Is this the right way of determining the Jordan normal form and Jordan basis?

Question

For the linear map $Tv = Av$, T : $\mathbb{C}^{3} \rightarrow \mathbb{C}^{3}$, where

$A= \begin{bmatrix} 3 & -2 & 2 \\ 0 & 1 & 0 \\ -1 & 2 & 0 \end{bmatrix} $

We are given the characteristic polynomial of T, $ch(x) = (x-2)(x-1)^2$

So far I have found that by solving:

$(I_3 - A)v = 0$

$V_1(1) = span( \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} )$

$(I_3 - A)^2v = 0$

$V_2(1) = span( \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} )$

and

$(2I_3 - A)v = 0$

$V_1(2) = span( \begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix} )$

for each eigenspace,

Does this mean that the Jordan basis is:

$B = ({ \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix},\begin{bmatrix} -2 \\ 0 \\ 1 \end{bmatrix}})$

and the JNF is

$J= \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix} $

due to the multiplicities of the eigenvalues.

Would this be correct?

Thanks very much in advance.

Answer 1

Learn how to verify your work. Let (by abuse of notation) $B$ be the matrix whose columns are the vectors of the basis $B$. Is it true that $BJB^{-1}=A$? If it is so (and if $J$ is in Jordan form, which is obvious to check), then your Jordan basis is correct.

Equivalently, you may check that $BJ=AB$ and that $B$ is invertible, which is easier.

Answer 2

The standard way to find a Jordan basis is not to begin by finding the eigenvectors for the eigenvalues with multiplicity $>1$, but finding first the generalised eigenvectors with highest level, then cascading until obtaining the eigenvectors as images of generalised eigenvectors.

Let's see in more detail what this procedure becomes in the present situation.

You have two eigenvalues: $1$ (multiplicity $2$) and $2$ (multiplicity $1$). Further more you can check the eigenspace $E_1$ has dimension $1$, so we have to consider the subspace $\ker (A-I)^2$ (generalised eigenspace), choose a vector $v_2\in \ker (A-I)^2\setminus\ker(A-I)$, and set $v_1=(A-I)v_2$. $v_1\in E_1$.

Add a vector $v_3\in E_2$. By construction, the basis $(v_1, v_2,v_3)$ is a Jordan basis.