Question 29838
 Some facts that you have to know:
 1) The Caley Hamilton Theorem:
 If f(x) is the characteristic polynomial for a given matrix A,
 then f(A)= 0. [Note deg of f(x) = size A = n)
 Now A is 4 x4 matrix, deg char poly of A = 4, so any high power 
 of A can be reduced to at most of exponeent 3 by division.

 2) To find the minnimal polynomial m(x) for A, i.e. the lowest
 deg (real heer)poy. such that m(A)= 0. Also, m(x) must divide
 char for A and contains all eigenvalues for A as roots.
 3) A is diagonalizable (in C)iff the min. ploy. for A is
 the product of distinct linear factors.

 Now, clearly,this given matrix has 1 as the unique eigenvalue.
 (actually, such matrix ialled unipotent, i.e. (A-I)^k = 0 for some I)
 
 to find A^2006 , we have to get its min. poly. first.
 
 Since A - I =  
 [0, 0, 0, 0
  0, 0, 0, 0 
  0, 1, 0, 0
  0, -1, 0, 0]
 by dirct computation (A-I)^2 = 0

 So, m_A(x) = (x-1)^2 (min. poly.)
 Then try to find the remainder of x^2006 divided by (x-1)^2:
 Since x^2006 -1 = (x^2 -1) q(x) 
 [ or sove x^= (x^2-1) q(x) + ax + b for a & b. Don't use brutal force 
 of long division]
 So,  x^2006 = 1 mod (x^2 -1) 
 Therefore A^2006 = I.

 Make sure you understand about each step.
 By the way, [since nullitt(A-I) = 3 & m(x)=(x-1)^2] 
 the Jordan Canonical Form for A is (i.e A is similar to)
 (1 0  0 0)
 (1 1  0 0)
 (0 0 1 0)
 (0 0 0 1)

 Kenny