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Suppose $A$, $B$, are symmetric, real valued matrices and $B-A$ is positive-semidefinite, i.e. $A≼B$. Does that imply $e^A ≼ e^B$? Would love some intuition here.

I know for instance that $A≼cI \iff e^A ≼ e^cI$ and also for diagonal $A$ and $B$ the inequality is easy to show, but I'm not convinced that it translates into the general case.

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Functions with this property are called "operator monotone". It is well-known that $\log$ is operator-monotone, but squaring or exponentiating ist not. –  Andreas Thom Jun 30 at 16:17
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seems like a reasonable question to me. –  Anthony Quas Jun 30 at 17:59
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To me as well. It might be "well-known" to the right people, but I doubt it's standardly known across the whole community. (I didn't know it, anyway.) –  Todd Trimble Jun 30 at 18:12
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@Dmitry: I did not say "It is well-known..." is the reason for the answer; my reason was the fact that Andreas has provided the magic keyword "operator monotone" googling which the answer becomes apparent (e.g., by reading the Wikipedia entry that shows up on searching "operator monotone matrix exponential"). Since it seems that others care about this question, I'll pen down an answer with some additional useful information that people may appreciate. –  Suvrit Jun 30 at 18:35
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@Suvrit: I am glad that you have written an answer, which makes it unlikely that the question will be closed, and so makes my comment less pressing. But I will say for the record: responses that simply provide a missing "magic keyword" can be extremely useful, and questions should certainly not be closed just because they can be answered in that way. –  Neil Strickland Jun 30 at 19:39

2 Answers 2

up vote 15 down vote accepted

To supplement Robert's counterexample, let me mention below some interesting facts about the matrix exponential, along with what may be regarded as the "correct" way of obtaining matrix exponential like operator inequalities.

Assume throughout that $A \ge B$ (in Löwner order).

The map $X \mapsto X^r$ for $0 \le r \le 1$ is operator monotone, i.e., $A^r \ge B^r$. This result is called Löwner-Heinz inequality (it was apparently originally discovered by Löwner). Now using \begin{equation*} \lim_{r\to 0} \frac{X^r-I}{r} = \log X, \end{equation*} we can conclude monotonicity of $\log X$, so that $\log A \ge \log B$.

A quick experiment reveals that $A^2 \not\ge B^2$ in general (in fact $X^r$ for $r > 1$ is not monotone), which severely diminishes hopes of $e^X$ being monotone.

However, though $e^A \not\ge e^B$, T. Ando (On some operator inequalities, Math. Ann., 1987) showed that \begin{equation*} e^{-tA} \# e^{tB} \le I,\qquad\forall t \ge 0. \end{equation*} Here, $X \# Y := X^{1/2}(X^{-1/2}YX^{-1/2})^{1/2}X^{1/2}$ denotes the matrix geometric mean (so basically, operator inequalities go along better with the "right" notion of a geometric mean)


Additional remarks

Although $A^2 \ge B^2$ does not hold, it turns out that a slightly modified version $(BA^2B)^{1/2} \ge B^2$ does hold, as does $A^2 \ge (AB^2A)^{1/2}$. These inequalities are special cases of a family of such results proved by T. Furuta, and these are called Furuta inequalities, for example, we have \begin{equation*} (B^rA^sB^r)^{1/q} \ge (B^{s+2r})^{1/q},\quad 0 \le s \le 1, r \ge 0, q \ge 1. \end{equation*}

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For example, try $$A = \pmatrix{1 & 0\cr 0 & 0\cr},\ B = A + t C\ \text{where}\ C = \pmatrix{1 & 1\cr 1 & 1\cr}$$ where $t>0$ is small. Then $A ≼ B$ but $$ e^B - e^A = t \pmatrix{e & e-1 \cr e-1 & 1 \cr} + O(t^2)$$ so that $$ \det(e^B - e^A) = (e - (e-1)^2) t^2 + O(t^3) < 0$$

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