Finding the Jordan Form and basis

Question

$$A= \begin{pmatrix} 2&1&2\\ -1&0&2 \\ 0&0&1 \end{pmatrix}$$

I found that $$f_A(x)=m_A(x) = (x-1)^3.$$

So the Jordan form must be:

$$J= \begin{pmatrix} 1&0&0\\ 1&1&0 \\ 0&1&1 \end{pmatrix}$$

Now,

$$A-I= \begin{pmatrix} 1&1&2\\ -1&-1&2 \\ 0&0&0 \end{pmatrix} $$

$$(A-I)^2 = \begin{pmatrix} 0&0&4\\ 0&0&-4 \\ 0&0&0 \end{pmatrix}$$

$$(A-I)^3 = 0$$

We choose $x_3$ such that $(A-I)^3 x_3= 0$ but $(A-I)^2 x_3\ne 0$. We may choose for example $(0,0,1)^T$. Then we choose $x_2 = (A-I)x_3 = (2,2,0)^T$ and $x_1 = (A-I)x_2 = (4,-4,0)$.

We denote $$Q = \begin{pmatrix} 4&2&0\\ -4&2&0 \\ 0&0&1 \end{pmatrix}$$

What I got is: $$Q^{-1}AQ = \begin{pmatrix} 1&1&0\\ 0&1&1 \\ 0&0&1 \end{pmatrix} \ne J$$

Questions:

1. I want the $1$'s to be below the main-diagonal. Why did I get them above? (And how do I change it?)
2. Is this algorithm "bullet-proof"? Could I always find the Jordan form by observing the characteristic polynomial and minimal polynomial?

Answer 1

1. If you choose a basis such that $Ab_2 = b_1 + \lambda b_2$ the ones will be above the diagonal. If you choose a basis such that $A b_1 = b_2 + \lambda b_1$ they will be below. It's a matter of whether the chains go up or down your indexing.

2. No. These two matrices have different Jordan forms, but the same characteristic and minimal polynomial

$$A = \left(\begin{array}{cccc} 0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0 \end{array}\right)$$

$$B = \left(\begin{array}{cccc} 0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0 \end{array}\right)$$

The characteristic polynomial is $t^4$ and minimal polynomial $t^2$.