You can
put this solution on YOUR website! 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
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