# Jordan Normal Form Problem

1. Feb 1, 2009

### springo

1. The problem statement, all variables and given/known data
f, g endomorphisms on R3.
f non-diagonalizable and g injective.
h = g o f
dim h-1 { (0,0,0) } = 2
f(1,0,1) = f(1,1,2)
f(1,1,-1) = (1,-1,0)

I have to find:
a.- J, Jordan normal form of f
b.- A, canonical basis matrix for f and transition matrix P (J = P-1AP)
c.- inverse of (A20 + I)

2. Relevant equations

3. The attempt at a solution
a.- dim Ker h = 2 with g injective so dim Ker f = 2.
Therefore 0 is an eigenvalue with multiplicity 2.
I don't know how to find the other eigenvalue.
(I mean without using A, which is found in the next question)

b.- dim Ker f = 2 so dim Im f = 1
f(1,1,-1) = (1,-1,0)
Therefore: Im f = L { (1,-1,0) }
f(1,0,1) - f(1,1,2) = f(0,-1,-1) = (0,0,0)
Therefore Ker f = L { (1,0,-1) (0,1,1) }
(I thought any vector that's linearly independent with (0,1,1) would be OK)
f(1,1,-1) - f (1,0,-1) = f(0,1,0) = (1,-1,0)
f(0,1,1) - f(0,1,0) = f(0,0,1) = (-1,1,0)
f(1,0,-1) + f(0,0,1) = f(1,0,0) = (-1,1,0)
Therefore:
$$A=$\left( \begin{array}{ccc} -1 & 1 & -1 \\ 1 & -1 & 1 \\ 0 & 0 & 0 \end{array} \right)$$$
But this can't be OK, because then dim Ker A2 = 2 and then 0's multiplicity would be 1.

c.- I suppose it's done using the Caley-Hamilton theorem once I have the eigenvalues, which I don't...