Diagonalization of Leslie matrix

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I am trying to show that this matrix:

$$L=\begin{bmatrix} f_0 & f_1& \cdot & \cdot & f_n\\ p_0 & 0 & \cdot & \cdot & 0\\ 0 & p_1& \cdot & \cdot & 0\\ \cdot & \cdot& \cdot & \cdot & \cdot\\ 0 & 0& \cdot & p_{n-1} & 0 \end{bmatrix}\mbox{ , } 0\leq f_i \mbox{ and }0<p_i<1.$$

commonly known as the Leslie matrix is diagonalizable.

I know a $n \times n$ matrix is diagonalizable if there exist $n$ linearly independent eigenvectors. I also know two different eigenvalues have linearly independent eigenvectors. So my approach was to show the characteristic equation of the Leslie matrix has $n$ different solutions, but here I got stuck. Can somebody help me?

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Here is a simple counterexample with a $3 \times 3$ matrix:

$$L=\begin{bmatrix} 1 & 5& 3\\ 1 & 0& 0\\ 0 & 1& 0 \end{bmatrix}$$

with eigenvalues $-1$ (order of multiplicity $2$) and $3$.

The eigenspace associated with eigenvalue $-1$ is obtained as solution of

$$\begin{cases}2x&+&5y&+&3z&=&0\\x&+&y&&&=&0\\&&y&+&z&=&0\end{cases}$$

which gives $\begin{bmatrix}x\\y\\z\end{bmatrix} \ = \ k \ \begin{bmatrix}\ \ 1\\-1\\ \ \ 1\end{bmatrix}.$

Thus, this eigenspace is only one dimensional. As this dimension is less than the order of multiplicity of the corresponding eigenvalue, matrix $L$ is not diagonalizable.

Remark 1: $kL$, with any positive real $k$, is a counterexample as well.

Remark 2: matrix $L$ is a "companion" matrix : see the Wikipedia article (http://www.math.wsu.edu/faculty/watkins/seminar/pdfiles/fiedler_companion.pdf) where it is explicitly said: "In general, the companion matrix may be non-diagonalizable.". But the difficulty is to find such a matrix with positive coefficients.

Remark 3: a companion matrix can have different equivalent forms: one can find the (opposite of the) list of polynomial coefficients on

Remark 4: the counterexample given by @quasi is, up to a multiplication by 2 (which does not affect the "non-diagonalizability" property), a "companion matrix".

If you are interested by companion matrices, here is a very interesting discussion by Cleve Moler, the founder of Matlab: (http://blogs.mathworks.com/cleve/2013/12/23/fiedler-companion-matrix/)