaystyle x\in\mathbb{R}\,,\,\,A=\begin{pmatrix}1&\frac{x}{n }\\\!\!\!-\frac{x}{n}&1\end{pmatrix}$:

The char. polynomial of this matrix is $\displaystyle t^2-2t+1+\frac{x^2}{n^2}$ . Assuming $\displaystyle x\neq 0$ (if this is the case the solution is trivial), we get that this quadratic's

roots are $\displaystyle 1\pm \frac{xi}{n}$ , so there's an invertible matrix $\displaystyle P\,\,\,s.t.\,\,\,P^{-1}AP=D=\begin{pmatrix}1+\frac{xi}{n}&0\\0&1-\frac{xi}{n}\end{pmatrix}$ $\displaystyle \Longrightarrow A=PDP^{-1}\Longrightarrow A^n=PD^nP^{-1}$

After getting the eigenvectors we get that $\displaystyle P=\begin{pmatrix}1&\!\!\!-1\\i&i\end{pmatrix}$ , so $\displaystyle A^n=\,.\,.\,.\,=\begin{pmatrix}\cos x&\sin x\\\!\!\!-\sin x&\!\!\!-\cos x\end{pmatrix}$

Tonio

It seems, the answer has a little typo.

$\displaystyle A^n=\,.\,.\,.\,=\begin{pmatrix}\cos x&\sin x\\\!\!\!-\sin x&\!\!\!\cos x\end{pmatrix}$
• May 1st 2010, 05:29 AM
tonio
Quote:

Originally Posted by DeMath
It seems, the answer has a little typo.

$\displaystyle A^n=\,.\,.\,.\,=\begin{pmatrix}\cos x&\sin x\\\!\!\!-\sin x&\!\!\!\cos x\end{pmatrix}$

Of course: the 22 entry in the matrix must be $\displaystyle ie^{xi}\left(-\frac{i}{2}\right)+ie^{-xi}\left(-\frac{i}{2}\right)=-\frac{i^2}{2}(e^{xi}+e^{-xi})=\frac{1}{2}(e^{xi}+e^{-xi})=\cos x$ .

Thanx
Tonio
• May 1st 2010, 05:38 AM
NonCommAlg
there's also an elementary solution which doesn't use any linear algebra concept. we clearly have these identities:

i) $\displaystyle 1 - \binom{n}{2}z^2 + \binom{n}{4}z^4 - \binom{n}{6}z^6 + \cdots = \frac{(1+iz)^n +(1-iz)^n}{2},$

ii) $\displaystyle \binom{n}{1}z - \binom{n}{3}z^3 + \binom{n}{5}z^5 - \cdots = \frac{(1+iz)^n-(1-iz)^n}{2i}.$

now let $\displaystyle I$ be the $\displaystyle 2\times2$ identity matrix and $\displaystyle J=\begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}.$ let $\displaystyle \frac{x}{n}=z_n.$ then $\displaystyle A_n=\begin{pmatrix}1 & z_n \\ -z_n & 1 \end{pmatrix}=I+z_nJ.$ note that $\displaystyle J^2=-I$ and hence, using binomial theorem, we have:

$\displaystyle A_n^n = (I+z_nJ)^n = \left[ 1 - \binom{n}{2}z_n^2 + \binom{n}{4}z_n^4 - \cdots \right]I \ + \ \left[\binom{n}{1}z_n - \binom{n}{3}z_n^3 + \cdots \right]J.$ thus by i) and ii) we have: $\displaystyle A_n^n = \frac{(1+iz_n)^n +(1-iz_n)^n}{2}I + \frac{(1+iz_n)^n-(1-iz_n)^n}{2i}J.$

now $\displaystyle \lim_{n\to\infty} (1+iz_n)^n=\lim_{n\to\infty} \left(1+ \frac{ix}{n} \right)^n=e^{ix}$ and $\displaystyle \lim_{n\to\infty} (1-iz_n)^n=\lim_{n\to\infty} \left(1- \frac{ix}{n} \right)^n=e^{-ix}.$ therefore $\displaystyle \lim_{n\to\infty} A_n^n = \frac{e^{ix}+e^{-ix}}{2}I + \frac{e^{ix}-e^{-ix}}{2i}J=(\cos x) I + (\sin x) J=\begin{pmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{pmatrix}.$
• May 1st 2010, 08:52 AM
DeMath
Greatly, NonCommAlg (Rock)

It's really an elementary solution!
• May 1st 2010, 08:04 PM
simplependulum
I have a quite stupid method :

We know :

$\displaystyle \begin{pmatrix} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \end{pmatrix}^n = \begin{pmatrix} \cos(n\theta) & \sin(n\theta) \\ -\sin(n\theta) & \cos(n\theta) \end{pmatrix}$

The matrix $\displaystyle \!\left(\!\begin{array}{*{20}c}1&\dfrac{x}{n}\\-\dfrac{x}{n}&1\\\end{array}\!\right)^n$

