# Math Help - Nice matrix limit))

1. ## Nice matrix limit))

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

P.S. I've got not very elegant solution

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

P.S. I've got not very elegant solution

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

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

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

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

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

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

Tonio

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

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

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

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

Tonio

It seems, the answer has a little typo.

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

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

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

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

Thanx
Tonio

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

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

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

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

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

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

6. Greatly, NonCommAlg

It's really an elementary solution!

7. I have a quite stupid method :

We know :

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

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

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

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

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

As $n \to \infty$

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

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

Therefore , the limit of your matrix is

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

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

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

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

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

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

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

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

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

9. Originally Posted by simplependulum
I have a quite stupid method :

We know :

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

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

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

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

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

As $n \to \infty$

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

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

Therefore , the limit of your matrix is

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

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

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

10. Originally Posted by simplependulum
It is really a great idea and an applicable method in solving other matrix problems , but i think we must put a matrix $A = I + kJ$ where $I$ is the identity matrix , because the product of any other is not commutative : $AB \neq BA$ unless one of them is $I$ ...
as long as the matrices $X$ and $Y$ commute with each other, you can always use binomial theorem to expand $(X+Y)^n.$ in our problem $X=I,$ which commutes with any matrix.

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

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

12. Originally Posted by simplependulum
So if we cannot evaluate $A^n$ very quickly but for $(A - rI )^k$ we can , using binomial theorem to expand $[( A-rI )+ rI]^n$ seems to be a great method .
that is correct! another application is finding powers of some low dimensional upper (lower) triangular matrices. for example if $A=\begin{pmatrix}a & x & y \\ 0 & a & z \\ 0 & 0 & a \end{pmatrix}.$ then $A=aI + B,$ where $B=\begin{pmatrix}0 & x & y \\ 0 & 0 & z \\ 0 & 0 & 0 \end{pmatrix}.$

now, look at $B^2$ and then $B^3,$ 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 $A^n=a^nI + na^{n-1}B + \frac{n(n-1)}{2}a^{n-2}B^2.$

13. Edited

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

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

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