Strict concavity of log determinant on positive-definite matrices

Put $H:=A_1-A_0$ and $A_s:=A_0+sH$ for $0\le s\le 1$. For every nonzero $v$, $v^{\mathsf{T}}{A}_{s}v=(1-s)v^{\mathsf{T}}{A}_{0}v+s{v}^{\mathsf{T}}{A}_{1}v>0$, so each $A_s$ is symmetric positive definite. By [F1], choose an orthogonal eigenbasis $A_s=U\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_p)U^{\mathsf{T}}$ with every ${\lambda }_j>0$. The determinant laws [F9], including the triangular formula for the diagonal matrix, give $\det A_s=(\det U{)}^2{\prod }_j{\lambda }_j$ and $(\det U{)}^2=\det (U^{\mathsf{T}}U)=\det I=1$, so $\det A_s={\prod }_j{\lambda }_j>0$. Since the determinant is a polynomial by [F2] and $\log$ is differentiable, hence continuous, on $(0,\mathrm{\infty })$ by [F4], $g(s):=\log \det A_s$ is real and continuous on $[0,1]$.

On $0<s<1$, Jacobi's formula [F3], the scalar derivative [F4], and the chain rule [F6] give $g^{\mathrm{\prime }}(s)=\mathrm{tr}(A_s^{-1}H)$.

Differentiate the expression in step 2.1 along the same affine path. The inverse derivative [F5] and linearity of the finite trace [F7] give $g^{\mathrm{\prime }\mathrm{\prime }}(s)=-\mathrm{tr}(A_s^{-1}H{A}_s^{-1}H)$.

By [F1], form the symmetric positive-definite inverse square root $A_s^{-1\mathrm{/}2}$, and set $B:=A_s^{-1\mathrm{/}2}H{A}_s^{-1\mathrm{/}2}$. Since $H$ is symmetric, $B$ is symmetric; if $A_0\mathrm{\ne }{A}_1$, then $H\mathrm{\ne }0$ and invertibility of $A_s^{-1\mathrm{/}2}$ gives $B\mathrm{\ne }0$. Expanding the finite trace sum and interchanging its indices gives $\mathrm{tr}(XY)=\mathrm{tr}(YX)$; applying this identity yields $\mathrm{tr}(A_s^{-1}H{A}_s^{-1}H)=\mathrm{tr}(B^2)={\sum }_{i,j}{B}_{ij}^2>0$. Thus $g^{\mathrm{\prime }\mathrm{\prime }}(s)<0$ for every $0<s<1$. For any $0<a<b<1$, $g^{\mathrm{\prime }}$ is continuous on $[a,b]$ and differentiable on $(a,b)$ because $g$ is twice differentiable on $(0,1)$. By [F8] applied to $g^{\mathrm{\prime }}$, there is $c\in (a,b)$ with $g^{\mathrm{\prime }}(b)-g^{\mathrm{\prime }}(a)=g^{\mathrm{\prime }\mathrm{\prime }}(c)(b-a)<0$. Hence $g^{\mathrm{\prime }}$ is strictly decreasing on $(0,1)$.

For $0<t<1$, step 1.1 shows that $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$, so [F8] applies on both $[0,t]$ and $[t,1]$. Their secant slopes are $g^{\mathrm{\prime }}(a)$ and $g^{\mathrm{\prime }}(b)$ for some $a\in (0,t)$ and $b\in (t,1)$. Step 4.1 makes $g^{\mathrm{\prime }}$ strictly decreasing, so the first slope is strictly larger than the second; rearranging yields $g(t)>(1-t)g(0)+tg(1)$. If $A_0=A_1$, the path is constant and equality holds.