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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

as=randn(4,4);
bs=randn(4,4);
ts=as*as';
rs=bs*bs';
ls=trace(inv(ts^2+ts*rs^2*ts)-inv(ts^2+rs*ts^2*rs))

end

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

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2  
Is there any motivation you hope that it is true? –  Tran Chieu Minh Apr 10 '10 at 18:19
    
@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. –  Sunni Apr 10 '10 at 19:53
    
Nice inequality, nice proof! –  Tran Chieu Minh Apr 11 '10 at 0:53

2 Answers 2

up vote 14 down vote accepted

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 $p\in(-1,1)$.

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4  
Le plus court chemin entre deux vérités dans le domaine réel passe par le domaine complexe. ---Jacques Hadamard –  Sunni Apr 11 '10 at 3:43
    
Sometimes I can be a complete idiot. See the revised version for the general case. –  fedja Apr 11 '10 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=Diag($p_1,p_2,...$). Then $$Tr(BPB)=\Sigma p_j |B_{ij}|^2=\Sigma |B_{ij}|^2 (p_i+p_j)/2 $$ and $$Tr\(PBP^{-1}BP\)=\Sigma |B_{ij}|^2 p_i^2 p_j^{-1}=\Sigma |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$$ for positive $p_{i,j}$. By homogeneity we may take $p_i=1$. Multiplying through by $p_j$, the inequality now follows from the identity $$1+p^3-p-p^2=(p-1)^2(1+p)\ge 0.$$ $\square$

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Can't understand anything. If the inequality is scale-invariant, how on Earth can you make B small? –  fedja Apr 11 '10 at 1:38
    
Is your 'proof of weaker conjecture' means that the proof is under the assumption '$B$ is sufficiently small'? –  Sunni Apr 11 '10 at 3:10
    
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. –  Jon Apr 11 '10 at 3:20
    
minwalin: I clarified the weaker conjecture, it is as stated just above "PROOF." –  Jon Apr 11 '10 at 3:21
    
[deleted earlier comment] –  Yemon Choi Apr 11 '10 at 3:48

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