Lemma 7.8.1.BB19C89586B1 Let $A$ be a positive semidefinite Hermitian $n\times n$ matrix. Then
\[ (\det A)^{1/n}=\inf \left\{ \frac{\operatorname {tr}(BA)}n:\ B\text{ positive definite Hermitian},\ \det B=1\right\} . \]
If $C$ is another positive semidefinite Hermitian $n\times n$ matrix, define the numbers $M_ k(A,C)$, $0\le k\le n$, by
\[ \det (tA+sC)=\sum _{k=0}^ n\binom nk M_ k(A,C)\, t^ ks^{n-k}\qquad (t,s\in \mathbb R). \]
Then $M_ n(A,C)=\det A$, $M_0(A,C)=\det C$, and for each integer $1\le k\le n-1$,
\[ M_ k(A,C)\ge (\det A)^{k/n}(\det C)^{(n-k)/n}. \]
Proof.
Let $B$ be positive definite with $\det B=1$. The positive semidefinite matrix $B^{1/2}AB^{1/2}$ has trace $\operatorname {tr}(BA)$ and determinant $\det A$, so the arithmetic–geometric mean inequality for its eigenvalues gives $\operatorname {tr}(BA)/n\ge (\det A)^{1/n}$. If $A$ is positive definite, equality holds for $B=(\det A)^{1/n}A^{-1}$. If $A$ is singular, the matrices $B_\varepsilon =\det (A+\varepsilon I)^{1/n}(A+\varepsilon I)^{-1}$, $\varepsilon >0$, are admissible and
\[ 0\le \operatorname {tr}(B_\varepsilon A)\le \operatorname {tr}\bigl (B_\varepsilon (A+\varepsilon I)\bigr )=n\det (A+\varepsilon I)^{1/n}\longrightarrow 0 \]
as $\varepsilon \downarrow 0$.
The formulas for $M_ n$ and $M_0$ follow from $\det (tA)=t^ n\det A$ and $\det (sC)=s^ n\det C$. Suppose first that $A$ and $C$ are positive definite, and let $\lambda _1,\ldots ,\lambda _ n>0$ be the eigenvalues of $A^{-1/2}CA^{-1/2}$. Since $\det (tA+sC)=\det A\prod _{i=1}^ n(t+s\lambda _ i)$, comparison of coefficients gives
\[ M_ k(A,C)=\frac{\det A}{\binom nk}\sum _{|J|=n-k}\prod _{j\in J}\lambda _ j, \]
the sum being over the subsets $J\subset \{ 1,\ldots ,n\} $ with $n-k$ elements. Each eigenvalue occurs in exactly $\binom {n-1}{n-k-1}$ of these $\binom n{n-k}$ products, so the geometric mean of the products is $(\lambda _1\cdots \lambda _ n)^{(n-k)/n}$. Since $\lambda _1\cdots \lambda _ n=\det C/\det A$, the arithmetic–geometric mean inequality gives
\[ M_ k(A,C)\ge \det A\left(\frac{\det C}{\det A}\right)^{(n-k)/n}=(\det A)^{k/n}(\det C)^{(n-k)/n}. \]
For semidefinite $A$ and $C$, apply this to $A+\varepsilon I$ and $C+\varepsilon I$ and let $\varepsilon \downarrow 0$; both sides depend continuously on the matrices.
$\square$
Pome
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