# Finding positive definite scalar-product for Jordan block

I want to solve the following problem: Given $\lambda \in (-1,1)$ and the corresponding Jordan block $J=\begin{pmatrix} \lambda & 1 & \dots & 0 \\ 0 & \ddots & \ddots & \vdots \\ \vdots & \ddots & \ddots & 1 \\ 0 & \dots & 0 & \lambda \end{pmatrix}$.

I want to find a positive definite scalar product $< \cdot, \cdot >_M$ such that $< Jv, Jv >_M \ < \ <v,v>_M$ for all $v \in \mathbb{R}^n \setminus \{ 0 \}$. Unfortunately I don't really know how to tackle this problem. I tried to find an idea for $M$ by solving the smaller problems such as $n=1,2,3$. Furthermore since I know that $M$ has to be positive definite, it can be written as $M=A^T A$ with a specific invertible matrix $A$ (Cholesky decomposition).

From the $n=1$ case I couldn't learn anything as everything works out nicely, no matter how the matrix $A$ is chosen.

For the $n=2$ case, expressions get big very quickly. I tried to first of all solve the $\lambda=0$ case to get some idea how $A$ should look like. Using some "interesting" vectors $v=\begin{pmatrix} v_1 & v_1 \end{pmatrix}$, $v=\begin{pmatrix} v_1 & -v_1 \end{pmatrix}$ I got: \begin{align*} -2 (a_{11} a_{12} + a_{21} a_{22}) &< a_{12}^2+a_{22}^2, \\ 2 (a_{11} a_{12} + a_{21} a_{22}) &< a_{12}^2+a_{22}^2. \end{align*} Values for $A$ which solve that system are for instance $a_{12}=a_{21}=0$ and $a_{11}^2 < a_{22}^2$. Choosing $a_{12}=a_{21}=\lambda$ I went back to the general $n=2$ case but unfortunately couldn't find any values for the other entries of the matrix.

Do you have any ideas what I could try next? Is there any nice way to solve that system of inequalities? That high number of terms in the expressions (even for small $n$) makes it very hard for me to make good estimations about the matrix $A$...

Regards!

• Nobody an idea what to try next? I already added what I found out so far, unfortunately it does not really help me. – blablablup Dec 4 '16 at 14:37