Jordan Normal Form / Jordan basis

[Maybe_Memorie]

Homework Statement

Determine the Jordan Normal form and find some Jordan basis of the matrix
3 -3 1
A = 2 -2 1
2 -3 2

Homework Equations

dim(A) = rk(A) + dimKer(A)

The Attempt at a Solution

My problem here is that my lecturer seems to be doing completely different things with every question and it's getting confusing.

So, I calculated the characteristic polynomial of the matrix, and got one
eigenvalue of t = 1.

So I'm now dealing with the matrix A - tI, in this case A - I.

rk(A - I) = 1, so dimKer(A - I) = 2.

(A - I)^2 = 0 = B

So rk(B) = 0, then dimKer(B) = 3

Continually raising the powers of A - I will results in 0, so the kernels of powers stabilize at the second step, so we should expect a thread of length 2.

The kernel of A - I is spanned by the vectors ( 3/2, 1, 0) and (-1/2, 0, 1)
I should be using columns instead of rows, but I don't know latex so this was the easiest way to write it. Just imagine they were written as columns..


Now, I'm lost. Can you please explain what to do, step by step, and please state where something would be done differently in a different scenario.

Thank you!

 

[Maybe_Memorie]

Here's the main issue, when I find out what vectors span the kernel, and use those as columns of a new matrix, I then reduce that matrix to Reduced Column Echelon Form.

That's fine, but, in some of my lecturers examples he takes a vector corresponding to the missing leading one as a basis, and it others he takes one of the columns of the RCEF.

My question: why the differences and does it matter which of the columns are taken?

 

[Maybe_Memorie]

Anyone?