# Show that $\log(\det(H_1)) ≤ \log(\det(H_2)) + \operatorname{tr}[H^{-1}_2H_1]−N$ for all positive semidefinite matrices $H_1,H_2 \in C^N$

Show that $\log(\det(H_1)) ≤ \log(\det(H_2)) + \operatorname{tr}[H^{-1}_2H_1]−N$ for all positive semidefinite matrices $H_1,H_2 \in C^N$.

We know that all positive semidefinite matrices are singular and so the determinant is zero and as such they are not invertible. It is clear from the expression that $\log(\det(H_1)) = \log(\det(H_2)) = \log(0) = -\infty$.

Also in the right hand side, inverse of $H_2$ does not exist.

I would be grateful if someone can through some light how to proceed with this proof. Is there any specific property of positive semidefinite matrix to handle this?

• It is not true that all positive semidefinite matrices are singular. Positive semidefinite means $\det H \ge 0$, which leaves open the possibility that $\det H = 0$ or $\det H > 0$. You are correct however that if $H_{i}$ are singular then both sides are meaningless. What do you think about the case $\det H >0$? Commented Jun 15, 2018 at 12:34
• @DanielLittlewood Thanks for the reply Daniel. Basically, for a positive semidefinite matrix the eigenvalues $\lambda_i >= 0$. If we say that the eigenvalues are strictly $>0$, then it will become a positive definite matrix. Also the determinant of a matrix is equal to product of its eigenvalues. So for a positive semidefinite matrix, the det $=\prod \lambda_i = 0$ because at least one of the eigenvalues have to be zero. But in the case of positive definite since all the eigenvalues are greater than 0, so the determinant if positive. Commented Jun 15, 2018 at 12:38
• I notice that the $\text{tr}(H_{2}^{-1} H_{1})$ is bound up in the trace and can't be split up easily. Perhaps the $H_{1}$ and $H_{2}$ can be brought together inside the determinant? (I recommend restricting to the case where $H_{1},H_{2}$ are positive definite at this point). Commented Jun 15, 2018 at 13:09

If $$H_1$$ and $$H_2$$ are positive (i.e., $$>0$$), the inequality is equivalent to $$\log(\det(H_2^{-1}H_1))\leq \mathrm{Tr}(H_2^{-1}H_1) - N.$$ Notice that $$\det(H_2^{-1}H_1) = \det(H_2^{-\frac12}H_1 H_2^{-\frac12})$$ and $$\mathrm{Tr}(H_2^{-1}H_1) = \mathrm{Tr}(H_2^{-\frac12}H_1 H_2^{-\frac12})$$. Denote $$H = H_2^{-\frac12}H_1 H_2^{-\frac12}$$, the inequality is equivalent to $$\log \det(H) \leq \mathrm{Tr}(H) -N.$$ $$H$$ is positive, let $$a_1, \ldots, a_N$$ denote its eigenvalues. Then $$\det(H) = a_1 \cdots a_N$$ and $$\mathrm{Tr}(H) = a_1+ \cdots + a_N$$. The inequality is followed from the elementary inequaly $$\log x \leq x-1$$ for $$x >0$$.

• How did u combine the two determinant function? Did not get that part @nguyeno610 Commented Jun 18, 2018 at 14:25
• You have $\log \det H - \log \det G = \log \det(H)/\det(G)$. Using $\det(AB)=\det(B) \det(A)$ with $B = G^{-1} H, A = G$ yields $\log \det G^{-1} H$. Commented May 28 at 20:00

We consider that $H_1,H_2\in S_n^{>0}$ (they are $>0$); the other cases have not any interest.

Let $f:Z\in S_n^{>0}\rightarrow tr(H_2^{-1}Z)-\log(\det(Z))+\log(\det(H_2)-n$. We show that the minimum of $f$ is $0$.

The derivative is $Df_{Z}:K\in S_n\rightarrow tr(H_2^{-1}K)-tr(KZ^{-1})=tr(K(H_2^{-1}-Z^{-1}))$ (indeed, the tangent space to $S_n^{>0}$ in $Z$ is $S_n$, the space of symmetric matrices). Thus $Df_{Z}=0$ iff for every symmetric $K$, $tr(K(H_2^{-1}-Z^{-1}))=0$.

Finally, $H_2^{-1}-Z^{-1}$ is in the orthogonal of $S_n$, that is the space of skew symmetric matrices. That implies $H_2^{-1}=Z^{-1}$ or $Z=H_2$.

Then, if $f$ admits a local extremum, it is necessarily $f(H_2)=0$.

It suffices to show that $f$ is convex (note that $S_n^{>0}$ is convex).

The second derivative is

$D^2f_Z(K,L)=tr(KZ^{-1}LZ^{-1})$, where $K,L\in S_n$.

Then $D^2f_Z(K,K)=tr((KZ^{-1})^2)$. Since $Z^{-1}>0$ and $K\in S_n$, $KZ^{-1}$ has only real eigenvalues. Consequently $tr((KZ^{-1})^2)\geq 0$, $D^2f_Z(K,K)\geq 0$ and we are done. $\square$

This is really just a consequence of the concavity of $$H \mapsto \log \det H$$. Indeed, for a differentiable concave function $$\phi$$, we have $$\phi(H) \leq \phi(H') + \nabla \phi(H')^T(H- H').$$ Using $$\phi(H) = \log \det H$$, we have $$\nabla \phi(H) = \nabla_H \log \det H = H^{-1},$$ from which it immediately follows that \begin{align*} \log \det H_1 &= \phi(H_1) \\ &\leq \phi(H_2) + \nabla \phi(H_2) ^T (H_1 - H_2) \\ &= \log \det H_2 + \mathrm{tr}(H_2^{-1}(H_1 - H_2)). \end{align*} Using $$H_2^{-1}H_2 = I_N$$, we have $$\log \det H_1 \leq \log \det H_2 + \mathrm{tr}(H_2^{-1}H_1) - N,$$ as required.