I am trying to recover the Jordan normal form of a matrix given a list of invariant factors and was wondering if I am proceeding correctly in constructing the Jordan blocks.
Let $F = \mathbb{C}$ and let $V$ be a finite dimensional vector space over $F$. Let $T:V\to V$ be a linear operator and give $V$ the structure of a module over the polynomial ring $F[x]$ by defining $x \alpha = T(\alpha) \alpha \in V$
let
$$
A = \left( \begin{array}{ccc}
x^2(x-1)^2 & 0 & 0 \\
0 & x(x-1)(x-2)^2 & 0 \\
0 & 0 & x(x-2)^3 \end{array} \right)
$$
be a relation matrix for V with respect to $\{v_1, v_2, v_3\}$ generators of $V$.
Then $d_1 = x$, $d_2 = x(x-1)(x-2)^2$ and $d_3 = x^2(x-1)^2(x-2)^3$ are the invariant factors of $T$. Then we know $ V = F[x] / (x) \oplus F[x]/ (x(x-1)(x-2)^2) \oplus F[x]/(x^2(x-1)^2(x-3)^3)$. Further we know that the minimal polynomial of $T$ is the largest of the invariant factors so that $m_T(x) = (x^2(x-1)^2(x-2)^3)$ and the characteristic polynomial will be the product of $d_1 d_2 d_3$.
Question: what is the appropriate Jordan normal form of T?
Since 0, 1 and are repeated roots and 2 is repeated 3 times.
Does that give me Jordan blocks $$ J_1 = \begin{pmatrix}0 & 1 \\0 & 0 \end{pmatrix}$$
$$ J_2 = \begin{pmatrix}1 & 1 \\0 & 1 \end{pmatrix}$$
and $$ J_3 = \begin{pmatrix}2 & 1 \\0 & 2 \end{pmatrix}$$
Best Answer
Edit: I'm sorry, but my first answer was definitely incorrect and I really hope I didn't cause any confusion. Won't speedread problems in the future :)
Since we have $V = F[x] / (x) \oplus F[x]/ (x(x-1)(x-2)^2) \oplus F[x]/(x^2(x-1)^2(x-3)^3)$, we can look at our three summands separately.
Our first summand, $F[x]/(x)$, has a single eigenvalue of zero (of multiplicity one), so our first Jordan block is simply $$J_1 = \begin{pmatrix}0 \end{pmatrix}$$
Next, we look at our second summand, $F[x]/(x(x-1)(x-2)^2)$, which has eigenvalues 0, 1, and 2 of multiplicities 1, 1, and 2 (respectively), so our Jordan blocks are now $$ J_2 = \begin{pmatrix}0 \end{pmatrix}, J_3 = \begin{pmatrix} 1 \end{pmatrix}, J_4 = \begin{pmatrix}2 & 1 \\0 & 2 \end{pmatrix}$$
For our final summand of $F[x]/(x^2(x-1)^2(x-3)^3)$ the eigenvalues are 0, 1, and 2 (with multiplicities 2, 2, and 3 respectively), so the Jordan blocks will be of the form $$ J_5 = \begin{pmatrix}0 & 1 \\0 & 0 \end{pmatrix}, J_6 = \begin{pmatrix}1 & 1 \\0 & 1 \end{pmatrix}, J_7 = \begin{pmatrix}2 & 1 & 0 \\0 & 2 & 1 \\0 & 0 & 2\end{pmatrix}$$
Putting the 7 blocks together gives our answer.