# Cayley Hamilton Theorem

1. May 25, 2014

### negation

Given a Matrix A = [a,b;c,d] and it's characteristic polynomial, why does the characteristic polynomial enables us to determine the result of the Matrix A raised to the nth power?

2. May 25, 2014

### HallsofIvy

Staff Emeritus
The scalar "$\lambda$" is an "eigenvalue" for matrix A if and only if there exist a non-zero vector, v, such that $Av= \lambda v$. It can be shown that $\lambda$ is an eigenvalue for A if and only if it satisfies A's characteristic equation. A vector v, satisfying $Av= \lambda v$ is an eigenvector for A corresponding to eigenvalue $\lambda$ (some people require that v be non-zero to be an "eigenvector" but I prefer to include the 0 vector as an eigenvector for every eigenvalue).

Further, if we can find n independent eigenvectors for A (always true if A has n distinct eigenvalues but often true even if the eigenvalues are not all distinct) then the matrix, P, having those eigenvectors as columns is invertible and $P^{-1}AP=D$ where D is the diagonal matrix having the eigenvalues of A on its diagonal. Then it is also true that $PDP^{-1}= A$ and $$A^n= (PDP^{-1})^n= (PDP^{-1})(PDP^{-1})\cdot\cdot\cdot(PDP^{-1})= PD(P^{-1}P)(D)(P^{-1}P)\cdot\cdot\cdot(P^{-1}P)DP= PD^nP^{-1}$$

Of course, $D^n$ is easy to calculate- it is the diagonal matrix having the nth power of the entries in D on its diagonal.

Notice that this is "if we can find n independent eigenvectors for A". (Such a matrix is said to be "diagonalizable" matrix.) There exist non-diagonalizable matrices. They can be put in what is called "Jordan normal form" which is slightly more complicated than a diagonal matrix and it is a little more complicated to find powers.

3. May 25, 2014

### AlephZero

What HallsofIvy said is true, but I think it's not quite the point of the question.

The Cayley-Hamilton theorem says that the matrix $A$ satisfies its own characteristic equation. For an $n \times n$ matrix, the characteristic equation is of order $n$, so $A^n$ is a linear combination of $I, A, \dots, A^{n-1}$. It follows that every power $A^k$ where $k > n$ is also a linear combination of the first $n-1$ powers.

For A $2 \times 2$ matrix, that means every power of $A$ is a linear combination of $A$ and $I$. That is true even if the eigenvectors are not independent, and you can't diagonalize $A$.