Is there a general form for the determinant of this matrix?

Question

This came up in trying to deal with small oscillations of an $N$-pendulum. I obviously want to calculate the characteristic polynomial in $\omega^2$ to see if I can deal with the equation even in special cases. Is it the first time you see this kind of matrix? Does the characteristic polynomial have a reasonably nice form? Oh, $N$ is the dimension of the matrix. You can suppose $m,g,l>0.$

$$A=\begin{pmatrix} N(mgl-\omega^2ml^2) & -\omega^2ml^2(N-1)& -\omega^2ml^2(N-2)&...& -\omega^2ml^2 \\ -\omega^2ml^2(N-1)& (N-1)(mgl-\omega^2ml^2) &-\omega^2ml^2(N-2) &... &-\omega^2ml^2 \\ -\omega^2ml^2(N-2) &-\omega^2ml^2(N-2) &(N-2)(mgl-\omega^2ml^2) & ...& -\omega^2ml^2 \\ \vdots&\vdots&\vdots&\vdots&\vdots \end{pmatrix}$$

Thank you in advance.

EDIT: As pointed out by Omnomnomnom (I had to copy paste that name) the problem reduces to finding the determinant of:

$$A = -\omega^2ml^2 \pmatrix{ N&N-1&N-2&\cdots&1\\ N-1&N-1&N-2&\cdots&1\\ N-2&N-2&N-2&\cdots&1\\ \vdots&\vdots&\ddots } + mgl \pmatrix{ N\\&N-1\\&&N-2\\&&&\ddots }=-\omega^2ml^2T+mglV$$ Dividing by $mgl$ and calling $$\tilde\omega=\frac{\omega^2l}{g}$$ We have $$\operatorname{Det}\bigg(\frac{A}{mgl}\bigg)=\operatorname{Det}B=\operatorname{Det}(V-\tilde\omega T)$$

I computed the first few polynomials in $\tilde \omega$.

For $N=2$ $$P_2(\tilde\omega)= \tilde\omega^2-4\tilde\omega+2$$

For $N=3$:

$$P_3(\tilde\omega)=-\tilde\omega^3+9\tilde\omega^2-18\tilde\omega+6$$

For $N=4$

$$P_4(\tilde\omega)=\tilde\omega^4-16\tilde\omega^3+72\tilde\omega^2-96\tilde\omega+24$$

For $N=5$

$$P_5(\tilde\omega)= -\tilde\omega^5+25\tilde\omega^4-200\tilde\omega^3+600\tilde\omega^2-600\tilde\omega+120$$ Do you see a pattern? The first coefficient might be $(-1)^N$, the second $N^2(-1)^{N+1}$, the last one is probably $N!$, the second to last one might be $-N!N$.

Answer 1

Not a complete answer, but a helpful note: we can write this matrix as $$ A = -\omega^2m\ell^2 \pmatrix{ N&N-1&N-2&\cdots&1\\ N-1&N-1&N-2&\cdots&1\\ N-2&N-2&N-2&\cdots&1\\ \vdots&\vdots&\ddots } + mg\ell \pmatrix{ N\\&N-1\\&&N-2\\&&&\ddots } $$ After we factor out the second matrix, I'd imagine we could say something nice about the eigenvalues of the first.

Answer 2

Consider the matrix $$ P^{-1}Q = \pmatrix{ N\\&N-1\\&&N-2\\&&&\ddots }^{-1} \pmatrix{ N&N-1&N-2&\cdots&1\\ N-1&N-1&N-2&\cdots&1\\ N-2&N-2&N-2&\cdots&1\\ \vdots&\vdots&\ddots } = \\ \pmatrix{ 1&1-1/N&1-2/N&\cdots&1/N\\ 1&1&1 - 1/(N-1)&\cdots&1/(N-1)\\ 1&1&1&\cdots&1/(N-2)\\ \vdots&\vdots&\ddots } $$ Let $q_N(x)$ denote the characteristic polynomial $\det(P^{-1}Q - xI)$. The answer to the determinant we seek will be $$ \det(A) = N! \cdot a^N q_N(-b) = N! (-\omega^2 m \ell^2)^N q_N(- mg \ell) $$ Hopefully you find this expression useful.