Jordan Normal Form / Jordan basis

Maybe_Memories

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

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 this was the easiest way to write it. Just imagine they were written as columns..


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?