Given a positive sequence $\{a_n\}_{n=1}^N$, please consider the matrix
$$ \begin{bmatrix} a_1 & -a_1 & & & & \\ -a_1 & a_2 + a_1 & -a_2 & & & \\ & - a_2 & a_3 + a_2 & \ddots & & \\ & & \ddots& \ddots & \ddots & \\ & & & \ddots & a_{N-1} + a_{N-2}& -a_{N-1} \\ & & & & -a_{N-1}& a_{N-1} \end{bmatrix}$$
where blank areas denote null matrix elements.
Does this type of matrix have a name, or resembles others from contexts you know of?
It is real, symmetric, tridiagonal. All its rows sum to zero. All its columns sum to zero as well. Can anything nontrivial be said on its eigenvalues?
I believe (correct me if I'm wrong) that the kernel is one-dimensional, generated by the vector $(1,1,\ldots$).
In the 3d case, I see that the nonzero eigenvalues are simply $a_1 + a_2 \pm \sqrt{a_1^2 - a_1 a_2 + a_2^2}$. So I am trying to guess a generalization of this formula - of which several come to mind. But there may be smarter way to go about it?
Thank you very much for any tip you may have.
as far as the rank, that is indeed $n-1.$
Your matrix is of the form $Q^T D Q$ where
$$ D= \left( \begin{array}{rrrr} a & 0 & 0 & 0 \\ 0 & b & 0 & 0 \\ 0 & 0 & c & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) $$
and
$$ Q = \left( \begin{array}{rrrr} 1 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) $$
This relationship is called congruence rather than similarity. Sylvester's Law of Inertia says that there are thus $n-1$ positive eigenvalues and a single zero eigenvalue. Note that $Q^T DQ$ is the proper way to change basis for a quadratic form.