Let $A, B\in M_{n}(\mathbb{R})$ be symmetric positive definite matrices. It is easy to see $Tr(A^2+AB^2A)=Tr(A^2+BA^2B)$. Numerical experiments indicate $$Tr[(A^2+AB^2A)^{-1}]\ge Tr[(A^2+BA^2B)^{-1}],~~(1)$$ but it seems difficult to show it.

Remark. When $n=2,3$, by direct computation, (1) is true. Here is an expriment done by matlab:

for s=1:1000



{\bf Updated.} What about $A, B\in M_{n}(\mathbb{C})$ be positive definite Hermitian matrices.

  • 2
    $\begingroup$ Is there any motivation you hope that it is true? $\endgroup$
    – abcdxyz
    Apr 10, 2010 at 18:19
  • $\begingroup$ @Minh: This is a very special case for a result (I wish to establish) in quantum information, but it is still difficult to prove this special case. $\endgroup$
    – Sunni
    Apr 10, 2010 at 19:53
  • $\begingroup$ Nice inequality, nice proof! $\endgroup$
    – abcdxyz
    Apr 11, 2010 at 0:53

2 Answers 2


Note first that $A^2+AB^2A=(A+iAB)(A-iBA)$. The reverse product is $(A-iBA)(A+iAB)=A^2+BA^2B-i(BA^2-A^2B)=X-iC$. Thus, the quantity on the left is $\operatorname{Tr} (X-iC)^{-1}$ and that on the right is $\operatorname{Tr} X^{-1}$. Moreover, the self-adjoint complex matrix $X-iC$ is positive definite (as the product of an invertible operator and its adjoint). Similarly, considering the factorization $A^2+AB^2A=(A-iAB)(A+iBA)$, we can write the quantity on the left as $\operatorname{Tr} (X+iC)^{-1}$. Symmetrizing, we see that it will suffice to show that $(X-iC)^{-1}+(X+iC)^{-1}\ge 2X^{-1}$ in the sense of quadratic forms (then the inequality for traces will hold too). We can multiply by $X^{1/2}$ from both sides to reduce it to $(I-iD)^{-1}+(I+iD)^{-1}\ge 2I$ where $D=X^{-1/2}CX^{-1/2}$ and both operators on the left are positive definite. Diagonalizing the self-adjoint operator $iD$, we see that the inequality reduces to $(1+p)^{-1}+(1-p)^{-1}\ge 2$ for $p\in(-1,1)$.

  • 7
    $\begingroup$ Le plus court chemin entre deux vérités dans le domaine réel passe par le domaine complexe. ---Jacques Hadamard $\endgroup$
    – Sunni
    Apr 11, 2010 at 3:43
  • $\begingroup$ Sometimes I can be a complete idiot. See the revised version for the general case. $\endgroup$
    – fedja
    Apr 11, 2010 at 4:22

For what it is worth, a weaker conjecture is proved below.

Applying the formula for the derivative of the inverse $$d(M^{-1}) = -M^{-1}\ dM\ M^{-1},$$ to compute the t=0 derivative of the LHS of $$Tr(A^2+A(t^{1/2}B)^2A)^{-1}-Tr(A^2+(t^{1/2}B)A^2(t^{1/2}B))^{-1} \ge 0$$ gives $$Tr(A^{-2}BA^2BA^{-2})\ge Tr(A^{-1}B^2A^{-1})=Tr(BA^{-2}B).$$ Replacing $A^{-2}$ by $P$ gives the weaker conjecture that $$Tr(PBP^{-1}BP)\ge Tr(BPB)$$ for positive B and P.

PROOF OF WEAKER CONJECTURE: By the spectral theorem, we may take $P=\operatorname{Diag}(p_1,p_2,\dotsc)$. Then $$Tr(BPB)=\sum p_j |B_{ij}|^2=\sum |B_{ij}|^2 (p_i+p_j)/2 $$ and $$Tr(PBP^{-1}BP)=\sum |B_{ij}|^2 p_i^2 p_j^{-1}=\sum |B_{ij}|^2 (p_i^2 p_j^{-1}+p_i^{-1}p_j^2)/2.$$ It remains to show that $$p_i^2 p_j^{-1}+p_i^{-1}p_j^2\ge p_i+p_j$$ for positive $p_i, p_j$. By homogeneity we may take $p_i=1$. Multiplying through by $p_j$, the inequality now follows from the identity $$1+p^3-p-p^2=(p-1)^2(1+p)\ge 0.$$ $\square$

  • $\begingroup$ Can't understand anything. If the inequality is scale-invariant, how on Earth can you make B small? $\endgroup$
    – fedja
    Apr 11, 2010 at 1:38
  • $\begingroup$ Is your 'proof of weaker conjecture' means that the proof is under the assumption '$B$ is sufficiently small'? $\endgroup$
    – Sunni
    Apr 11, 2010 at 3:10
  • $\begingroup$ fedja: Sorry, there was a gap in the part where I reduced consideration to small B. I took it out, I don't currently know how to extend to prove the full conjecture. $\endgroup$
    – Jon
    Apr 11, 2010 at 3:20
  • $\begingroup$ minwalin: I clarified the weaker conjecture, it is as stated just above "PROOF." $\endgroup$
    – Jon
    Apr 11, 2010 at 3:21
  • $\begingroup$ [deleted earlier comment] $\endgroup$
    – Yemon Choi
    Apr 11, 2010 at 3:48

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.