## Jordan basis

Hi Sir

I was looking over my Algebra 3 notes and I have a few questions regarding Jordan bases. I was looking at an example where
$A=\left(\begin{array}{ccc}2&1&-2\\1&2&-2\\1&1&-1\end{array}\right)$

on page 25 of the notes. I could find the pre-Jordan basis and from this I was able to find the Jordan basis, but only by checking LI and spanning. Is there an easy way of finding out which vector to replace so that you still have a basis without having to check if it is LI and spans? I found my Jordan Basis to be B_1 = {(-1,1,0), (1,1,1)} B_2 = {(1,0,0)} which is slightly different to the notes, is this possible?

How do I obtain the Jordan Canonical Form from this?

$\left(\begin{array}{ccc}1&1& 0\\ 0&1& 0\\ 0& 0&1\end{array}\right)$
?

Thank you very much for taking the time to read this.

Kind Regards

Ramiz

The final answer is correct. Once you have a Jordan basis, you can put them into the columns of a matrix $P$. Then, as in diagonalization,

$P^{-1}AP$

will be the Jordan normal form. But it’s often not necessary to have the Jordan basis to compute this form. For example, up to $6 \times 6$ matrices, it suffices to know the characteristic polynomial, the minimal polynomial, and the dimension of the eigenspaces. Up to $3 \times 3$ matrices (your case) it even suffices to know the characteristic and minimal polynomials. Look at the discussion of this in Tom Leinster’s notes on the course web page.

Regarding the question of turning a pre-Jordan basis into a Jordan basis, in general, you definitely have to just check which ones are OK to replace. There’s no `clever’ way to do this for a general matrix. But for reasonably sized matrices, this is not such a difficult procedure.

Finally, a Jordan basis is not unique, so it’s certainly possible to have an answer different from that of the notes.

MK