A special case: determinant of a $n\times n$ matrix

Question

I would like to solve for the determinant of a $n\times n$ matrix $V$ defined as:

$$ V_{i,j}= \begin{cases} v_{i}+v_{j} & \text{if} & i \neq j \\[2mm] (2-\beta_{i}) v_{i} & \text{if} & i = j \\ \end{cases} $$

Here, $v_i, v_j \in(0,1)$, and $\beta_i>0$. Any suggestions or thoughts are highly appreciated.

Answer 1

By investigating the structure of the result I come to the nice general formula

$$ \left|V\right| = (-1)^n \left(1-\sum _{i=1}^n \left(\frac{2}{\beta _i}+\sum _{j=1}^{i-1} \frac{\left(v_i-v_j\right){}^2}{v_i v_j \beta _i \beta _j}\right)\right) \prod _{k=1}^n v_k \beta _k $$

If you have Mathematica you can use the following code to check the result

n = 4;
V = Table[If[i == j, (2 - β[i]) v[i], v[i] + v[j]], {i, n}, {j, n}];
(-1)^n (1 - Sum[2/β[i] + Sum[(v[i] - v[j])^2/(
   v[i] v[j] β[i] β[j]), {j, i - 1}], 
   {i, n}]) Product[v[k] β[k], {k, n}] == Det[V] // Expand
(* True *)

I believe that there is a proof for this formula but I don't find it yet.

Answer 2

Let me consider $W:=-V$; we have $$ \det(V)=(-1)^n\det(W). $$ As already noted in the comments, $W=D-(ve^T+ev^T)$, where $v:=[v_1,\ldots,v_n]^T$, $e:=[1,\ldots,1]^T$, and $D:=\mathrm{diag}(\beta_i v_i)_{i=1}^n$. Since $D>0$, we can take $$ D^{-1/2}WD^{-1/2}=I-(\tilde{v}\tilde{e}^T+\tilde{e}\tilde{v}^T), $$ where $\tilde{v}:=D^{-1/2}v$, $\tilde{e}:=D^{-1/2}e$. Note that $\det(D^{-1/2}WD^{-1/2})=\det(D)^{-1}\det(W)$ and hence $$ \det(W)=\det(D)\det(I-(\tilde{v}\tilde{e}^T+\tilde{e}\tilde{v}^T)) =\det(I-(\tilde{v}\tilde{e}^T+\tilde{e}\tilde{v}^T))\prod_{i=1}^n v_i\beta_i. $$ Since $I-(\tilde{v}\tilde{e}^T+\tilde{e}\tilde{v}^T)$ is the identity plus a symmetric rank-at-most-two matrix, its determinant can be written as $$ \det(I-(\tilde{v}\tilde{e}^T+\tilde{e}\tilde{v}^T))=(1-\lambda_1)(1-\lambda_2), $$ where $\lambda_1$ and $\lambda_2$ are the two largest (in magnitude) eigenvalues of $BJB^T$ with $$B=[\tilde{v},\tilde{e}]\quad\text{and}\quad J=\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}.$$

Since the eigenvalues of $BJB^T$ are same as the eigenvalues of $JB^TB$ (up to some uninteresting zero eigenvalues), we have that $$ \begin{split} \det(I-BJB^T) &= \det(I-JB^TB) = \det \left( \begin{bmatrix} 1-\tilde{e}^T\tilde{v} & -\tilde{e}^T\tilde{e} \\ -\tilde{v}^T\tilde{v} & 1-\tilde{v}^T\tilde{e} \end{bmatrix} \right)\\ &= \det \left( \begin{bmatrix} 1-e^TD^{-1}v & -e^TD^{-1}e \\ -v^TD^{-1}v & 1-v^TD^{-1}e \end{bmatrix} \right)\\ &= (1-e^TD^{-1}v)^2-(e^TD^{-1}e)(v^TD^{-1}v)\\ &= \left(1-\sum_{i=1}^n\frac{1}{\beta_i}\right)^2 - \left(\sum_{i=1}^n\frac{v_i}{\beta_i}\right) \left(\sum_{i=1}^n\frac{1}{v_i\beta_i}\right). \end{split} $$ Putting everything together, $$ \det(V) = (-1)^n\left[\left(1-\sum_{i=1}^n\frac{1}{\beta_i}\right)^2 - \left(\sum_{i=1}^n\frac{v_i}{\beta_i}\right) \left(\sum_{i=1}^n\frac{1}{v_i\beta_i}\right)\right]\prod_{i=1}^n\beta_iv_i. $$ I guess that after some manipulation you could arrive to the formula given in the other answer (probably by playing further with the term $(1-e^TD^{-1}v)^2-(e^TD^{-1}e)(v^TD^{-1}v)$).