Formula for matrix raised to power n?

JohnfromTampa

New member
Joined
Feb 4, 2008
Messages
4
What is the general formula for raising a square 2x2 matrix to a power such as 10 or 20?
 

pka

Elite Member
Joined
Jan 29, 2005
Messages
7,805
Re: Formula for matrix raised to power n

JohnfromTampa said:
What is the general formula for raising a square 2x2 matrix to a power such as 10 or 20?
If fear that the answer that you expect does not exist.
That is not to say that there are no general algorithms for performing the operations.
But they are not simple is the way a general formula is.
If A is a square matrix then \(\displaystyle A^4\) is found this way:
\(\displaystyle \begin{array}{l} A^4 = A^3 A \\ A^3 = A^2 A \\ A^2 = AA \\ \end{array}\).
 

JohnfromTampa

New member
Joined
Feb 4, 2008
Messages
4
The solution is the product of three matrices: eigenvectors as the columns of a non-singular matrix T, diag the diagonal matrix of eigenvalues, and the inverse of T.

A[sup:2lj8ud76]k[/sup:2lj8ud76]=Tdiag[sup:2lj8ud76]k[/sup:2lj8ud76]T[sup:2lj8ud76]-1[/sup:2lj8ud76]

diag[sup:2lj8ud76]k[/sup:2lj8ud76]=[\(\displaystyle a\)[sup:2lj8ud76]k[/sup:2lj8ud76], 0; 0, \(\displaystyle d\)[sup:2lj8ud76]k[/sup:2lj8ud76]]

For example, a matrix A=[2,1;0,3] has eigenvalues \(\displaystyle a\), \(\displaystyle d\) of 2,3 and eigenvectors of [1;0] and [1;1]. Then T=[1,1;0,1]

A[sup:2lj8ud76]k[/sup:2lj8ud76]=[1,1;0,1][2,0;0,3][sup:2lj8ud76]k[/sup:2lj8ud76][1,1;0,1][sup:2lj8ud76]-1[/sup:2lj8ud76]

A[sup:2lj8ud76]k[/sup:2lj8ud76]=[1,1;0,1][2[sup:2lj8ud76]k[/sup:2lj8ud76],0;0,3[sup:2lj8ud76]k[/sup:2lj8ud76]][1,-1;0,1], and

A[sup:2lj8ud76]k[/sup:2lj8ud76]=[2[sup:2lj8ud76]k[/sup:2lj8ud76],3[sup:2lj8ud76]k[/sup:2lj8ud76]-2[sup:2lj8ud76]k[/sup:2lj8ud76];0,3[sup:2lj8ud76]k[/sup:2lj8ud76]]
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
9,778
Well, let's not get all arrogant and decide that we have THE WAY to solve the problem. Here's another...

Using a little Cayley-Hamilton Theorem (A square matrix satisfies its own Characteristic Equation)...
Throwing in a little Division Algorithm for Polynomials (or Remainder Theorem, if you like)...

We get this remarkable result:

If A is a 2x2 matrix, and If \(\displaystyle \lambda_{1}\) and \(\displaystyle \lambda_{2}\) are the distinct Eigenvalues, then we have

\(\displaystyle A^{n}\;=\;\frac{\lambda_{2}(\lambda_{1})^{n}-\lambda_{1}(\lambda_{2})^{n}}{\lambda_{2}-\lambda_{1}}I_{2}\;+\;\frac{(\lambda_{2})^{n}-(\lambda_{1})^{n}}{\lambda_{2}-\lambda_{1}}A\)

It's a little different if the Eigenvalues are equal. Obviously, it takes more than that for 3x3 or larger.

Never think you have the ONLY way unless you have PROVEN it so.
 

jwpaine

Full Member
Joined
Mar 10, 2007
Messages
723
tkhunny said:
Well, let's not get all arrogant and decide that we have THE WAY to solve the problem. Here's another...

Using a little Cayley-Hamilton Theorem (A square matrix satisfies its own Characteristic Equation)...
Throwing in a little Division Algorithm for Polynomials (or Remainder Theorem, if you like)...

We get this remarkable result:

If A is a 2x2 matrix, with elements a, b, c, and d, and

If \(\displaystyle \lambda_{1}\) and \(\displaystyle \lambda_{2}\) are the distinct Eigenvalues, then we have

\(\displaystyle A^{n}\;=\;\frac{\lambda_{2}(\lambda_{1})^{n}-\lambda_{1}(\lambda_{2})^{n}}{\lambda_{2}-\lambda_{1}}I_{2}\;+\;\frac{(\lambda_{2})^{n}-(\lambda_{1})^{n}}{\lambda_{2}-\lambda_{1}}A\)

It's a little different if the Eigenvalues are equal. Obviously, it takes more than that for 3x3 or larger.

Never think you have the ONLY way unless you have PROVEN it so.
THAT is why I want to be a math major - can't wait for linear algebra :p
 

o_O

Full Member
Joined
Oct 20, 2007
Messages
396
Haha, I was at the bookstore and randomly opened up a book on tensor calculus. You'd think I'd be scared but it looks so interesting o_O. Although I'd love to major in mathematics, it wouldn't be ideal for me ... *sigh*
 

JohnfromTampa

New member
Joined
Feb 4, 2008
Messages
4
No arrogance here. I'm just a student trying to learn. I thought I'd post a solution rather than leaving it hanging open.
 

tkhunny

Moderator
Staff member
Joined
Apr 12, 2005
Messages
9,778
To be fair, I was just picking on your choice of words, not your personal attitude. Good work.
 
Top