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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?

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  • $\begingroup$ Is this sum over $j$, or over $i$ and $j$, s.t they're different? $\endgroup$ – enedil Mar 6 at 18:35
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No, here is a counter example:

  • 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 ${\rm det}(A^2-D).$

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