The Limit of a Matrix Sequence as n Approaches Infinity

[geoffrey159]

Homework Statement

[/B]
Find the limit as $n \to \infty$ of $U_n(a) =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & a/n \\ 0 & -a/n & 1 \end{pmatrix}^n$, for any real $a$.

Homework Equations

The Attempt at a Solution

I find $U =\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos a & \sin a \\ 0 & -\sin a & \cos a \end{pmatrix}$ but I'm not too sure. Do you think it is correct ?

I wrote $U_n(a) = \begin{pmatrix} 1 & 0 \\ 0 & (C_n(a))^n \end{pmatrix}$, where $C_n(a) = \begin{pmatrix} 1 & a/n \\ -a/n & 1 \end{pmatrix}$

Then I diagonalized $C_n(a)$ in $M_2(\mathbb{C})$ so that

$U_n(a) = \begin{pmatrix} 1 & 0 \\ 0 & P \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & (D_n(a))^n \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & P^{-1} \end{pmatrix}$

with $D_n(a) = \begin{pmatrix} 1+ia/n & 0 \\ 0 & 1 - ia/n \end{pmatrix}$, $P = \begin{pmatrix} i & i \\ -1 & 1 \end{pmatrix}$, and $P^{-1} = \frac{1}{2i}\begin{pmatrix} 1 & -i \\ 1 & i \end{pmatrix}$

and then $(D_n(a))^n \to \begin{pmatrix} e^{ia} & 0 \\ 0 & e^{-ia} \end{pmatrix}$

 

[fresh_42]

How did you get the limits? Beside the $1$ all terms include increasing powers of $\frac{1}{n}$.

 

[geoffrey159]

Honestly I have extrapolated to the complex numbers the fact that $(1+x/n)^n \to e^x$ for any real $x$. I didn't think too much about it. But now you ask I would say something like

$| (1+ ia/n)^n - e^{ia}| = | \sum_{k=0}^n (\frac{n(n-1)...(n-k+1)}{n^k} - 1) i^k \frac{a^k}{k!} - \sum_{k> n}i^k \frac{a^k}{k!}|$

and by the triangle inequality
$| (1+ ia/n)^n - e^{ia}| \le \sum_{k=0}^n |\frac{n(n-1)...(n-k+1)}{n^k} - 1| \frac{|a|^k}{k!} + \sum_{k> n} \frac{|a|^k}{k!}$

The second term converges to 0 as $n\to \infty$ as it is the rest of a convergent sum
The first term also converges to 0 as for any $\epsilon >0$ there exist $N\in\mathbb{N}$ such that $n\ge N \Rightarrow |\frac{n(n-1)...(n-k+1)}{n^k} - 1| \le \epsilon$ and the term $\sum_{k=0}^n \frac{|a|^k}{k!}$ is bounded by $e^{|a|}$.

Hope it is correct

 

[fresh_42]

You cannot take the real sequence and pretend the complex would behave in the same way.
Btw $\begin{pmatrix} n \\ k \end{pmatrix} = \frac{n(n-1) \cdot ... \cdot (n-k+1)}{1 \cdot ... \cdot k} ≠ \frac{n(n-1) \cdot ... \cdot (n-k+1)}{n^k}$.
But you can take your formulas and calculate $C_n(a)^n = P \cdot D_n(a)^n \cdot P^{-1}$.
This results in sums of $(1+iα)^n$ and $(1-iα)^n$ (with some $±i$ coefficients) and $α=\frac{a}{n}$.
Now expand both with the binomial formula and look what cancels out due to $i^n$.
What's left should converge and since $U$ is real the limit should as well be real.

My first approach was simply calculating some powers of $U, C$, resp.
If I made no mistakes there are many terms of increasing powers of $α$.

 

[Ray Vickson]

Homework Statement

[/B]
Find the limit as $n \to \infty$ of $U_n(a) =\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & a/n \\ 0 & -a/n & 1 \end{pmatrix}^n$, for any real $a$.

Homework Equations

The Attempt at a Solution

I find $U =\begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos a & \sin a \\ 0 & -\sin a & \cos a \end{pmatrix}$ but I'm not too sure. Do you think it is correct ?

I wrote $U_n(a) = \begin{pmatrix} 1 & 0 \\ 0 & (C_n(a))^n \end{pmatrix}$, where $C_n(a) = \begin{pmatrix} 1 & a/n \\ -a/n & 1 \end{pmatrix}$

Then I diagonalized $C_n(a)$ in $M_2(\mathbb{C})$ so that

$U_n(a) = \begin{pmatrix} 1 & 0 \\ 0 & P \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & (D_n(a))^n \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & P^{-1} \end{pmatrix}$

with $D_n(a) = \begin{pmatrix} 1+ia/n & 0 \\ 0 & 1 - ia/n \end{pmatrix}$, $P = \begin{pmatrix} i & i \\ -1 & 1 \end{pmatrix}$, and $P^{-1} = \frac{1}{2i}\begin{pmatrix} 1 & -i \\ 1 & i \end{pmatrix}$

and then $(D_n(a))^n \to \begin{pmatrix} e^{ia} & 0 \\ 0 & e^{-ia} \end{pmatrix}$

Well done, clearly presented and a pleasure to read. To satisfy #2 you might mention that $(1 + ia/n)^n \to e^{ia}$, etc.

 

[fresh_42]

You cannot take the real sequence and pretend the complex would behave in the same way.
Btw $\begin{pmatrix} n \\ k \end{pmatrix} = \frac{n(n-1) \cdot ... \cdot (n-k+1)}{1 \cdot ... \cdot k} ≠ \frac{n(n-1) \cdot ... \cdot (n-k+1)}{n^k}$.
But you can take your formulas and calculate $C_n(a)^n = P \cdot D_n(a)^n \cdot P^{-1}$.
This results in sums of $(1+iα)^n$ and $(1-iα)^n$ (with some $±i$ coefficients) and $α=\frac{a}{n}$.
Now expand both with the binomial formula and look what cancels out due to $i^n$.
What's left should converge and since $U$ is real the limit should as well be real.

My first approach was simply calculating some powers of $U, C$, resp.
If I made no mistakes there are many terms of increasing powers of $α$.

Edit: I've seen it now. Sorry. (Where has my delete option gone to?)

 

[geoffrey159]

Thank you !