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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    $\begingroup$ The eigenvalues are the roots of the characteristic polynomial, right? And the eigenvalues should show up on the diagonals of the Jordan blocks, shouldn't they? Also, the only non-zero entries off the diagonal of a Jordan block should be 1, right? $\endgroup$ Aug 11, 2011 at 3:32
  • $\begingroup$ Thank you I just edited the post because I had not written the eigenvalues on the diagonal of the blocks. $\endgroup$
    – user7980
    Aug 11, 2011 at 3:37
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    $\begingroup$ Three generators are needed to get $V$ as a module over $F[x]$. Now, how many generators are needed to get $V$ as a vector space over $F$? What's $\dim_FV$? IOW, why do you think that $T$ is a $3\times3$ matrix? $\endgroup$ Aug 11, 2011 at 6:45
  • $\begingroup$ @Jyrki Thank you I have updated the post with a more complete problem statement. The motivation for this problem is described in more detail here math.stackexchange.com/questions/52859/… $\endgroup$
    – user7980
    Aug 11, 2011 at 6:56
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    $\begingroup$ 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!! $\endgroup$ Aug 11, 2011 at 7:38

1 Answer 1


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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    $\begingroup$ If the characteristic polynomial is what OP says it is, $x^4(x-1)^3(x-2)^5$, won't the multiplicities be 4, 3, and 5? Won't there be some smaller Jordan blocks? $\endgroup$ Aug 11, 2011 at 6:10
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    $\begingroup$ @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 ... $\endgroup$ Aug 11, 2011 at 6:42

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