Jordan Canonical form 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?
 A: 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:


*

*All the entries along the main diagonal must be equal.

*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 $a_1$.
A: 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 $X^n$.
