Let $A$ and $B$ be real (symmetric) and positive definite. It follows that $AB+BA$ is not necessarily positive definite (it can be indefinite, negative definite or positive definite).

But now suppose $A\geq B$. Can one always say $2A^2 \geq AB+BA$?

Since this ordering implies $2A^2-AB-BA\geq0$ and adding indefinite/neg-def/pos-def to a pos-def matrix may still be pos-def I am supposing the question is well-defined.

  • $\begingroup$ How do you define $A \ge B$? $\endgroup$ – user114628 Jan 11 '14 at 21:44
  • $\begingroup$ $A\geq B$ implies $A-B\geq0$ implies $x^T(A-B)x\geq0$ for all $x$ $\endgroup$ – John U Jan 11 '14 at 21:45
  • $\begingroup$ Try it with a=(5/4, 1; 1 19/10) and b=(3/5 7/10; 7/10 17/10). Sorry, but too lazy to type it in latex format. Oh I see someone beat me to it :) $\endgroup$ – user114628 Jan 11 '14 at 22:32

No. Let me reuse an old counterexample, where $\epsilon>0$ is small: \begin{align*} A&=\pmatrix{2+\epsilon&1\\ 1&3},\\ B&=\pmatrix{1\\ &2},\\ 2A^2-AB-BA&=\pmatrix{2(1+\epsilon)(2+\epsilon)+2&7+2\epsilon\\ 7+2\epsilon&8}\approx\pmatrix{6&7\\ 7&8}\not\succeq0. \end{align*}

  • $\begingroup$ Thanks.. so we now have it that one can claim $2A^2-AB-BA\geq0$ if $A^2\geq B^2$.. but not necessarily if only $A\geq B$.. $\endgroup$ – John U Jan 11 '14 at 22:34
  • $\begingroup$ You can find this kind of results in the book: Marshall Olkin Arnold: "Inequalities: Theory of majorization and its applications." (second edition) Search for "Loewner order" which this ordering is called. $\endgroup$ – kjetil b halvorsen Jan 11 '14 at 22:51

Note that $(A-B)^2 \ge 0$ in the sense that you defined earlier. Now $(A-B)^2 = A^2 - AB - BA + B^2$ and thus $$ A^2 + B^2 \ge AB + BA\, . $$ Now use $A^2 \ge B^2$ to arrive at the conclusion.

  • $\begingroup$ $A\geq B$ does not imply $A^2\geq B^2$.. though i wonder if this realisation implies that the inequality in my question holds only when $A\geq B$ does imply $A^2 \geq B^2$ ? $\endgroup$ – John U Jan 11 '14 at 22:18
  • $\begingroup$ You're right, I assumed implicitly that $A$ and $B$ are jointly diagonalizable. $\endgroup$ – Hans Engler Jan 25 '14 at 14:10

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