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)?

  • $\begingroup$ 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). $\endgroup$ – John U Jan 14 '14 at 18:59
  • $\begingroup$ 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 $I-SS^T\geq0$. $\endgroup$ – John U Jan 15 '14 at 11:28
  • $\begingroup$ Please re ask the question and I'll look at it in the morning. $\endgroup$ – JPi Jan 22 '14 at 10:32
  • $\begingroup$ Re-asked - looking really for a generic result that really considers the case in which the matrix is never negative definite. $\endgroup$ – 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).

  • $\begingroup$ The square root of the identity matrix has an infinite number of solutions (if we use $\mathbb{C}$ as our field): en.wikipedia.org/wiki/Square_root_of_a_matrix#Properties 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. $\endgroup$ – Squirtle Jan 20 '14 at 20:34
  • $\begingroup$ 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. $\endgroup$ – John U Jan 24 '14 at 0:34
  • $\begingroup$ Thanks! I will consider the new constraints. I will leave this answer up and try to post another.... $\endgroup$ – Squirtle Jan 24 '14 at 5:05
  • $\begingroup$ 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? $\endgroup$ – Squirtle Jan 24 '14 at 5:10
  • $\begingroup$ 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. $\endgroup$ – Squirtle Jan 24 '14 at 6:01

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.