Jordan Normal Form Problem

[springo]

Homework Statement

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)

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:
$$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...

 

[mighty2000]

a.- To find the Jordan normal form of f, you need to find the eigenvalues of f and their corresponding eigenvectors. Since you know that 0 is an eigenvalue with multiplicity 2, you can use the characteristic polynomial to find the other eigenvalues. The characteristic polynomial is given by det(f-λI) = 0, where λ is the eigenvalue and I is the identity matrix. So in this case, you would have det(f-λI) = (λ-0)^2 = λ^2 = 0. This means that the only other eigenvalue is 0, with multiplicity 2. Therefore, the Jordan normal form of f would be \[ \left( \begin{array}{ccc}
0 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \end{array} \right)\]

b.- To find the canonical basis matrix for f, you can use the eigenvectors corresponding to the eigenvalues you found in part a. In this case, since 0 is the only eigenvalue, the canonical basis matrix would be the zero matrix. As for the transition matrix P, you can use the eigenvectors as columns to form P. So in this case, P would be \[ \left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)\]

c.- To find the inverse of (A20 + I), you can use the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic polynomial. In this case, the characteristic polynomial is (λ-0)^2 = λ^2 = 0. So (A20 + I)^2 = (0)^2 + I = I. Therefore, the inverse of (A20 + I) is (A20 + I)^-1 = I.