# Determinant of Hadamard product / sum of matrices (one diagonal)

I am trying to compute the determinant of $$\boldsymbol{W}\odot \boldsymbol{S}$$, where $$\boldsymbol{S} \in PD(p)$$ positive semidefinite matrix and $$\boldsymbol{W}$$ is a matrix whose diagonal entries $$w_ {jj}\geq 1$$, and off diagonal entries $$w_ {ij}=1$$, $$\forall i\neq j$$; $$i,j=1,\ldots,p$$. Alternatively, the problem can be recast as follows: compute the determinant of $$\boldsymbol{A} + \boldsymbol{S}$$, where $$\boldsymbol{A}=diag(a_{1},\ldots,a_{p})$$ with $$a_j=s_{jj}(w_{jj}-1)$$. Is there a way the sought determinant can be expressed as a function of the quantities involved? Thanks.

• This is just a determinant of an arbitrary symmetric positive semidefinite matrix (whose entries just happen to be called $s_{ij}$ when they are off-diagonal and $s_{jj} w_{jj}$ when they are on-diagonal), right? I don't think there is a formula for that. – darij grinberg Feb 19 at 17:33
• Thanks for your answer. Exactly, I wanted to know if with such a decomposition it may be possible to rewrite and/or simplify the results. – AndreaC Feb 19 at 17:41
• If you had any formula like that, you could always set $w_{jj} = 1$ and get a formula for the determinant of an arbitrary symmetric positive semidefinite matrix. Since (I assume) you want an algebraic formula, the positive semidefiniteness doesn't actually matter, so it would be a formula for the determinant of an arbitrary symmetric matrix. Over 1 century of linear algebra and combinatorics hasn't managed to produce such a formula that would be better than the classical "sum over permutations" (in particular, it does not factor as a polynomial). – darij grinberg Feb 19 at 17:43