1. 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
Originally Posted by tonio
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
Maybe some motivation? I have a plausible reason for the answer by showing that if $M$ is the limit then $M\in\text{SO}(2)$, but it requires me interchanging the limit and transpose and I'm not sure if that's legal!

2. Originally Posted by Drexel28
Maybe some motivation? I have a plausible reason for the answer by showing that if $M$ is the limit then $M\in\text{SO}(2)$, but it requires me interchanging the limit and transpose and I'm not sure if that's legal!

Neither do I and possibly nobody else if you don't tell us what's in your mind. Of course, if there's a valid reason to deduce that the limit matrix is in $SO(2)$ then we're almost done since all such matrices have the form of the already known limit matrix...

Tonio

3. Originally Posted by tonio
Neither do I and possibly nobody else if you don't tell us what's in your mind. Of course, if there's a valid reason to deduce that the limit matrix is in $SO(2)$ then we're almost done since all such matrices have the form of the already known limit matrix...

Tonio
Well, now that I'm less tired I'm not even sure if what I was thinking makes sense, but I've at least answered my question.

So, let me get as far as I can and see if either A) it works or B) I run into problems and give up.

Let $M=\displaystyle \begin{bmatrix}1 & \frac{x}{n} \\ \frac{-x}{n} & 1\end{bmatrix}$ and let $A=\lim_{n\to\infty}M^n$.

Since, $\text{Trans}:\text{GL}(2,\mathbb{R})\to\text{GL}(2 ,\mathbb{R}):B\mapsto B^T$ is continuous we may say that $A^T=\lim_{n\to\infty}\left(M^n\right)^T$ and so $AA^T=\lim_{n\to\infty}M^n\cdot\lim_{n\to\infty}\le ft(M^n\right)^T=\lim_{n\to\infty}M^n\left(M^T\righ t)^n$. But, it's pretty easy to show that $MM^T=I$ and so $M^n\left(M^T\right)^n=M\cdots M\cdot M^T\cdots M^T=I$ and so $AA^T=\lim_{n\to\infty}I=I$ and so $A\in\text{O}(2)$ and also since $\det:\text{GL}(2,\mathbb{R})\to\mathbb{R}$ is continuous we see that $\det A=\det\left(\lim_{n\to\infty}M^n\right)=\lim_{n\to \infty}\det M^n=\lim_{n\to\infty}\left(\det M\right)^n$, but it is fairly easy to see that $\det M=\frac{x^2}{n^2}+1$ and so $\det A=\lim_{n\to\infty}\left(\frac{x^2}{n^2}+1\right)^ n=1$. Thus, $A\in\text{SO}(2)$.

Ta-da?

4. Originally Posted by Drexel28
Well, now that I'm less tired I'm not even sure if what I was thinking makes sense, but I've at least answered my question.

So, let me get as far as I can and see if either A) it works or B) I run into problems and give up.

Let $M=\displaystyle \begin{bmatrix}1 & \frac{x}{n} \\ \frac{-x}{n} & 1\end{bmatrix}$ and let $A=\lim_{n\to\infty}M^n$.

Since, $\text{Trans}:\text{GL}(2,\mathbb{R})\to\text{GL}(2 ,\mathbb{R}):B\mapsto B^T$ is continuous we may say that $A^T=\lim_{n\to\infty}\left(M^n\right)^T$ and so $AA^T=\lim_{n\to\infty}M^n\cdot\lim_{n\to\infty}\le ft(M^n\right)^T=\lim_{n\to\infty}M^n\left(M^T\righ t)^n$. But, it's pretty easy to show that $MM^T=I$

Uh? $MM^t=\begin{pmatrix}1&x/n\\\!\!\!-x/n&1\end{pmatrix}\begin{pmatrix}1&\!\!\!-x/n\\x/n&1\end{pmatrix}=\begin{pmatrix}1+x^2/n^2&0\\0&1+x^2/n^2\end{pmatrix}\neq I$ ...

Tonio

and so $M^n\left(M^T\right)^n=M\cdots M\cdot M^T\cdots M^T=I$ and so $AA^T=\lim_{n\to\infty}I=I$ and so $A\in\text{O}(2)$ and also since $\det:\text{GL}(2,\mathbb{R})\to\mathbb{R}$ is continuous we see that $\det A=\det\left(\lim_{n\to\infty}M^n\right)=\lim_{n\to \infty}\det M^n=\lim_{n\to\infty}\left(\det M\right)^n$, but it is fairly easy to see that $\det M=\frac{x^2}{n^2}+1$ and so $\det A=\lim_{n\to\infty}\left(\frac{x^2}{n^2}+1\right)^ n=1$. Thus, $A\in\text{SO}(2)$.

