A curious determinantal inequality

Question

In my study, I come across the following curious inequality, which I do not know a proof yet (so I am asking it here).

Let $A, B$ be $n\times n$ (Hermitian) positive definite matrices. It is very likely true that $$\det \left(A^{\frac{1}{2}}(A+B)A^{\frac{1}{2}}+B^{\frac{1}{2}}(A+B)B^{\frac{1}{2}}\right) \ge \det(A+B)^2. $$ Here $A^{\frac{1}{2}}$ is the unique positive definite square root of $A$. I am able to confirm the $3\times 3$ case.

Comments: Only recently did I notice that the majorization $\lambda\left(A^{\frac{1}{2}}(A+B)A^{\frac{1}{2}}+B^{\frac{1}{2}}(A+B)B^{\frac{1}{2}}\right) \prec \lambda(A+B)^2$ follows immediately by THEOREM 2 of [R.B. Bapat, V.S. Sunder, On majorization and Schur products, Linear Algebra Appl. 72 (1985) 107–117.] http://www.sciencedirect.com/science/article/pii/0024379585901478

Answer 1

Let $C := A^{1/2} (A+B) A^{1/2} + B^{1/2} (A+B) B^{1/2}$; this is a positive semi-definite matrix with the same trace as $(A+B)^2$. We show that the eigenvalues of $C$ are majorised by the eigenvalues of $(A+B)^2$, that is to say that the sum of the top $k$ eigenvalues of $C$ is at most the sum of the top $k$ eigenvalues of $(A+B)^2$ for any $k$. By the Schur concavity of $(\lambda_1,\dots,\lambda_n) \mapsto \lambda_1 \dots \lambda_n$, this gives the claimed determinantal inequality.

The sum of the top $k$ eigenvalues of $C$ can be written as $$ \hbox{tr}( C P_V )$$ where $V$ is the $k$-dimensional space spanned by the top $k$ eigenvectors of $C$. This can be rearranged as $$ \hbox{tr}( (A+B) (A^{1/2} P_V A^{1/2} + B^{1/2} P_V B^{1/2}) ). \quad\quad (*)$$

We can conjugate $A+B$ to be a diagonal matrix $\hbox{diag}(\lambda_1,\dots,\lambda_n)$ with $\lambda_1 \geq \dots \geq \lambda_n \geq 0$. In particular we have $A+B \leq \lambda_k I + D$ in the sense of positive definite matrices, where $D := \hbox{diag}(\lambda_1-\lambda_k, \dots, \lambda_{k-1}-\lambda_k, 0, \dots, 0)$. Using the fact that $\hbox{tr}(XZ) \leq \hbox{tr}(YZ)$ whenever $X,Y,Z$ are positive semi-definite with $X \leq Y$, we can bound (*) by

$$ \hbox{tr}( (\lambda_k I + D) (A^{1/2} P_V A^{1/2} + B^{1/2} P_V B^{1/2}) )$$

which rearranges as

$$ \lambda_k \hbox{tr}( (A+B) P_V ) + \hbox{tr}( P_V (A^{1/2} D A^{1/2} + B^{1/2} D B^{1/2}) ).$$ Using $A+B \leq \lambda_k I + D$ for the first term and $P_V \leq I$ for the second term, this is bounded by $$ \lambda_k^2 \hbox{tr}( P_V ) + \lambda_k \hbox{tr}( D P_V ) + \hbox{tr}( A^{1/2} D A^{1/2} + B^{1/2} D B^{1/2} ).$$ For the second term we use $P_V \leq 1$, and for the third term we use the cyclic property of trace to bound by $$ \lambda_k^2 \hbox{tr}( P_V ) + \lambda_k \hbox{tr}( D ) + \hbox{tr}( (A+B) D ).$$ For the first term we write $\hbox{tr}(P_V) = k = \hbox{tr}(P_W)$, where $W$ is the span of the first $k$ basis vectors $e_1,\dots,e_k$. For the third term we use $A+B \leq \lambda_k I + D$ to bound the above by $$ \lambda_k^2 \hbox{tr}( P_W ) + \lambda_k \hbox{tr}( D ) + \hbox{tr}( (\lambda_k I + D) D ).$$ Since $D = P_W D P_W$, we can collect terms to obtain $$ \hbox{tr}( P_W (\lambda_k I + D)^2 P_W ).$$ But by construction, $P_W (\lambda_k I + D)^2 P_W = \hbox{diag}( \lambda_1^2, \dots, \lambda_k^2, 0, \dots, 0 )$, so we have bounded (*) by the sum of the top $k$ eigenvalues of $(A+B)^2$, as required.

