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springo

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## Homework Statement

f, g endomorphisms on R

^{3}.

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

^{-1}AP)

c.- inverse of (A

^{20}+ I)

## Homework Equations

## 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 I am f = 1

f(1,1,-1) = (1,-1,0)

Therefore: I am 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:

[tex]A=\[ \left( \begin{array}{ccc}

-1 & 1 & -1 \\

1 & -1 & 1 \\

0 & 0 & 0 \end{array} \right)\] [/tex]

__But this can't be OK, because then dim Ker A__.

^{2}= 2 and then 0's multiplicity would be 1c.- I suppose it's done using the Caley-Hamilton theorem once I have the eigenvalues, which I don't...