#### JohnfromTampa

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What is the general formula for raising a square 2x2 matrix to a power such as 10 or 20?

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- Thread starter JohnfromTampa
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What is the general formula for raising a square 2x2 matrix to a power such as 10 or 20?

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If fear thatJohnfromTampa said:What is the general formula for raising a square 2x2 matrix to a power such as 10 or 20?

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:

\(\displaystyle \begin{array}{l} A^4 = A^3 A \\ A^3 = A^2 A \\ A^2 = AA \\ \end{array}\).

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http://mathworld.wolfram.com/Eigenvector.html

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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

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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.

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

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