# Show the form of $J$ and $P$ for Leslie matrix $A$ when $A = PJP^{-1}$

I'm trying to solve this for a homework assignment.

The Jordan Normal form theorem states that every complex $n \times n$ matrix $A$ van be written as $A=PJP^{-1}$, where $J$ is the diagonal matrix formed by Jordan blocks. $P$ is the matrix whose columns are the base formed by the (generalised) eigenvectors of A. I need to show what the form of $J$ and $P$ is when $A$ is a Leslie matrix.

My thoughts

A Leslie matrix is of the form: \begin{pmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \ddots & \vdots\\ \vdots & \vdots & \ddots & \ddots & 0 \\ 0&0&\cdots&0&1\\ -a_n &-a_{n-1}&\cdots& -a_2&-a_1 \end{pmatrix}

Now I've calculated the characteristic polynomial, and found it is of the form $$h(\lambda) = \lambda^{n-1}a_1+\lambda^{n-2}a_2+...+\lambda a_{n-1}+a_n$$ and that by solving $h(\lambda) = 0$ I can find the eigenvalues (and thus corresponding eigenvectors). But I'm stuck..

Any help?

I think I'm supposed to use a theorem which states that the n-th order differential equation $$\frac{d^n u}{dx^n} + a_1 \frac{d^{n-1} u}{dx^{n-1}}+....+a_n u =0$$ has a base formed by the functions $$u_{l,j} (x) = x^l e^{\lambda_jx}, \ \ \ \ 0\leq l \leq m_j-1$$ where $\lambda_j$ are the complex roots of the characteristic polynomial $h(\lambda)$ with multiplicity $m_j$

• You mean, you want to write down explicitely J and P for every Leslie matrix A as above? This seems to depend very much on the entries a_i of A, in a way that makes the result too diverse to be described easily. – Did Mar 6 '17 at 10:08
• Whizkid95 = Di-lemma? – Did Mar 6 '17 at 10:09
• @Did Well, I want to write down some general form of J and P – Di-lemma Mar 6 '17 at 10:19
• @Did Not the same person, but follow a lot of the same classes and are working on the problem together now. – Di-lemma Mar 6 '17 at 10:20