Jordan matrix of $A$ and $A^{-1}$ Suppose I have the Jordan normal form of a matrix $A$. I need to find the the Jordan normal form of $A^{-1}$.
I have the following suggestion:
$$J_{\lambda,n}\rightarrow J_{1/\lambda,n} $$
where $J_{\lambda,n}$ is a Jordan block of $A$, $J_{1/\lambda,n}$ is a Jordan block of $A^{-1}$. 
We can see that if $\lambda $ is an eigenvalue of $A$ then $1/\lambda$ is an eigenvalue of $A^{-1}$ because $$Av=\lambda v \Leftrightarrow   A^{-1}\lambda v=v\Rightarrow \mu=1/\lambda$$ but how can we make sure that $J_{\lambda,n}$ has the same dimension as $J_{1/\lambda,n}$?
Thank you in advance.
 A: First of all, given the matrix $A$, due to primary decomposition theorem we can restrict our view to a single generalized eigenspace $V_{\lambda} := Ker(A-\lambda Id)^{n_{\lambda}}$,where $n_{\lambda}$ is the algebraic multiplicity of $\lambda$ in the characteristical polynomial $P_{A}(t)$. 
In this case the job is much easier, since we can work just with one eigenvalue and repeat this process with all of them.
Secondly we are going to do a further semplification; we additionally restrict our view to a single Jordan block of size $n$ with eigenvalue $\lambda \ne 0$.
Now we are able to determine the Jordan form of $J^{k}$, $\hspace{0.1cm} k \in \mathbb{Z}$. In other words we can easily answer the question as long as we have a single Jordan block of a given size.
We just need the following : 

$Proposition :$ Let $A$ be the upper triangular matrix 
$$ A = \begin{pmatrix}\mu & a_{1,2} & a_{1,3} & \cdots & a_{1,n} \\ & \mu & a_{2,3} & \cdots & a_{2,n} \\ & & \ddots & \ddots & \vdots \\ & & & \mu & a_{n-1,n} \\ & & & & \mu\end{pmatrix}$$
With $a_{1,2}\cdot a_{2,3}\cdots \cdots a_{n-1,n} \ne 0$, then the jordan form of $A$ is 
$$J_{\mu,n} = \begin{pmatrix} \mu & 1 & & & \\ & \mu & \ddots \\
& & \ddots \\ & & & & \mu \end{pmatrix}$$ 

Proof of Proposition : 
The matrix $A$ has just one eigenvalue $\mu$ of algebraic multiplicity $n$. 
Moreover, $A-\mu Id$ is nilpotent and is index of nilpotencty is $n$. Since the index of nilpotency of $A-\mu Id$ is the biggest size of a block in Jordan form associated to the eigenvalue $\mu$ in the Jordan form of $A$, 
It follows that the Jordan of $A$ has just one block of order $n$, hence it's $J_{\mu,n} \hspace{0.2cm}\Box$.
Now we can answer our original problem. For the assumpution made 
(Note that this is not exactly the Jordan of $A$, so the notation $J$ is improper, this would represent the Jordan form of a block of size $n$ related to the eigenvalue $\lambda$,one of whom we restricted our view)
now we have:
$$J = \begin{pmatrix} \lambda & 1 & & & \\ & \lambda & \ddots \\
& & \ddots \\ & & & & \lambda \end{pmatrix}$$
By induction on $k$ we can see that 
$$ J^{k} = \begin{pmatrix}\lambda^{k} & k\lambda^{k-1} &  \\ & \lambda^{k} & \ddots & \\ & & \ddots & k\lambda^{k-1} \\ & & & \lambda^{k}\end{pmatrix}$$
So, $J^{k}$ is an upper triangular matrix with eigenvalue $\lambda^{k}$ of algebraic multiplicity $n$, with non-zero elements on the above parallel diagonal (here we are using $\lambda \ne 0$).
Thanks to the proposition, the Jordan form of $J^{k}$ has only one block of size $n$, which means it is $J_{\lambda^{k},n}$.

Edit : Using induction we can at most extend this argument for every $k \in \mathbb{N}$.
Since we want to work even with $k=-1$ we can use the following result to determine negative powers :
Why does the $n$-th power of a Jordan matrix involve the binomial coefficient? taking $f(x) = x^{k}$. 
Now it works with the same argument for every value of $k \in \mathbb{Z}$.

Note : If $\lambda = 0$, it no longer holds that the Jordan form of $J^{k}$ has just one block. 
Let's take $n=3,k=2$ as example with
$$J = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix} \hspace{0.5cm} J^{2} = \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$$ 
The jordan form of $J^{2}$ is : 
$$J(J^{2}) = \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}$$
Observe that if $k=n, J^{n} = 0$.
A: Hint: Note that an $n \times n$ matrix $M$ is similar to the Jordan block $J_{\lambda,n}$ if and only if it has $\lambda$ as its only eigenvector and the eigenspace associated with eigenvalue $\lambda$ is one-dimensional.
A: The first thing is to see the appearance of the inverse of a Jordan block. With eigenvalue $\lambda \neq 0,$
we begin with $n=2$ and
$$
W =
\left(
\begin{array}{rr}
\lambda & 1 \\
0 & \lambda \\
\end{array}
\right)
$$
$$
W^{-1} =
\left(
\begin{array}{rr}
\frac{1}{\lambda} & -\frac{1}{\lambda^2} \\
0 &  \frac{1}{\lambda}\\
\end{array}
\right)
$$
CAREFULLY find the Jordan form of this block!
Note that we could have written a (finite) Taylor series, using $N^2 = 0$ we have
$$ (\lambda I + N)^{-1} =  \frac{1}{\lambda} I -\frac{1}{\lambda^2} N $$
we continue with $n=3$ and $N^3 = 0$ but $N^2 \neq 0.$
$$
W =
\left(
\begin{array}{rrr}
\lambda & 1 &0\\
0 & \lambda &1 \\
0 & 0 & \lambda
\end{array}
\right)
$$
$$
W^{-1} =
\left(
\begin{array}{rrr}
\frac{1}{\lambda} & -\frac{1}{\lambda^2} &\frac{1}{\lambda^3} \\
0 &  \frac{1}{\lambda}  & -\frac{1}{\lambda^2}\\
0 & 0 &  \frac{1}{\lambda}
\end{array}
\right)
$$
Find the Jordan form of this block!
Note that we could have written a (finite) Taylor series, using $N^3 = 0$ we have
$$ (\lambda I + N)^{-1} =  \frac{1}{\lambda} I -\frac{1}{\lambda^2} N + \frac{1}{\lambda^3} N^2 $$
A: It is clear by looking at the inverse of a matrix in block form that each Jordan block of $A$ gives rise to at least one Jordan block of $A^{-1}$; the same is true of $A^{-1}$, so that each Jordan block of $A^{-1}$ gives rise to at least one Jordan block of $(A^{-1})^{-1}=A$. That is, $A$ and $A^{-1}$ have the same number of Jordan blocks. Hence $(J_{\lambda,n})^{-1}$ has one Jordan block $J_{1/\lambda,n}$.