Ta-da?
.

5. Originally Posted by tonio
.
Oops! I knew I would make a stupid error!

$MM^T=\left(1+\frac{x^2}{n^2}\right)I$ and so

$M^n\left(M^T\right)^n=M\cdots MM^T\cdots M^T=\left(1+\frac{x^2}{n^2}\right)^nI^n=\left(1+\f rac{x^2}{n^2}\right)^nI\to I$ and the result still holds true that $A$ is orthogonal!

6. Drexel28, the large hole in your proof is that you assumed that $A=\lim_{n\to\infty}M^n$ exists, which is not obvious to me.

7. Originally Posted by NonCommAlg
Drexel28, the large hole in your proof is that you assumed that $A=\lim_{n\to\infty}M^n$ exists, which is not obvious to me.
I made no claim that it was a proof. I merely gave incentive to believe what the answer should be.

That said, I don't think (I haven't tried) it would be very hard to show that it converges.

EDIT: Embarrassing as it is to say, it isn't even obvious to me what "converge" would mean here. What metric are we talking about with $\text{GL}\left(n,\mathbb{R}\right)$ or I guess more properly $\mathfrak{M}^{2\times 2}(\mathbb{R})$? The only thing that immediately comes to mind is the one induced by $\|\cdot\|_{\text{op}}$

8. Excuse me , i only know about elementary linear algebra , what does $SO(2)$ mean ? It doesn't refer to sulfur dioxide , right ?

9. Originally Posted by Drexel28
I made no claim that it was a proof. I merely gave incentive to believe what the answer should be.

That said, I don't think (I haven't tried) it would be very hard to show that it converges.

EDIT: Embarrassing as it is to say, it isn't even obvious to me what "converge" would mean here. What metric are we talking about with $\text{GL}\left(n,\mathbb{R}\right)$ or I guess more properly $\mathfrak{M}^{2\times 2}(\mathbb{R})$? The only thing that immediately comes to mind is the one induced by $\|\cdot\|_{\text{op}}$
for a sequence of $m \times m$ matrices $M_n=[x_{ij}^{(n)}]$ we say that $\lim_{n\to\infty}M_n=M=[a_{ij}]$ if $\lim_{n\to\infty}x_{ij}^{(n)}=a_{ij},$ for all $1 \leq i,j \leq m.$ in our problem i really don't see any way to prove that the limit exists

without actually finding the limit! haha

10. I'd like to point out that the limit can be written $\lim_{n\rightarrow \infty}\left(I+\frac{A}{n}\right)^n$, which is just $e^A$, which converges for any square matrix $A$.

What's more, in this case the matrix $A$ is skew-symmetric, and it follows from the theory of Lie groups that $e^A$ is orthogonal.

11. Originally Posted by Bruno J.
I'd like to point out that the limit can be written $\lim_{n\rightarrow \infty}\left(I+\frac{A}{n}\right)^n$, which is just $e^A$, which converges for any square matrix $A$.

What's more, in this case the matrix $A$ is skew-symmetric, and it follows from the theory of Lie groups that $e^A$ is orthogonal.
Why does it follow from Lie groups? I don't see how the differentiable structure of a manifold comes into play here, or do you mean the fact that the common matrix groups are also Lie groups and the result is lumped in there?

12. Originally Posted by Drexel28
Why does it follow from Lie groups? I don't see how the differentiable structure of a manifold comes into play here, or do you mean the fact that the common matrix groups are also Lie groups and the result is lumped in there?
No; the Lie algebra of the orthogonal group is the vector space of skew-symmetric matrices, and the exponential map takes an element of the Lie algebra to its corresponding Lie group.

For example, the Lie algebra of the unit circle is the imaginary line (it's the tangent space to the identity), and the exponential takes the imaginary line to the unit circle.

13. Interesting thread, four proofs so far. Here's mine (which is essentially Tonio's but without much use of linear algebra):

Simply note that the matrix defines a $\mathbb{C}$-linear transformation $A_n: \mathbb{C} \rightarrow \mathbb{C}$, namely multiplication by $\left( 1+i\frac{x}{n} \right) ^n$ and this converges to $e^{ix}$ trivially.

14. Originally Posted by simplependulum
Excuse me , i only know about elementary linear algebra , what does $SO(2)$ mean ? It doesn't refer to sulfur dioxide , right ?
Orthogonal group - Wikipedia, the free encyclopedia

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