# Difference equation and diagonal block matrix

1. Oct 14, 2015

### pondzo

1. The problem statement, all variables and given/known data

Compute $A^j~\text{for} ~~j=1,2,....,n$ for the block diagonal matrix$A=\begin{bmatrix} J_2(1)& \\ &J_3(0) \end{bmatrix}$,

And show that the difference equation $x_{j+1}=Ax_{j}$ has a solution satisfying $|x_{j}|\rightarrow\infty~\text{as}~j\rightarrow\infty$
2. Relevant equations

3. The attempt at a solution

So $A^1=\begin{bmatrix} 1&1&0&0&0 \\ 0&1&0&0&0 \\ 0&0&0&1&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0 \end{bmatrix} ,~~A^2=\begin{bmatrix} 1&2&0&0&0 \\ 0&1&0&0&0 \\ 0&0&0&0&1 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \end{bmatrix} ,~~A^j=\begin{bmatrix} 1&j&0&0&0 \\ 0&1&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \end{bmatrix}\forall ~~j\geq 3$

I am certain that this isn't a difficult question, but I am not sure how to apply this to the difference equation. Which is probably due to my lack of experience with them. Help would be appreciated, thanks.

Last edited by a moderator: Oct 14, 2015
2. Oct 14, 2015

### BvU

For my sake: apparently $J_2$ and $J_3$ have a special meaning. Could you explain ?

3. Oct 14, 2015

### pondzo

Yes sorry about that, I should have put it in the question.

$J_2(1) = ~\text{the 2x2 jordan block matrix with 1 as eigenvalues}~=\begin{bmatrix} 1&1 \\ 0&1 \end{bmatrix}$
And $J_3(0) = ~\text{the 3x3 jordan block matrix with 0 as eigenvalues}~=\begin{bmatrix} 0&1&0 \\ 0&0&1 \\0&0&0 \end{bmatrix}$

4. Oct 14, 2015

### BvU

Well, I learned something !

But now I can't be of much further help, because I can't imagine what they want to count as a 'solution' for $x_{j+1}=Ax_{j}$, which to me is just a recipe to get $x_{j+1}$ from $x_{j}$.

I mean, is $x_0 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ a 'solution' ? (I ignore the other dimensions that go to 0 anyway)
Because $x_{j+1} = A x_j$ is satisfied (always !?) and $x_j = \begin{bmatrix} 1 + j \\ 1 \end{bmatrix}$ so that
$|x_{j}|\rightarrow\infty~\text{as}~j\rightarrow\infty$

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5. Oct 14, 2015

### pondzo

I am glad I was involved in helping someone learn something new!

Mod note: Fixed typo in post #1
I think I should point out that there is a typo in the original question, it should be:

$A^j=\begin{bmatrix} 1&j&0&0&0 \\ 0&1&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \end{bmatrix}\forall ~~j\geq 3 ~\text{rather than}~ A^j=\begin{bmatrix} 1&n&0&0&0 \\ 0&1&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \\ 0&0&0&0&0 \end{bmatrix}\forall ~~j\geq 3$

And yes what you say makes sense.. I think as long as $x_0$ has a positive value in the 2nd row it will be satisfied (since the second column in $A^j$ is unbounded in $j$) and $|x_ j| =\sqrt{j^2+1}$ which tends to infinity as $j$ does. Do you agree ?

Last edited by a moderator: Oct 14, 2015
6. Oct 14, 2015

### BvU

I agree with you, but that doesn't mean much (real knowledge trumps common sense). Would appreciate it very much if you would post if it turns out we are both wrong

Here they call $a_n = r^n$ the solution for the difference equation $a_n = r a_{n-1}$ so in that jargon I have inadvertently posted the actual solution,

which is in fact against PF rules. May I be forgiven by O2 and all the other good spirits that watch over us, because I did not know what I was doing

7. Oct 14, 2015

### Ray Vickson

Try writing out the first few terms $x_2, x_3 x_4 \ldots$, to see how they relate to $x_1$.