# Determinant of the sum of a positive semi-definite matrix and a diagonal matrix

Is it possible to compute $$\text{det} (A^2 - D)$$ in terms of $$\text{det}\, A$$ and $$\det\, D$$? Here, $$A$$ is a positive semidefinite matrix with the property that each of its diagonal entries is equal to the negative of the sum of its off-diagonal entries $$a_{ii} = -\sum_{j \neq i} a_{ij}$$. Also, $$D$$ is a diagonal positive semi-definite matrix.

If it is not possible, can the determinant be simplified using the above information?

• Is this sum over $j$, or over $i$ and $j$, s.t they're different? – enedil Mar 6 at 18:35

• Consider $$A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix},$$ which satisfies the condition on $$A$$ you had. Note that the condition implies $${\rm det}(A)=0$$, since $$A \cdot [1, ..., 1]^T = 0$$, which implies $$A$$ is singular.
• Now consider $$D_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$ and $$D_2 = \begin{bmatrix} 1/2 & 0 \\ 0 & 2 \end{bmatrix}.$$ Clearly, $${\rm det}(D_1) = {\rm det}(D_2) = 1$$.
• However, $${\rm det}(A^2-D_1) = -3$$ and $${\rm det}(A^2-D_2) = -4$$.
Therefore, only knowing $${\rm det}(A)$$ and $${\rm det}(D)$$, you cannot infer $${\rm det}(A^2-D).$$