I first let $\displaystyle \frac{x}{n} = \tan{\theta}$

I obtain $\displaystyle \!\left(\!\begin{array}{*{20}c}1&\tan{\theta}\\-\tan{\theta}&1\\\end{array}\!\right)^n$

$\displaystyle = \frac{1}{\cos^n{\theta}} \!\left(\!\begin{array}{*{20}c}\cos{\theta}&\sin{\ theta}\\-\sin{\theta}&\cos{\theta}\\\end{array}\!\right)^n$

$\displaystyle = \frac{1}{\cos^n{\theta}} \!\left(\!\begin{array}{*{20}c}\cos(n\theta)&\sin( n\theta)\\-\sin(n\theta)&\cos(n\theta)\\\end{array}\!\right)$

As $\displaystyle n \to \infty$

$\displaystyle \lim_{n\to\infty} \cos^n{\theta} = \frac{1}{\sqrt{1 + \frac{x^2}{n^2} }^n} \to e^{-\frac{x^2}{2n} } = 1$

While $\displaystyle \lim_{n\to\infty} n \theta = \lim_{n\to\infty} n \tan^{-1}(\frac{x}{n}) = x$

Therefore , the limit of your matrix is

$\displaystyle 1 \!\left(\!\begin{array}{*{20}c}\cos{x}&\sin{x}\\-\sin{x}&\cos{x}\\\end{array}\!\right)$

$\displaystyle = \!\left(\!\begin{array}{*{20}c}\cos{x}&\sin{x}\\-\sin{x}&\cos{x}\\\end{array}\!\right)$
• May 1st 2010, 08:10 PM
simplependulum
Quote:

Originally Posted by NonCommAlg
there's also an elementary solution which doesn't use any linear algebra concept. we clearly have these identities:

i) $\displaystyle 1 - \binom{n}{2}z^2 + \binom{n}{4}z^4 - \binom{n}{6}z^6 + \cdots = \frac{(1+iz)^n +(1-iz)^n}{2},$

ii) $\displaystyle \binom{n}{1}z - \binom{n}{3}z^3 + \binom{n}{5}z^5 - \cdots = \frac{(1+iz)^n-(1-iz)^n}{2i}.$

now let $\displaystyle I$ be the $\displaystyle 2\times2$ identity matrix and $\displaystyle J=\begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}.$ let $\displaystyle \frac{x}{n}=z_n.$ then $\displaystyle A_n=\begin{pmatrix}1 & z_n \\ -z_n & 1 \end{pmatrix}=I+z_nJ.$ note that $\displaystyle J^2=-I$ and hence, using binomial theorem, we have:

$\displaystyle A_n^n = (I+z_nJ)^n = \left[ 1 - \binom{n}{2}z_n^2 + \binom{n}{4}z_n^4 - \cdots \right]I \ + \ \left[\binom{n}{1}z_n - \binom{n}{3}z_n^3 + \cdots \right]J.$ thus by i) and ii) we have: $\displaystyle A_n^n = \frac{(1+iz_n)^n +(1-iz_n)^n}{2}I + \frac{(1+iz_n)^n-(1-iz_n)^n}{2i}J.$

now $\displaystyle \lim_{n\to\infty} (1+iz_n)^n=\lim_{n\to\infty} \left(1+ \frac{ix}{n} \right)^n=e^{ix}$ and $\displaystyle \lim_{n\to\infty} (1-iz_n)^n=\lim_{n\to\infty} \left(1- \frac{ix}{n} \right)^n=e^{-ix}.$ therefore $\displaystyle \lim_{n\to\infty} A_n^n = \frac{e^{ix}+e^{-ix}}{2}I + \frac{e^{ix}-e^{-ix}}{2i}J=(\cos x) I + (\sin x) J=\begin{pmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{pmatrix}.$

It is really a great idea and an applicable method in solving other matrix problems (Clapping) , but i think we must put a matrix $\displaystyle A = I + kJ$ where $\displaystyle I$ is the identity matrix , because the product of any other is not commutative : $\displaystyle AB \neq BA$ unless one of them is $\displaystyle I$ ...
• May 1st 2010, 08:11 PM
tonio
Quote:

Originally Posted by simplependulum
I have a quite stupid method :

We know :

$\displaystyle \begin{pmatrix} \cos{\theta} & \sin{\theta} \\ -\sin{\theta} & \cos{\theta} \end{pmatrix}^n = \begin{pmatrix} \cos(n\theta) & \sin(n\theta) \\ -\sin(n\theta) & \cos(n\theta) \end{pmatrix}$

The matrix $\displaystyle \!\left(\!\begin{array}{*{20}c}1&\dfrac{x}{n}\\-\dfrac{x}{n}&1\\\end{array}\!\right)^n$

I first let $\displaystyle \frac{x}{n} = \tan{\theta}$

I obtain $\displaystyle \!\left(\!\begin{array}{*{20}c}1&\tan{\theta}\\-\tan{\theta}&1\\\end{array}\!\right)^n$

$\displaystyle = \frac{1}{\cos^n{\theta}} \!\left(\!\begin{array}{*{20}c}\cos{\theta}&\sin{\ theta}\\-\sin{\theta}&\cos{\theta}\\\end{array}\!\right)^n$

