Second derivative of the determinant ($D^2{\det }_I(H,K)=\mathrm{tr}H\mathrm{tr}K-\mathrm{tr}(HK)$)

The derivative of the determinant at an invertible matrix is $D{\det }_A(K)=\det A\mathrm{tr}(A^{-1}K)$. At $A=I+H$, use $\det (I+H)=1+\mathrm{tr}H+O(\Vert H{\Vert }^2)$ and $(I+H{)}^{-1}=I-H+O(\Vert H{\Vert }^2)$. Differentiating the first derivative gives the displayed symmetric bilinear map. More generally $D^2{\det }_A(H,K)=\det A[\mathrm{tr}(A^{-1}H)\mathrm{tr}(A^{-1}K)-\mathrm{tr}(A^{-1}H{A}^{-1}K)]$.