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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.

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  • $\begingroup$ Have you tried to compute the Hessian? $\endgroup$ Oct 23, 2021 at 20:00
  • $\begingroup$ 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 $H, K \mapsto \text{ tr } \Gamma^{-1} H \Gamma^{-1} K$ $\endgroup$
    – medihde
    Oct 23, 2021 at 20:06
  • $\begingroup$ But then $(\ln \circ \det)''(\Gamma) \cdot (H,H) = tr((\Gamma^{-1}H)^2) $ and you only have to check this is negative. $\endgroup$ Oct 23, 2021 at 20:12
  • $\begingroup$ 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. $\endgroup$ Oct 23, 2021 at 20:15
  • $\begingroup$ What does the "lemma that allows me to restrict myself to a line" in your question refer to? $\endgroup$
    – user1551
    Oct 24, 2021 at 19:26

2 Answers 2

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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 $C$.

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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.

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