I need to construct a matrix of the following form.

Question

Let a matrix $\theta = $

$ \begin{bmatrix} 1 & 1 &\cdots \cdots \cdots & 1\\ \theta_{11} & \theta_{12} &\cdots & \theta_{1n} \\ \theta_{21} & \theta_{22} &\cdots & \theta_{2n} \\ \theta_{31} & \theta_{32} &\cdots & \theta_{3n} \\ \vdots & \vdots &\ddots& \vdots\\ \theta_{n-1,1} & \theta_{n-1,2} &\cdots & \theta_{n-1,n} \end{bmatrix} $

such that

• $ \theta_{i1}+\theta_{i2} +\cdots +\theta_{in} = 1,$

• matrix $\theta$ is invertible,

• And all the entries of the matrix are non zero and non negative.

$\theta$ is a $n \times n$ matrix.

For a given $n$, I need an algorithm to follow to get a matrix of the above form.

Answer 1

Take the matrix:

$$ \begin{bmatrix} 1 & 1 & 1 & \cdots & 1 & 1\\ 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 1 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 1 & \cdots & 0 & 0\\ \vdots & \vdots & \vdots& \vdots & \vdots\\ 0 & 0 & 0 & \cdots & 0 & 0\\ 0 & 0 & 0 & \cdots & 1 & 0\\ \end{bmatrix}$$

which is a so-called "companion matrix" with determinant $(-1)^n \neq 0$ and characteristic polynomial :

$x^{n}-x^{n-1}-x^{n-2}- \cdots - x -1$

Edit: new answer with your new constraint about strict positivity of the entries:

For the case $n=4$, take

$$ \begin{bmatrix} 1 & 1 & 1 & 1 \\ 2/5 & 1/5 & 1/5 & 1/5 \\ 1/5 & 2/5 & 1/5 & 1/5 \\ 1/5 & 1/5 & 2/5 & 1/5 \end{bmatrix}$$

For the case $n=5$, take: $$ \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 2/6 & 1/6 & 1/6 & 1/6 & 1/6\\ 1/6 & 2/6 & 1/6 & 1/6 & 1/6\\ 1/6 & 1/6 & 2/6 & 1/6 & 1/6\\ 1/6 & 1/6 & 1/6 & 2/6 & 1/6 \end{bmatrix}$$

Do you see the general pattern ? (It is rather similar to the first proposition with the companion matrix).

Answer 2

\begin{bmatrix} 1 & 1 & \cdots & 1 & 1\\ 1 & 0 &\cdots & 0 & 0 \\ 0 & 1 &\cdots & 0 & 0 \\ \vdots & \vdots &\ddots& \vdots & \vdots\\ 0& 0 &\cdots & 1 & 0 \end{bmatrix} this matrix is invertible and all the entries are non negative.