Answer 2

EDIT: the argument below is not correct, but I am leaving it here in case it is of use in locating a better solution.

By a limiting argument we may assume that $C := A+B$ is invertible. If we write

$$ D := C^{-1/4} A^{1/2} C^{-1/4} $$ and $$ E := C^{-1/4} B^{1/2} C^{-1/4} $$

then $D,E$ are positive semi-definite with $D^2+E^2=1$ EDIT: as pointed out in comments, this is not correct, so in particular $D,E$ commute. The inequality can now be written in terms of $C,D,E$ as $$ \det( C^{1/4} D C^{3/2} D C^{1/4} + C^{1/4} E C^{3/2} E C^{1/4} ) \geq \det( C^2 )$$ which on multiplying on left and right by $C^{-1/4}$ and setting $F := C^{3/2}$ becomes $$ \det( D F D + E F E ) \geq \det( F ).$$ Now observe that the matrix $$ \begin{pmatrix} D & E \\ -E & D \end{pmatrix} \begin{pmatrix} F & 0 \\ 0 & F \end{pmatrix} \begin{pmatrix} D & -E \\ E & D \end{pmatrix} = \begin{pmatrix} DFD + EFE & EFD-DFE \\ DFE-EFD & DFD+EFE \end{pmatrix}$$ is positive semi-definite and has determinant $\det(F)^2$ (the first and last matrices on the LHS are orthogonal). Passing to the block-diagonal matrix $$ \begin{pmatrix} DFD + EFE & 0 \\ 0 & DFD+EFE \end{pmatrix},$$ which is still positive semi-definite, the eigenvalues here are majorized by the previous matrix (by the Schur-Horn theorem), and so (by the Schur concavity of the product function $(\lambda_1,\dots,\lambda_n) \mapsto \lambda_1 \dots \lambda_n$), the determinant of the latter matrix must be at least as large as the determinant of the former. (This inequality can also be established using Schur complements.) Thus $$ \det( DFD + EFE )^2 \geq \det(F)^2 $$ and the claim follows.

Answer 3

Here is a complementary approach without using majorization. The answer is partial because it has an open "TODO". I am writing it down here already in case someone wishes to complete the argument.

---

Let $A, B, X, Y > 0$. It is easy to show using Schur complements that \begin{equation*} \tag{$*$} AX^{-1}A + BY^{-1}B \ge (A+B)(X+Y)^{-1}(A+B). \end{equation*} From $(*)$ it follows that $\det(AX^{-1}A + BY^{-1}B)\det(X+Y)\ge \det(A+B)^2$.

Let $C=A^{1/2}(A+B)A^{1/2}$ and $D=B^{1/2}(A+B)B^{1/2}$. If we can find (TODO) $X$ and $Y$ such that \begin{equation*} X \gets A(X+Y)^{1/2}C^{-1}(X+Y)^{1/2}A,\quad Y \gets B(X+Y)^{1/2}D^{-1}(X+Y)^{1/2}B, \end{equation*} then we will obtain $$(X+Y)^{1/2}(AX^{-1}A + BY^{-1}B)(X+Y)^{1/2} = C+D = A^{1/2}(A+B)A^{1/2} + B^{1/2}(A+B)B^{1/2}.$$ Combining this identity with the above inequality immediately implies the desired inequality.

Notice that in particular, if $A$ and $B$ commute, then $X=A(A+B)^{-1}$ and $Y=B(A+B)^{-1}$ is a solution.