Difference equation and diagonal block matrix

[pondzo]

Homework Statement

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##

Homework Equations

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.

 

[BvU]

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

 

[pondzo]

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

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}##

 

[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##

--

 

[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 ?

 

[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​

 

[Ray Vickson]

Homework Statement

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##

Homework Equations

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.

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