2
$\begingroup$

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 $$ J_3 = \begin{pmatrix}2 & 1 \\0 & 2 \end{pmatrix}$$

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – Gerry Myerson Aug 11 '11 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 '11 at 3:37
  • 1
    $\begingroup$ 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? $\endgroup$ – Jyrki Lahtonen Aug 11 '11 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 '11 at 6:56
  • 1
    $\begingroup$ 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!! $\endgroup$ – Jyrki Lahtonen Aug 11 '11 at 7:38
4
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • 3
    $\begingroup$ 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? $\endgroup$ – Gerry Myerson Aug 11 '11 at 6:10
  • 2
    $\begingroup$ @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 ... $\endgroup$ – Jyrki Lahtonen Aug 11 '11 at 6:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.