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Suppose $A$ and $B$ are positive definite (symmetric) real matrices.

In a previous post (An inequality on the root of matrix products) I asked whether

$(AB)^{1/2}+(BA)^{1/2} \geq A^{1/2}SB^{1/2}+B^{1/2}S^TA^{1/2}$ ?

where $S$ is a square contractive matrix (i.e. a square matrix that obeys $I-SS^T\geq 0$).

This was shown by counterexample to be false in some cases (it is true in others of course). I have since run numerous simulations on random matrices (obeying the stated properties) and found in those simulations that

$(AB)^{1/2}+(BA)^{1/2} - A^{1/2}SB^{1/2}-B^{1/2}S^TA^{1/2}$

always has at least one positive eigenvalue. In other words while the first inequality above may be false I have not been able to confirm that

$A^{1/2}SB^{1/2}+B^{1/2}S^TA^{1/2}\geq (AB)^{1/2}+(BA)^{1/2}$

is ever possible. Thus my question is: Does

$(AB)^{1/2}+(BA)^{1/2} - A^{1/2}SB^{1/2}-B^{1/2}S^TA^{1/2}$

always have at least one positive eigenvalue (or in other words is this construct either positive definite or indefinite for all $A,B,S$ obeying their respective property - i.e. is it never negative definite)?

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The scalar case is "positive definite" for all $s=[-1,1]$ and $a,b>0$. I am really looking for confirmation it is never negative definite (and preferably never negative semi-definite). – John U Jan 14 '14 at 18:59
This inequality arises when looking at partitioned covariance matrices $[A,C; C,B]\geq0$. Note $C=A^{1/2}SB^{1/2}$ in such partitions where $I-SS^T\geq0$. – John U Jan 15 '14 at 11:28
Please re ask the question and I'll look at it in the morning. – JPi Jan 22 '14 at 10:32
Re-asked - looking really for a generic result that really considers the case in which the matrix is never negative definite. – John U Jan 24 '14 at 0:36


Let $A=B=S=Id$

Clearly we get

$$(II)^{\frac{1}{2}}+ (II)^\frac{1}{2} - (I)^\frac{1}{2} I (I)^\frac{1}{2} -(I)^\frac{1}{2} I^T (I)^\frac{1}{2}$$

$$= \sqrt{I} + \sqrt{I} - I - I$$ $$=\mathbb{0}$$

And the zero $n$ square matrix has $n$ zero eigenvalues (not one positive).

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The square root of the identity matrix has an infinite number of solutions (if we use $\mathbb{C}$ as our field): But.... your question was a general one, so in particular if I take the $\sqrt{I} = I$ as one such solution, then I have come up with a counter-example. – Squirtle Jan 20 '14 at 20:34
OK. Thank you.. apologies for the delay in replying to this post - I have been travelling with very little internet access. This is a kind of degenerative situation. I am considering only real matrices and the zero matrix is almost like an indefinite matrix (well it is neither pos-def, neg-def, indef etc by any standard definition). Apologies as the question was a little ambiguous in this regard - though I am really looking for confirmation it is never negative definite as noted also in the question. – John U Jan 24 '14 at 0:34
Thanks! I will consider the new constraints. I will leave this answer up and try to post another.... – Squirtle Jan 24 '14 at 5:05
I want to clarify a few things please: 1) Are you only interested in matrices in $M_n(\mathbb{F})$ with emphasis on finite matrices (i.e. should my answer involve functional analysis)? 2) You seem to have two questions that are not (entirely) mutually intelligible... you say: a) Does ... always have at least one positive eigenvalue b) i.e. is it never negative definite? You didn't like my 0 matrix response... so my question is, do you care about the difference between negative definite and negative SEMI-definite? – Squirtle Jan 24 '14 at 5:10
My initial guess is that your intuition is correct.... Here's just a thought and there are a ton of assumptions. Let $C=(\sqrt{AB} + \sqrt{BA})$ and $D= (I-S^2)$. Now suppose that $S=S^T$ and that it commutes with $\sqrt{A}$ and $\sqrt{B}$. Then we have the following: $\sqrt{AB}+\sqrt{BA} - S\sqrt{AB} - S\sqrt{BA} = CD$. If $CD=DC$ then surely $CD\ge$ but otherwise it might still (even after all those assumptions) be the case that we get no positive eigenvalues. – Squirtle Jan 24 '14 at 6:01

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