Proof of Minkowski determinant inequality

Question

I wonder where can I find the proof for the Minkowski determinant inequality?

( i.e., given two positive definite n x n symmetric matricies A and B, $\det(A+B)^{1/n}\ge \det(A)^{1/n}+\det(B)^{1/n}$ )

Thanks very much!

Answer 1

We may relax the assumption slightly. Suppose $A\in M_n(\mathbb C)$ is positive definite and $B\in M_n(\mathbb C)$ is positive semidefinite. Let $x_1,x_2,\ldots,x_n$ be the eigenvalues of the positive semidefinite matrix $X=A^{-1/2}BA^{-1/2}$. Then \begin{align} &\det(A+B)^{1/n}\ge\det(A)^{1/n}+\det(B)^{1/n}\\ &\iff\det(I+X)^{1/n}\ge1+\det(X)^{1/n}\\ &\iff\prod_i(1+x_i)^{1/n}\ge1+\prod_ix_i^{1/n}\\ &\iff1\ge\left(\prod_i\frac{1}{1+x_i}\right)^{1/n}+\left(\prod_i\frac{x_i}{1+x_i}\right)^{1/n}\\ \end{align} and the last line follows easily from the AM-GM inequality. Equality holds if and only if all $x_i$s are equal, i.e., if and only if $B$ is a nonnegative multiple of $A$.

By a continuity argument, Minkowski’s determinant inequality also holds when both $A$ and $B$ are positive semidefinite. However, the necessary and sufficient condition for the equality case no longer applies. E.g. neither $A=\operatorname{diag}(1,0,0)$ nor $B=\operatorname{diag}(0,1,0)$ is a scalar multiple of the other, but tie occurs here in Minkowski’s determinant inequality.

Answer 2

Because $A$ and $B$ are symmetric, so is $A+B$ and by the spectral Theorem it is diagonalizable. Let $\{e_i\}$ be the orthonormal basis for $A+B$. In this case, the determinant is the product of the eigenvalues. Also, it must be positive as all eigenvalues are positive and:

$$\det(A+B)^{1/n}=\left(\prod_{i=1}^n \langle (A+B)(e_i),e_i\rangle\right)^{1/n}=\left(\prod_{i=1}^n(\langle Ae_i,e_i\rangle+\langle Be_i,e_i \rangle)\right)^{1/n}$$

If $\langle (A+B)(e_i),e_i\rangle=0$, then the product is zero. Otherwise, it is clear from the geometric-arithmetic mean inequality that:

$$1=\frac{1}{n}\sum_{i=1}^n \frac{\langle Ae_i,e_i\rangle}{\langle (A+B)e_i,e_i\rangle}+\frac{1}{n}\sum_{i=1}^n \frac{\langle Be_i,e_i\rangle}{\langle (A+B)e_i,e_i\rangle} \geq $$ $$ \left(\prod_{i=1}^n \frac{\langle Ae_i,e_i\rangle}{\langle (A+B)e_i,e_i\rangle}\right)^{1/n}+\left(\prod_{i=1}^n \frac{\langle Be_i,e_i\rangle}{\langle (A+B)e_i,e_i\rangle}\right)^{1/n}$$

This in turn is equivalent to the following:

$$\left(\prod_{i=1}^n(\langle Ae_i,e_i\rangle+\langle Be_i,e_i \rangle)\right)^{1/n}\geq \left( \prod_{i=1}^n \langle Ae_i,e_i \rangle\right)^{1/n}+\left( \prod_{i=1}^n \langle Be_i,e_i \rangle\right)^{1/n}$$

Which also holds if $\langle (A+B)(e_i),e_i\rangle=0$ as this implies both $\langle A e_i,e_i\rangle$ and $\langle Be_i,e_i\rangle=0$ because the matrices involved are positive definite.

By Hadamard's Inequality for symmetric positive definite matrices:

$$\det(A) \leq \prod_{i=1}^n \langle A e_i,e_i\rangle$$

Using that $f(x)=x^{1/n}$ is increasing for positive $x$, then:

$$\left( \prod_{i=1}^n \langle Ae_i,e_i \rangle\right)^{1/n}+\left( \prod_{i=1}^n \langle Be_i,e_i \rangle\right)^{1/n}\geq \det(A)^{1/n}+\det(B)^{1/n}$$

This finishes the proof.

Answer 3

Thm: The function ${ X \mapsto \det(X) ^{\frac{1}{n}} }$ is concave on the cone of ${ n \times n }$ symmetric positive definite matrices ${ \text{SPD} _n . }$

Equivalently,

Thm: Let ${ A, B \in \mathbb{R} ^{n \times n} }$ be symmetric positive definite matrices, and ${ \alpha \in [0, 1] . }$ Then

$${ \det(\alpha A + (1-\alpha) B) ^{\frac{1}{n}} \geq \alpha \det(A) ^{\frac{1}{n}} + (1-\alpha) \det(B) ^{\frac{1}{n}} . }$$

Pf: Note that by simultaneous diagonalisation, there is an invertible matrix ${ P }$ such that

$${ {\begin{aligned} &\, P ^T AP = \text{diag}(\lambda _1, \ldots, \lambda _n) , \\ &\, P ^T BP = \text{diag}(\mu _1, \ldots, \mu _n) . \end{aligned}} }$$

Hence we are to show

$${ \text{To show: } \quad \left( \prod _{i=1} ^{n} ( \alpha \lambda _i + (1- \alpha) \mu _i ) \right) ^{\frac{1}{n}} \geq \alpha \left( \prod _{i=1} ^{n} \lambda _i \right) ^{\frac{1}{n}} + (1- \alpha) \left( \prod _{i=1} ^{n} \mu _i \right) ^{\frac{1}{n}} . }$$

Hence consider the function

$${ {\begin{aligned} &\, F : [0, \infty) ^n \longrightarrow \mathbb{R}, \\ &\, F(x) := (x _1 \ldots x _n) ^{\frac{1}{n}} . \end{aligned}} }$$

Note that ${ F }$ is concave, as needed.

Note that the ${ \alpha = 1/2 }$ case is the Minkowski determinant inequality.