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

Is it true that $(AB)^{1/2}+(BA)^{1/2} \geq A^{1/2}B^{1/2}+B^{1/2}A^{1/2}$ ?

EDIT: Shown to be false below. Extension sought: Consider now a square strictly contractive matrix $S$ (such that $I-SS^T>0$) and consider the inequality $(AB)^{1/2}+(BA)^{1/2}\geq A^{1/2}SB^{1/2}+B^{1/2}S^TA^{1/2}$ ? Does this contraction change things, i.e. is the inequality true?

A related question but "in reverse" is asked here: An inequality on the root of matrix products (part 2 - the reverse case)

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What does $ A\le B $ mean for matrices? – Braindead Jan 11 '14 at 10:13
$\geq$ is in the positive-definite sense – John U Jan 11 '14 at 10:15
Sorry, can you clarify? I don't understand what you mean by that. – Braindead Jan 11 '14 at 10:19
$A\geq B$ implies $A-B\geq0$ or $x^T(A-B)x\geq0$ for all non-zero $x$ – John U Jan 11 '14 at 10:24
$AB$ with $A$ and $B$ both symmetric positive definite has positive eigenvalues.. the full result is $AB$ with $A$ positive definite has the same number of positive, negative and zero eigenvalues as $B$ - see Matrix Analysis (Horn et. al.) – John U Jan 11 '14 at 14:17
up vote 2 down vote accepted

It's not true. If it's true, by continuity, it's also true for positive semidefinite matrices. Yet we have a counterexample in the latter case: \begin{align*} &A=X^2,\ B=Y^2\ \text{ where }\ X=\sqrt{\frac85}\pmatrix{1&2\\ 2&4},\ Y=\pmatrix{1&0\\ 0&\tfrac12},\ \\ &AB=\pmatrix{8&4\\ 16&8}=R^2,\ \text{ where }\ R=\pmatrix{2&1\\ 4&2},\\ &R+R^T=\pmatrix{4&5\\ 5&4},\\ &XY+YX=\sqrt{\frac85}\pmatrix{2&3\\ 3&4}. \end{align*} Neither is $R+R^T\succeq XY+YX$ (the last entry of $R+R^T$ is smaller than the last entry of $XY+YX$) and nor is $XY+YX\succeq0$ (its determinant is negative).

It should be easy to generate by computer a random counterexample for the positive definite case, but obtaining one with nice matrix entries may require a bit more work.

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Thanks.. the positive definite part of my question is countered simply by noting that $x^T(S+S^T)x=2x^TSx$ which need not be $\geq 0$ when $S$ is non-symmetric but with strictly positive eigenvalues – John U Jan 11 '14 at 21:19
I don't know if I should ask another question but I will try here first (also editing the actual question). What if now we have a square strictly contractive matrix $S$ (such that $I-SS^T>0$) and we consider the inequality $(AB)^{1/2}+(BA)^{1/2}\geq A^{1/2}SB^{1/2}+B^{1/2}S^TA^{1/2}$ ? Does this contraction change things? – John U Jan 12 '14 at 5:07
I guess the answer is no by continuity also but I am also seeking an illustration (through algebra not counter-example) if possible.. – John U Jan 12 '14 at 5:11
@JohnU Your guess is correct: the answer is no. Again, if the hypothesis is true, then by a continuity argument, it will also be true for $I-SS^T=0$ and in particular true for $S=I$. So, the counterexample in the above answer still applies. If you are seeking an illustration "through algebra not counter-example" (what does that mean?), please feel free to post a new question. – user1551 Jan 12 '14 at 7:08

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