Suppose, $ M=\begin{bmatrix}\begin{array}{ccccccc} -x & a_2&a_3&a_4&\cdots &a_n\\ a_{1}+x & -x-a_2 & 0&0&\cdots &0\\ a_1+x&0 & -x-a_3 &0&\cdots &0\\ \vdots&\vdots&\vdots&\vdots &\ddots&\vdots\\ a_1+x&0 & 0&0&\cdots & -x-a_n\\ \end{array}\end{bmatrix}$, then how to find the $\det (M)$?
The empirical formula I got from considering $n=2,3,4$ in Wolfram Mathematica is $$ (-1)^n(x^n-\sum_{k=2}^n (k-1)\sigma_k x^{n-k}) $$ where $\sigma_k$ is the $k$-th elementary symmetric polynomial in $a_1,\dots,a_n$.

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    $\begingroup$ Laplace expansion along the last row gives a recurrence relation of the form $D_n = -(x+a_n)D_{n-1} + (-1)^{n-1} (x+a_1)\cdots(x+a_{n-1})a_n$. You may try to prove your conjecture using mathematical induction. $\endgroup$ – user1551 Jan 3 '16 at 12:53

I start with the well-known and easily proved fact that the $n\times n$ all $1$'s matrix $$ \begin{pmatrix} 1 & 1 & \dots &1\\ 1& 1 & \dots &1\\ \vdots &\vdots &\ddots&\vdots\\ 1 & 1 &\dots &1 \end{pmatrix} $$ has one eigenvalue $n$ corresponding to the obvious eigenvector $(1,1,\dots,1)^{T}$ and the eigenvalue $0$ occurring $n-1$ times corresponding to the obvious eigenvectors $(1,-1,0,\dots, 0)^{T}$, $(1,0,-1,\dots, 0)^{T}$, ... $(1,0,0,\dots, -1)^{T}$. We will then also know the eigenvlaues of the matrix got by subtracting the identity matrix from the all-$1$'s matrix, and so we will have that $$ \det \begin{pmatrix} 0 & 1 & \dots &1\\ 1& 0 & \dots &1\\ \vdots &\vdots &\ddots&\vdots\\ 1 & 1 &\dots &0 \end{pmatrix} = (n-1)(-1)^{n-1}. $$

Now turn to the matrix $M$ of the question. To evaluate the determinant we can replace it by the matrix we get by adding the first row to every other row, that is by $$ \begin{pmatrix} -x & a_2 &\dots &a_n\\ a_1& -x & \dots & a_n\\ \vdots & \vdots &\ddots &\vdots\\ a_1 & a_2 &\dots & -x \end{pmatrix}. $$

The coefficient of $(-x)^{n-k}$ in this expansion is clearly the sum of all the principal $k\times k$ minors of $$ \begin{pmatrix} 0 & a_2 &\dots &a_n\\ a_1& 0 & \dots & a_n\\ \vdots & \vdots &\ddots &\vdots\\ a_1 & a_2 &\dots & 0 \end{pmatrix}. $$

For $k=0$ that sum is $0$; otherwise it is the symmetric sum of terms like $$ \det \begin{pmatrix} 0 & a_2 & \dots &a_k\\ a_1& 0 & \dots &a_k\\ \vdots &\vdots &\ddots&\vdots\\ a_1& a_2 &\dots &0 \end{pmatrix} = a_1 a_2\dots a_k \det \begin{pmatrix} 0 & 1 & \dots &1\\ 1& 0 & \dots &1\\ \vdots &\vdots &\ddots&\vdots\\ 1& 1 &\dots &0 \end{pmatrix} = a_1 a_2\dots a_k (-1)^{k} (k-1). $$

These calculations will provide a proof of the OP's conjecture.


We can rewrite $M$ in 2-by-2 block form as

$$M=\begin{bmatrix}-x & u^T A \\ (x+a_1)u & -x I_{n-1}-A \end{bmatrix}$$ where $u$ is a column vector of 1's and $A=\text{diag}(a_2,a_3,\cdots,a_n).$ Since the lower corner is invertible, we can use the Schur determinant identity (see for instance here) to obtain \begin{align} \det M &=\det(-x I_{n-1}-A)\det(-x-(u^T A)(-x I_{n-1}-A)^{-1}(x+a_1)u)\\ &=(-1)^{n-1}\det(xI_{n-1}+A)\cdot(-1)\left(x-(x+a_1)u^T A(x I_{n-1}+A)^{-1}u\right)\\ &=(-1)^n\prod_{j=2}^n (x+a_j)\cdot\left(x-(x+a_1)\sum_{k=2}^n a_k(x+a_k)^{-1}\right)\\ &=(-1)^n x\prod_{j=2}^n (x+a_j)-(-1)^n\sum_{k=2}^na_k(x+a_1)\prod_{j=2\\j\neq k}^n(x+a_j)\\ &=(-1)^n\prod_{j=1}^n (x+a_j)-(-1)^na_1\prod_{j=2}^n (x+a_j)-(-1)^n\sum_{k=2}^na_k\prod_{j=1\\j\neq k}^n(x+a_j)\\ &=(-1)^n\prod_{j=1}^n (x+a_j)-(-1)^n\sum_{k=1}^na_k\prod_{j=1\\j\neq k}^n(x+a_j).\\ \end{align} I suspect one could simplify further, but I'm not seeing how at the moment so I'll stop here at present. For now, I'll note that the above result is explicitly symmetric under any permutation of the $a_k$, and that the leading term of this polynomial is $(-1)^n x^n$; both of these properties agree with the form proposed by the OP, so it seems plausible that they are in fact identical. (However, until this is done, I don't consider this a complete answer.)


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