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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}$

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    Ok. But now with the problem definition clarified I must ask the question: Your matrix $A$ already gives $V$ as a direct sum of three cyclic $F[x]$-modules! So why do you bother with the invariant factors? Just do the Jordan decomposition one cyclic summand at a time!!2011-08-11

1 Answers 1

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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.

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    @user7980: Agree with Gerry. With these blocks suggested by Michael along the diagonal you only get the summand $F[x]/\left(x^2(x-1)^2(x-2)^3\right)$. So ...2011-08-11