$\displaystyle = \frac{1}{\cos^n{\theta}} \!\left(\!\begin{array}{*{20}c}\cos(n\theta)&\sin( n\theta)\\-\sin(n\theta)&\cos(n\theta)\\\end{array}\!\right)$

As $\displaystyle n \to \infty$

$\displaystyle \lim_{n\to\infty} \cos^n{\theta} = \frac{1}{\sqrt{1 + \frac{x^2}{n^2} }^n} \to e^{-\frac{x^2}{2n} } = 1$

While $\displaystyle \lim_{n\to\infty} n \theta = \lim_{n\to\infty} n \tan^{-1}(\frac{x}{n}) = x$

Therefore , the limit of your matrix is

$\displaystyle 1 \!\left(\!\begin{array}{*{20}c}\cos{x}&\sin{x}\\-\sin{x}&\cos{x}\\\end{array}\!\right)$

$\displaystyle = \!\left(\!\begin{array}{*{20}c}\cos{x}&\sin{x}\\-\sin{x}&\cos{x}\\\end{array}\!\right)$

This is very cute, but...would it be more or less plausible to arrive at this if we didn't know what the answer is beforehand?

Tonio
• May 1st 2010, 08:17 PM
NonCommAlg
Quote:

Originally Posted by simplependulum
It is really a great idea and an applicable method in solving other matrix problems (Clapping) , but i think we must put a matrix $\displaystyle A = I + kJ$ where $\displaystyle I$ is the identity matrix , because the product of any other is not commutative : $\displaystyle AB \neq BA$ unless one of them is $\displaystyle I$ ...

as long as the matrices $\displaystyle X$ and $\displaystyle Y$ commute with each other, you can always use binomial theorem to expand $\displaystyle (X+Y)^n.$ in our problem $\displaystyle X=I,$ which commutes with any matrix.
• May 1st 2010, 08:20 PM
simplependulum
Quote:

Originally Posted by NonCommAlg
as long as the matrices $\displaystyle X$ and $\displaystyle Y$ commute with each other, you can always use binomial theorem to expand $\displaystyle (X+Y)^n.$ in our problem $\displaystyle X=I,$ which commutes with any matrix.

So if we cannot evaluate $\displaystyle A^n$ very quickly but for $\displaystyle (A - rI )^k$ we can , using binomial theorem to expand $\displaystyle [( A-rI )+ rI]^n$ seems to be a great method . (Happy)
• May 1st 2010, 08:39 PM
NonCommAlg
Quote:

Originally Posted by simplependulum
So if we cannot evaluate $\displaystyle A^n$ very quickly but for $\displaystyle (A - rI )^k$ we can , using binomial theorem to expand $\displaystyle [( A-rI )+ rI]^n$ seems to be a great method . (Happy)

that is correct! another application is finding powers of some low dimensional upper (lower) triangular matrices. for example if $\displaystyle A=\begin{pmatrix}a & x & y \\ 0 & a & z \\ 0 & 0 & a \end{pmatrix}.$ then $\displaystyle A=aI + B,$ where $\displaystyle B=\begin{pmatrix}0 & x & y \\ 0 & 0 & z \\ 0 & 0 & 0 \end{pmatrix}.$

now, look at $\displaystyle B^2$ and then $\displaystyle B^3,$ which is zero. can you find a general fact about powers of strictly triangular matrices (that is triangular matrices with all diagonal entries equal to 0)?

anyway, so $\displaystyle A^n=a^nI + na^{n-1}B + \frac{n(n-1)}{2}a^{n-2}B^2.$
• May 1st 2010, 08:54 PM
simplependulum
Edited
• May 1st 2010, 08:55 PM
simplependulum
It is a tricky thing indeed , we have $\displaystyle A^n$ must be zero for $\displaystyle A$ is $\displaystyle n \times n$ matrix that you mentioned here , because at first the ' length ' of the adjacent side of the right-angled isosceles triangle is $\displaystyle n-1$ , when we multiply it by $\displaystyle A$ , the 'length' decreases by 1 unit so after we multiply $\displaystyle n-1$ times , the length becomes zero or say $\displaystyle A^n = 0$ . I think we just need to find out $\displaystyle A^2 , A^3 ,... ,A^{n-1}$ .
• May 1st 2010, 08:56 PM
NonCommAlg
Quote:

Originally Posted by simplependulum
It is a tricky thing indeed , we have $\displaystyle A^n$ must be zero for $\displaystyle A$ is $\displaystyle nxn$ matrix that you mentioned here , because at first the ' length ' of the adjacent side of the right-angled isosceles triangle is $\displaystyle n-1$ , when we multiply it by $\displaystyle A$ , the '\length' decreases by 1 so after we multiply $\displaystyle n-1$ times , the length becomes zero or say $\displaystyle A^n = 0$ . I think we just need to find out $\displaystyle A^2 , A^3 ,... ,A^{n-1}$ .

absolutely! (Clapping)
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