What is the nth term of this matrix?

[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

 

[Hurkyl]

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

 

[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?

 

[micromass]

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

 

[DonAntonio]

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

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

 

[micromass]

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

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

 

[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" into 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?

 

[inverse]

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

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

It is possible with Cayley–Hamilton theorem?

 

[inverse]

you'd be dealing with a 50th degree polynomial right?

Exactly

 

[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??

 

[inverse]

No, you would end up with something like

Why don't you want to use diagonalization??

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

 

[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.

 

[Hurkyl]

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

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.

 

[inverse]

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.

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

Thanks to all!