|May4-11, 02:52 PM||#1|
Blog Entries: 1
Jordan Normal Form / Jordan basis
1. The problem statement, all variables and given/known data
Determine the Jordan Normal form and find some Jordan basis of the matrix
3 -3 1
A = 2 -2 1
2 -3 2
2. Relevant equations
dim(A) = rk(A) + dimKer(A
3. 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..
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?
I'm lost. Can you please explain what to do, step by step?
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