What is the nth term of this matrix?

1. May 27, 2012

inverse

Hello,
I would like to know what are the nth terms of the following matrix,

$\begin{pmatrix} 2 & 1 & 1\\ 2 & 3 & 2 \\ 1 & 1 & 2\end{pmatrix} = A^{1}$

I want find the general terms of the matrix, and then can find $A^{50}$

Thank you very much for your help
a greeting

Last edited: May 27, 2012
2. May 27, 2012

Hurkyl

Staff Emeritus
Re: how can I find the nth term of a matrix?

The standard method is to diagonalize, right? So have you done it? Where are you stuck?

3. May 27, 2012

inverse

Exactly, I have used diagonalization but as I'd like to know if there is also the possibility to find the nth term of the matrix calculating of this form: A ^ 1, A ^ 2, A ^ 3 ... until being able to find some relation.

you think is too long or too compiled by calculating power and is it not better to use diagonalization?

assuming that the matrix is not diagonizable, how you would calculate the nth term?

4. May 27, 2012

micromass

The easiest method without diagonalization is using the Cayley-Hamilton theorem. So, did you find the characteristic polynomial??

5. May 27, 2012

DonAntonio

Just for curiosity: what exactly do you call "nth terms of matrix" to?? Seems like other responders know, but I still have no idea.

DonAntonio

6. May 27, 2012

micromass

Yeah, I have no idea either. I just assumed he wanted to find the nth power of the matrix.

7. May 27, 2012

algebrat

Allow me to expand on what Hurkyl was thinking.

I put {{2,1,1},{2,3,2},{1,1,2}} into wolfram alpha (after typing "matrix" in to find grab the format) and got $A=SJS^{-1}$. Then $A^2=(SJS^{-1})^2=SJ^2S^{-1}$ (check why that is true). I want you to find what $J^2$ is. Then find what $A^n=(SJS^{-1})^n$ is using the same idea.

note: wolfram alpha will tell you what S and J are.

Cayley Hamilton theorem looks interesting, and probably has great applications, but I don't think you'd want to use it here, you'd be dealing with a 50th degree polynomial right?

8. May 27, 2012

inverse

Yes, I want to find the nth power of the matrix without diadiagonalization.

It is possible with Cayley–Hamilton theorem?

9. May 27, 2012

inverse

Exactly

10. May 27, 2012

micromass

No, you would end up with something like

$$A^{50}=aA^3+bA^2+cA+d$$

of course, to get it in that form, you'd have to calculate somewhat. Or you would have to be lucky and notice a pattern.

Why don't you want to use diagonalization??

11. May 27, 2012

inverse

I have no problem in using diagonalization, I have just curious as you'd calculate if the matrix was not diagonizable

12. May 27, 2012

HallsofIvy

Any matrix that is not diagonalizable can be put into "Jordan Normal Form". It is a little harder to find the nth power of a Jordan Normal Form Matrix but not impossibly so.

Or, as micromass suggested, since this is three by three matrix, its characteristic polynomial is of degree at most 3 and A itself must make that equal to 0. That is, the third power of A can be written as a linear combination of lower powers. And, of course, higher powers of A can be reduced successively to quadratic and lower powers.

Here, $A^3= 7A^2- 11A+ 5I$ so that $A^4= 7A^3- 11A^2+ 5A= 7(7A^2- 11A+ 5I)- 11A^2+ 5A$$= (49A^2- 77A+ 35I)- 11A^2+ 5A= 38A^2- 72A+ 35I$ etc.

13. May 27, 2012

Hurkyl

Staff Emeritus
You would write it as
A = P(D + N)P^(-1)​
where D is diagonal, N is nilpotent, and D commutes with N (e.g. the Jordan Normal Form that HoI mentioned). When you work out the algebra, you essentially end up with a "Taylor series" centered on D:

f(A) = P (f(D) + N f'(D) + (1/2) N^2 f''(D) + ...) P^(-1)​

which terminates, because N^k = 0 for some k.

14. May 28, 2012

inverse

I can already get A^50 whith this, right?

Thanks to all!