= Second derivative of the determinant
{title2=$D^2\det_I(H,K)=\operatorname{tr}H\operatorname{tr}K-\operatorname{tr}(HK)$}
= Determinant Hessian
{synonym}
The <derivative of the determinant> at an invertible <matrix> is $D\det_A(K)=\det A\operatorname{tr}(A^{-1}K)$. At $A=I+H$, use $\det(I+H)=1+\operatorname{tr}H+O(\|H\|^2)$ and $(I+H)^{-1}=I-H+O(\|H\|^2)$. Differentiating the first derivative gives the displayed symmetric <bilinear map>. More generally $D^2\det_A(H,K)=\det A[\operatorname{tr}(A^{-1}H)\operatorname{tr}(A^{-1}K)-\operatorname{tr}(A^{-1}HA^{-1}K)]$.
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