# 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?

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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 $a_1$.

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 $X^n$.