Jordan matrix of $A$ and $A^{-1}$

Question

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.

Answer 1

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 $k=n, J^{n} = 0$.

Answer 2

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.

Answer 3

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 $$ (\lambda I + N)^{-1} = \frac{1}{\lambda} I -\frac{1}{\lambda^2} N + \frac{1}{\lambda^3} N^2 $$

Answer 4

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 $J_{1/\lambda,n}$.