Formula for matrix raised to power n?

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?
Click to expand...
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 A4\displaystyle A^4 is found this way:
A4=A3AA3=A2AA2=AA\displaystyle \begin{array}{l} A^4 = A^3 A \\ A^3 = A^2 A \\ A^2 = AA \\ \end{array}.
 
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]=[a\displaystyle a[sup:2lj8ud76]k[/sup:2lj8ud76], 0; 0, d\displaystyle d[sup:2lj8ud76]k[/sup:2lj8ud76]]

For example, a matrix A=[2,1;0,3] has eigenvalues a\displaystyle a, d\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]]
 
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 λ1\displaystyle \lambda_{1} and λ2\displaystyle \lambda_{2} are the distinct Eigenvalues, then we have

An  =  λ2(λ1)n−λ1(λ2)nλ2−λ1I2  +  (λ2)n−(λ1)nλ2−λ1A\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.
 
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 λ1\displaystyle \lambda_{1} and λ2\displaystyle \lambda_{2} are the distinct Eigenvalues, then we have

An  =  λ2(λ1)n−λ1(λ2)nλ2−λ1I2  +  (λ2)n−(λ1)nλ2−λ1A\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.
Click to expand...

THAT is why I want to be a math major - can't wait for linear algebra :p
 
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*
 
No arrogance here. I'm just a student trying to learn. I thought I'd post a solution rather than leaving it hanging open.
 
To be fair, I was just picking on your choice of words, not your personal attitude. Good work.
 
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