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Is there a good way of describing the form the inverse matrix of a "n by n matrix in Jordan canonical form"? I know how it should look like, but I don't know how to describe it... As an example: here.

Also, is there a simple way of getting the JCF of this inverse?

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You can always find the Jordan form of a square matrix $A=V^{-1}JV$ and compute the inverse using the data given in the link. (assuming $A$ is invertible) –  user13838 Nov 18 '11 at 10:19
    
It's not clear to me what the first part of your question is asking. That may be partly due to the grammatical errors and ambiguities. Do you mean "the form of the inverse matrix"? And does "in Jordan canonical form" belong to "n by n matrix" or to "describing"? –  joriki Nov 18 '11 at 10:32
    
Since Jordan blocks form the diagonal of the Jordan canonical forms and diagonal matrices behave nicely when taking powers, we need only investigate the problem for Jordan Blocks. The inverse of an upper triangular matrix is also upper triangular, and if $A$ has eigenvalues $\lambda_i $ then $A^{-1}$ has eigenvalues $ \lambda_i^{-1} . $ So we already know the diagonal of the upper triangular matrix which is the inverse of a Jordan block. Perhaps there is an easy way to find the other entries. –  Ragib Zaman Nov 18 '11 at 10:33
    
@percusse: Thanks, I know that we can find the JCF by $A=V^{-1}JV$. But I am looking for a way to describe the form of the inverse of a JCF matrix, something like "you take the reciprocal of the diagonal elements then if there is a 1 in the corner, you take -1/ab..." and so on but in a more concise /symbolic way... –  cumin Nov 18 '11 at 10:35
    
@joriki: I have put in quote marks to group the words now. Sorry about the confusion. –  cumin Nov 18 '11 at 10:38

2 Answers 2

First of all, your example is not the inverse of a Jordan canonical form. If it was, all the coefficients along the main diagonal should be equal: $a = b = c= d$.

I've been doing some experiments with Matlab and I'm making some conjectures. Let

$$ J = \begin{pmatrix} \lambda & 0 & 0 & \dots & 0 & 0 \\ 1 & \lambda & 0 & \dots & 0 & 0 \\ 0 & 1 & \lambda & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & \lambda & 0 \\ 0 & 0 & 0 & \dots & 1 & \lambda \end{pmatrix} $$

be a Jordan block.

Conjecture 1.

$$ J^{-1} = \begin{pmatrix} 1/\lambda & 0 & 0 & \dots & 0 & 0 \\ -1/\lambda^2 & 1/ \lambda & 0 & \dots & 0 & 0 \\ 1/\lambda^3 & -1/\lambda^2 & 1/\lambda & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ (-1)^{n-2}1/\lambda^{n-1} & (-1)^{n-3}1/\lambda^{n-2} & (-1)^{n-4}1/\lambda^{n-3} & \dots & 1/\lambda & 0 \\ (-1)^{n-1}1/\lambda^n & (-1)^{n-2}1/\lambda^{n-1} & (-1)^{n-3}1/\lambda^{n-2} & \dots & -1/\lambda^2 & 1/\lambda \end{pmatrix} $$

Conjecture 2.

The Jordan canonical form of $J^{-1}$ is

$$ \begin{pmatrix} 1/\lambda & 0 & 0 & \dots & 0 & 0 \\ 1 & 1/\lambda & 0 & \dots & 0 & 0 \\ 0 & 1 & 1/\lambda & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \dots & 1/\lambda & 0 \\ 0 & 0 & 0 & \dots & 1 & 1/\lambda \end{pmatrix} $$

Conjecture 3.

The change of basis matrix (that is, the matrix of generalized eigenvectors) is, at least for $n=2, 3$ (I'm too lazy to write the general formula):

$$ S_2 = \begin{pmatrix} 1 & 0 \\ 0 & -1/\lambda \\ \end{pmatrix} \qquad \qquad S_3 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1/\lambda^2 & 0 \\ 0 & 1/\lambda^3 & 1/\lambda^4 \end{pmatrix} $$

EDIT. Ok, so "conjecture" 1 is true -and well-known, as J.M. pointed out. As for conjecture 2, I think it's also true. Here is my proof.

More generally, let's find the Jordan canonical form of a triangular matrix like

$$ A = \begin{pmatrix} a_1 & 0 & 0 & \dots & 0 & 0 \\ a_2 & a_1 & 0 & \dots & 0 & 0 \\ a_3 & a_2 & a_1 & \dots & 0 & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ a_{n-1} & a_{n-2} & a_{n-3} & \dots & a_1 & 0 \\ a_n & a_{n-1} & a_{n-2} & \dots & a_2 & a_1 \end{pmatrix} $$

That is, our $J^{-1}$. In fact in what follows the only two things that we need are:

  1. All the entries along the main diagonal must be equal.
  2. All the entries along the second main diagonal (those $a_2$'s) must be different from zero (but not necessarily equal).

The rest of the entries could be as you please.

Ok, so the characteristic polynomial clearly is

$$ Q_A(t) = \pm (t - a_1)^n $$

-so we have just one eigenvalue: $a_1$- and the rang of the matrix $A - a_1 I$ is $n-1$, because of that $a_2 \neq 0$. Hence de dimension of the null space of $A - a_1 I$ is $1$. So there is just one $n\times n$ Jordan block, namely the one with eigenvalue $a_1$.

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Conjecture 1 is certainly true. –  J. M. Nov 18 '11 at 13:02

The inverse of a Jordan block $x$ with parameters $$ (\lambda,n)\in\mathbb C^\times\times\mathbb Z_{ > 0 } $$ is a Jordan block with parameters $(\lambda^{-1},n)$, and its inverse is given as described below.

Put $A:=K[X]/(X^n)$, where $K$ is a field, $X$ an indeterminate, and $n$ a positive integer. Let $x$ be the canonical image of $X$ in $A$, and $\lambda$ a nonzero element of $K$. Put $$ y:=\sum_{i=1}^{n-1}\ \frac{x^i}{\lambda^{i+1}}\quad. %\tag1 $$ Then, clearly, $\lambda^{-1}+y$ is the inverse of $\lambda-x$, and the minimal polynomial of $y$ is $X^n$.

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