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  1. #1
    Senior Member DeMath's Avatar
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    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
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    Quote Originally Posted by DeMath View Post
    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
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    Senior Member DeMath's Avatar
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    Quote Originally Posted by tonio View Post
    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}
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    Quote Originally Posted by DeMath View Post
    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
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    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}.
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  6. #6
    Senior Member DeMath's Avatar
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    Greatly, NonCommAlg

    It's really an elementary solution!
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    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}<br />

    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)
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    Quote Originally Posted by NonCommAlg View Post
    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 ...
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    Quote Originally Posted by simplependulum View Post
    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}<br />

    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
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    Quote Originally Posted by simplependulum View Post
    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.
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    Quote Originally Posted by NonCommAlg View Post
    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 .
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    Quote Originally Posted by simplependulum View Post
    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.
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    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} .
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    Quote Originally Posted by simplependulum View Post
    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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