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 !