# Concavity of $\ln \det$

I'm trying to prove that function $$A \mapsto \ln\det A$$ is concave on the set of symmetric definite positive matrices $$\mathcal{S}_d^{++}(\mathbb{R})$$ but without using the lemma that allows me to restrict myself to a line.

Hence, I'm trying to prove that $$\forall t \in [0, 1]$$, $$\forall \Gamma, \Sigma \in \mathcal{S}_d^{++}(\mathbb{R})$$ we have

$$\ln\det(t\Gamma + (1 - t)\Sigma) \geq t \ln\det \Gamma + (1 - t)\ln\det\Sigma$$ i.e. $$\det(t\Gamma + (1 - t)\Sigma) \geq (\det \Gamma)^t (\det \Sigma)^{1 - t}$$

If $$\Gamma$$ and $$\Sigma$$ commute then they share a common basis of eigenvectors and the inequality reduces to proving

$$\displaystyle \prod_{i = 1}^d (t \gamma_i + (1 - t)\sigma_i) \geq \left(\prod_{i = 1}^d \gamma_i\right)^t \left(\prod_{i = 1}^d \sigma_i\right)^{1 - t}$$

which is a corollary of inequality $$\forall u, v > 0$$, $$\forall p, q$$ conjugates, $$uv \leq \frac{u^p}p + \frac{v^q}q$$. However, if they do not commute I'm lost.

• Have you tried to compute the Hessian? Oct 23, 2021 at 20:00
• I did try - I found the first differential to be, at point $\Gamma \in \mathcal{S}_d^{++}(\mathbb{R})$, $H \mapsto \text{ tr } \Gamma^{-1}H$ and the second one to be $H, K \mapsto \text{ tr } \Gamma^{-1} H \Gamma^{-1} K$ Oct 23, 2021 at 20:06
• But then $(\ln \circ \det)''(\Gamma) \cdot (H,H) = tr((\Gamma^{-1}H)^2)$ and you only have to check this is negative. Oct 23, 2021 at 20:12
• Which is weird because if you plug in $H = \Gamma$ you get $n > 0$. Since a function is concave if and only if its second derivative is negative everywhere I think either your function is not concave, either this second derivative is false, either I made a mistake. Oct 23, 2021 at 20:15
• What does the "lemma that allows me to restrict myself to a line" in your question refer to? Oct 24, 2021 at 19:26

For any two positive definite matrices $$A$$ and $$B$$, we want to prove that for every $$t\in[0,1]$$, $$\log\det\left(tA+(1-t)B\right) \ge t\log\det(A)+(1-t)\log\det(B).$$ Let $$C=A^{-1/2}BA^{-1/2}$$. The LHS above is equal to $$\log\det\left(A^{1/2}\left(tI+(1-t)C\right)A^{1/2}\right) =\log\det(A)+\log\det\left(tI+(1-t)C\right)$$ while the RHS is equal to $$t\log\det(A)+(1-t)\log\det(A^{1/2}CA^{1/2}) =\log\det(A)+(1-t)\log\det(C).$$ Therefore the inequality is equivalent to $$\log\det\left(tI+(1-t)C\right) \ge(1-t)\log\det(C),$$ which is true because by the concavity of the logarithm function on the positive real line, $$\log(t+(1-t)\lambda) \ge(1-t)\log(\lambda)$$ for every eigenvalue $$\lambda$$ of $$C$$.
It's enough to show that $$\det \left(\frac{A^2 + B^2}{2}\right) \ge \det A B$$ if $$A$$, $$B$$ are positive definite matrices. Now, if $$A$$, $$B$$ commute then we have $$A^2 + B^2 - 2 AB = (A-B)^2 \succeq 0$$, so we have the inequality $$\frac{A^2+B^2}{2}\succeq AB$$, so the corresponding inequality for the determinants. But we can still reduce to one of the matrices being $$I$$, using the idea of @user1551. Indeed we have $$\det\frac{A^2+B^2}{2} = \det A \det \frac{I+A^{-1} B^2 A^{-1}}{2} \det A = \det A^2 \cdot \det \frac{I+A^{-1} B^2 A^{-1}}{2}$$ while $$\det A B = \det A^2 \cdot \det \sqrt{A^{-1} B^2 A^{-1}}$$
Now we have indeed $$\frac{I + A^{-1} B^2 A^{-1}}{2} \succeq \sqrt{A^{-1} B^2 A^{-1} }$$ and so the inequality for determinants.