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

  • $\begingroup$ Does it read $(\det(A+B))^2$ or $\det((A+B)^2)$? $\endgroup$
    – Moritz
    Jun 26, 2015 at 20:59
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    $\begingroup$ @Moritz: It doesn't matter since $\det(X)^2=\det(X^2)$. $\endgroup$
    – GH from MO
    Jun 26, 2015 at 21:02
  • $\begingroup$ Notice equality occurs if A=I. Can you rewrite (A+B) as A(I + A^(-1)B) and find appropriate square roots of A^-1 B ? $\endgroup$ Jun 27, 2015 at 4:03
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    $\begingroup$ Equality occurs if A B commute. $\endgroup$
    – Russel
    Jun 27, 2015 at 4:05
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    $\begingroup$ @GerhardPaseman: I think there would be lots of similar inequalities. The one I am asking may be the "simplest" unknown case. $\endgroup$
    – M. Lin
    Jun 27, 2015 at 20:26

3 Answers 3


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.

  • $\begingroup$ Great, thank you for your detailed and clean proof. I use majorization a lot in my study, but still I do not play it at the same level as you do. After reading your proof, I started asking why I did not find a proof myself. Aha, I failed to observe the "key" step $A+B \leq \lambda_k I + D$. Has a similar construction $D$ been used before? Maybe in this context, such a trick is novel. $\endgroup$
    – M. Lin
    Jun 28, 2015 at 2:35
  • $\begingroup$ My original goal is to prove the majorization relation, then as a byproduct, the determinantal inequality. Could the determinantal inequality be proved without appealing to majorization? I guess other MO readers would like to have such attempts. $\endgroup$
    – M. Lin
    Jun 28, 2015 at 2:39
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    $\begingroup$ I found this argument after playing around with the k=1 and k=2 cases for a while. Roughly speaking, $\lambda_k I + D$ represents the "largest" or "worst" that $A+B$ can be if one only constrains the top $k$ eigenvalues of $A+B$, which is what one is doing when trying to prove majorisation. (When $k=1$, $D$ is not present, and when $k=2$, $D$ is a rank one operator; after seeing these two cases I was able to extrapolate to the general case.) $\endgroup$
    – Terry Tao
    Jun 28, 2015 at 2:51
  • $\begingroup$ Very nice proof! It would be interesting to see a "coordinate free" proof in the spirit of the Schur-Horn theorem, but probably it would not be any simpler than this one (especially that your proof is self-contained). $\endgroup$
    – GH from MO
    Jun 28, 2015 at 11:45

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.

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    $\begingroup$ Dear Terry, I don't see that $D^2+E^2=1$, can you please give more detail? $\endgroup$
    – GH from MO
    Jun 27, 2015 at 10:52
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    $\begingroup$ I doubt too that $D^2+E^2$ equals $1$. 14 votes pro without a verification ? $\endgroup$ Jun 27, 2015 at 13:29
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    $\begingroup$ I tried with the choice $A^{1/2}={\rm diag}(2,1)$ and $B^{1/2}=\begin{pmatrix} 1 & 1 \\ 1 & 2 \end{pmatrix}$, which fails to pass the test $D^2+E^2=1$. $\endgroup$ Jun 27, 2015 at 13:47
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    $\begingroup$ You are right, of course; I had mistakenly identified a vector space with its dual when thinking about the problem, which translated into the sign error here when converted back into matrices. I can establish the weaker inequality $\det(A^{1/2} (A^2+B^2) A^{1/2} + B^{1/2} (A^2+B^2) B^{1/2}) \geq \det(A^2+B^2) \det(A^2 + A B A^{-1} B)$ with this approach, but it does not appear strong enough to recover the full inequality. $\endgroup$
    – Terry Tao
    Jun 27, 2015 at 17:04
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    $\begingroup$ Apply Fischer's inequality to $\begin{pmatrix} A^{1/2} & B^{1/2} \\ -B^{1/2} & A^{1/2} \end{pmatrix} \begin{pmatrix} A + B & 0 \\ 0 & A+B \end{pmatrix} \begin{pmatrix} A^{1/2} & -B^{1/2} \\ B^{1/2} & A^{1/2} \end{pmatrix}$. By Schur complement, the first and last matrix have determinant $\det( A + A^{1/2} B^{1/2} A^{-1/2} B^{1/2} )$, giving $\det( A^{1/2} (A+B) A^{1/2} + B^{1/2} (A+B) B^{1/2} ) \geq \det(A+B) \det( A + A^{1/2} B^{1/2} A^{-1/2} B^{1/2} )$ (I had some typos in the previous inequality as I had changed notation in my computations by squaring $A,B$). $\endgroup$
    – Terry Tao
    Jun 27, 2015 at 17:14

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.


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