Calculating a matrix to a high power using Cayley-Hamilton theorem

• Apr 6th 2010, 04:22 AM
Jhooikist
Calculating a matrix to a high power using Cayley-Hamilton theorem
Hi,

Does anyone know how I can use the Cayley-Hamilton theorem to compute a high power of a matrix, for instance

A = [0, 2, 0; 1, 1, -1; -1, 1, 1]

A^1000 = ?

I know the theorem states that plugging the matrix A into its own characteristic polynomial will result in the 0 matrix, but I don't yet see how this can be used to solve this problem.
• Apr 6th 2010, 06:42 AM
HallsofIvy
Did you actually calculate the characteristic equation for this matrix? It is $-\lambda^3- 2\lambda^2= 0$. That is the same as saying that $\lambda^3= -2\lambda^2$ and so A^3= -2A^2. There is not that much difference between a cube and a square so this is tedious but doable.

1000/3= 333 and 1/3 so $A^{1000}= (A^3)^333(A)= -2^{333}(A^2)^{333}(A)= -2^{333} A^{334}$. Now 334/3= 111 and 1/3 so $A^{334}= (A^3)^{111}(A)= (-2A^2)^{111}(A)= -2^{111} A^{223}$.

So far we have that [tex]A^{1000}= 2^{444}A^{223}. 223/3= 74 and 1/3 so $A^{223}= (A^3)^{74}(A)= (-2A^2)^{74}(A)= 2^{74}A^{159}$: $A^{1000}= 2^{518}A^{159}$.

159/3= 53 so $A^{159}= (A^3)^{53}= (-2A^2)^{53}= -2^{53}A^{106}$. $A^{1000}= -2^{571}A^{106}$.

106/3= 35 and 1/3 so $A^{106}= (A^3)^{35}(A)= (-2A^2)^{35}(A)= -2^{35}A^{71}$. $A^{1000}= 2^{605}A^{71}$.

Keep doing that until you have reduced this to a number times $A^2$.