The positive semidefinite cone is self-dual and its log determinant barrier has parameter n

The set ${\mathbb{S}}_{+}^n$ is a closed convex cone because quadratic-form inequalities are preserved by nonnegative sums and limits. If both $X$ and $-X$ are positive semidefinite, their quadratic forms vanish on every vector; testing $e_i$ gives $X_{ii}=0$, then testing $e_i+e_j$ gives $X_{ij}=0$, so the cone is pointed. The identity is interior: diagonalize $E=E^T=U\mathrm{diag}({\lambda }_i)U^T$. From the entry definition, $\mathrm{\parallel }E{\mathrm{\parallel }}_F^2={\sum }_{i,j}{E}_{ij}^2=\mathrm{tr}(E^2)$; the finite sums give $\mathrm{tr}(AB)={\sum }_{i,j}{A}_{ij}{B}_{ji}={\sum }_{j,i}{B}_{ji}{A}_{ij}=\mathrm{tr}(BA)$, so $\mathrm{tr}(E^2)=\mathrm{tr}(\mathrm{diag}({\lambda }_i^2))={\sum }_i{\lambda }_i^2$. Thus $\mathrm{\parallel }E{\mathrm{\parallel }}_F<1$ implies $\mathrm{\mid }{\lambda }_i\mathrm{\mid }<1$ for every $i$, hence $I+E\succ 0$, proving that $I$ is interior and the cone is proper. If $X,Y⪰0$, their positive square roots exist by [F1]. The same trace identity gives $\mathrm{tr}(XY)=\mathrm{tr}(X^{1\mathrm{/}2}Y{X}^{1\mathrm{/}2})\ge 0$, since $X^{1\mathrm{/}2}Y{X}^{1\mathrm{/}2}⪰0$ and its diagonal entries are nonnegative. Thus ${\mathbb{S}}_{+}^n\subseteq ({\mathbb{S}}_{+}^n{)}^{\ast }$. Conversely, if $Y$ is in the dual cone under [F2], testing against $v{v}^T⪰0$ gives $v^{T}Yv=\mathrm{tr}(Yv{v}^T)\ge 0$ for every $v$, so $Y⪰0$.

For $X\succ 0$ and symmetric $H$, set $Z=X^{-1\mathrm{/}2}H{X}^{-1\mathrm{/}2}$. Then $\det (X+sH)=\det (X)\det (I+sZ)$. Diagonalizing the symmetric $Z$ and differentiating the scalar sum $-\log \det (X+sH)=-\log \det X-{\sum }_i\log (1+s{\lambda }_i(Z))$ gives $DF(X)[H]=-\mathrm{tr}(X^{-1}H)$ and $D^{2}F(X)[H,H]=\mathrm{tr}(Z^2)$. Polarization gives the displayed bilinear Hessian formula. Equivalently the first derivative follows from [F3], and differentiating it uses [F4]–[F5].

Differentiating the inverse formula once more gives $D^{3}F(X)[U,V,V]=-2\mathrm{tr}(Z_U{Z}_V^2)$, where $Z_U=X^{-1\mathrm{/}2}U{X}^{-1\mathrm{/}2}$ and $Z_V=X^{-1\mathrm{/}2}V{X}^{-1\mathrm{/}2}$. By [F8] after vectorizing matrices, $\mathrm{\mid }\mathrm{tr}(Z_U{Z}_V^2)\mathrm{\mid }\le \mathrm{\parallel }{Z}_U{\mathrm{\parallel }}_F\mathrm{\parallel }{Z}_V^2{\mathrm{\parallel }}_F$. Diagonalizing the symmetric $Z_V$ gives $\mathrm{\parallel }{Z}_V^2{\mathrm{\parallel }}_F^2={\sum }_i{\lambda }_i^4\le ({\sum }_i{\lambda }_i^2{)}^2=\mathrm{\parallel }{Z}_V{\mathrm{\parallel }}_F^4$. The latter Frobenius norms are exactly the local norms, proving the mixed self-concordance inequality. Also $\mathrm{\parallel }Z{\mathrm{\parallel }}_2={\max }_i\mathrm{\mid }{\lambda }_i(Z)\mathrm{\mid }\le ({\sum }_i{\lambda }_i(Z{)}^2{)}^{1\mathrm{/}2}=\mathrm{\parallel }Z{\mathrm{\parallel }}_F$, so if $\mathrm{\parallel }H{\mathrm{\parallel }}_X=\mathrm{\parallel }Z{\mathrm{\parallel }}_F<1$, then $I+Z\succ 0$ and $X+H\succ 0$.

Scaling gives $F(\alpha X)=F(X)-n\log \alpha$. The local dual norm satisfies $\mathrm{\parallel }DF(X){\mathrm{\parallel }}_{X,\ast }^2=\mathrm{tr}(I_n)=n$. If a sequence tends to a finite boundary matrix, its determinant tends to zero through positive values, so $F\to +\mathrm{\infty }$. These facts, [F6], and the positive-definite Hessian in step 1.2 establish the stated barrier properties.