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
  • 14
    $\begingroup$ @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
  • 2
    $\begingroup$ Equality occurs if A B commute. $\endgroup$
    – Russel
    Jun 27, 2015 at 4:05
  • 1
    $\begingroup$ @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
  • 8
    $\begingroup$ 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.

  • 5
    $\begingroup$ 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
  • 10
    $\begingroup$ I doubt too that $D^2+E^2$ equals $1$. 14 votes pro without a verification ? $\endgroup$ Jun 27, 2015 at 13:29
  • 8
    $\begingroup$ 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
  • 7
    $\begingroup$ 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
  • 2
    $\begingroup$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.