# An Operator Arithmetic-Geometric Mean Inequality

Bhatia and Kittaneh studied operator arithmetic-geometric mean inequalities over a twenty year period and wrote three papers on the subject. The following theorem answers a question that they posed.

THEOREM 1 For P and Q positive semidefinite n \times n matrices, the inequality
\sqrt{\sigma_r(PQ)} \leq \frac{1}{2} \lambda_r(P + Q)
holds for r=1,2,\ldots,n.

In fact, Bhatia and Kittaneh settled this question positively in the two special cases r=1 and r=n. This theorem has now been proved and the proof is to appear in Linear Algebra and its Applications. One of the referees complained that although he/she was able to follow each step of the proof, it was impossible to gain any insight. The purpose of this page is to put down the thought processes that led to the proof.

In fact, it is really quite hard to get started. The first step is to reduce the problem to the following.

THEOREM 2 Let A and B be r \times r positive definite matrices, and let Z be an r \times r matrix such that BA(I+ZZ^\star)AB = I, then
\lambda_r \pmatrix{A+B & AZ\cr Z^\star A& Z^\star A Z\cr} \geq 2.

It's very hard to see where to go from here. Even though it had been solved, I looked at the r=1 case, where the operators are scalars. Let
T = \pmatrix{A+B & AZ\cr Z^\star A& Z^\star A Z\cr}
then in case r=1, the characteristic polynomial {\rm det}(\lambda I - T) has the form (\lambda - b)(\lambda -x) - b^{-1}x^{-1} where the operator B has been replaced by b and x is a scalar version of B^{-1}A^{-1}B^{-1}. The key point is that the characteristic polynomial is symmetric in b and x. So, there was some hope that there might be an operator X, something like B^{-1}A^{-1}B^{-1} that would play a key role in the proof. Of course X = B^{-1}A^{-1}B^{-1} doesn't lead anywhere. Many months passed without any progress.
Then came a seemingly trivial observation, namely that in the case r=1, the matrices T and
\pmatrix{b & (bx)^{-\frac{1}{2}}\cr (bx)^{-\frac{1}{2}} & x\cr}
have the same characteristic polynomial. Perhaps something similar could be true in the operator setting? Since there was supposed to be symmetry between X and B, the natural choice for an operator to replace (bx)^{-\frac{1}{2}} was (B \# X)^{-1}. where B\#X denotes the geometric mean of B and X. But how to choose X? I set about writing down the equations in case r=2 for T and
\pmatrix{B & (B \# X)^{-1}\cr (B \# X)^{-1} & X\cr}
to have the same characteristic polynomial. This is really ugly. It would be much easier to figure the equations for S = B\#X and then come to X later. But then the equations all involve S^2 rather than S, so one should set Y = S^2 and solve for the entries of Y. I programmed this into Maple taking B to be diagonal, expecting a very complicated answer many hours later. But the answer
y_{{1,1}}={\frac {a_{{2,2}}}{b_{{1}} \left( a_{{1,1}}a_{{2,2}}-{a_{{1,2}}}^{2} \right) }},y_{{1,2}}={\frac {{\it RootOf} \left( {{\it \_Z}} ^{2}b_{{2}}b_{{1}}-1 \right) a_{{1,2}}}{a_{{1,1}}a_{{2,2}}-{a_{{1,2}}} ^{2}}},y_{{2,2}}={\frac {a_{{1,1}}}{b_{{2}} \left( a_{{1,1}}a_{{2,2}}- {a_{{1,2}}}^{2} \right) }}
came out instantly. I nearly fell off my chair! The answer was clearly Y = B^{-\frac{1}{2}}A^{-1}B^{-\frac{1}{2}} and it didn't take long to check that this also works for all r. I am still as mystified as the referee!
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