Are the off-diagonal elements of exp(At) log-concave in t, for nonnegative matrices?
The following Matlab code (unless I coded something wrong, which is well possible) finds a few random counterexamples each time I run it, even when restricted to off-diagonal entries:
n = 3;
for trie = 1:100
A = rand(n);
B1 = log(expm(A));
B2 = log(expm(2*A));
B3 = log(expm(3*A));
C = B2 - (B3+B1)/2; %its off-diagonal should be >= 0 if the claim holds
C = C - diag(diag(C));
if not(all(all(C >= 0)))
A
end
end
For instance:
A =
0.804449583613070 0.535664190667238 0.989144909700340
0.986104241895970 0.087077219900892 0.066946258397750
0.029991950269390 0.802091440555804 0.939398361884535
A =
0.018177533636696 0.534137567882728 0.625937626080496
0.683838613746355 0.885359450931142 0.137868992412558
0.783736480083219 0.899004898906140 0.217801593712125
A =
0.133503859661312 0.300819018069489 0.286620388894259
0.021555887203497 0.939409713873458 0.800820286951535
0.559840705872510 0.980903636046859 0.896111351432604
A =
0.108016694136759 0.559370572403004 0.848709226458282
0.516996758096945 0.004579623947323 0.916821270253738
0.143156022083576 0.766681998621487 0.986968274783658
A =
0.068357220470829 0.026107108154905 0.961558573103663
0.436327077480103 0.954678274080449 0.762414484002993
0.173853037365001 0.430596519859417 0.007348661102847
Quick remarks with some tips for numerical experimentation:
- For continuous functions, midpoint convexity is equivalent to convexity, so I only tested for that. The $t=1,2,3$ interval is chosen because it looked like the simplest thing to try.
- Out of habit, I first coded this with 1000 tries with $5\times 5$ matrices. $100$ and $3\times 3$ are here just for quicker display. It's so fast that it doesn't matter anyway, so it's better to err on the side of more and larger examples.
- One should be careful with instructions such as
C - diag(diag(C))
(which subtracts its diagonal from a matrix), which could hide numerical mistakes (what if a-1e-16
pops up on the diagonal?). In this case though the subtractions are of the forma - a
, which is guaranteed to return 0 even in double-precision arithmetic. Matlab does not have a simpler way to set the diagonal of a matrix to zero or ignore it, unfortunately. (I actually first wroteC(1:n+1:n^2) = 0
, but then I replaced it because it is hackish and difficult to read). - There are lots of factoids about matrices (and especially about monotonicity of eigenvalues of nonsymmetric matrices and the matrix exponential) that look true at first sight but have counterexamples. I suggest you to always try some random experiments like these ones. Once one gets in the habit, it's faster to write the code than to think about it. :)
For $2\times2$ matrices, the log-concavity is true. One has $e^{tA}=f(t)I_2+g(t)A$ by Cayley-Hamilton. Writing that the eigenvalues of $e^{tA}$ are the exponentials of those of $tA$, we find $$g(t)=\frac{e^{t\mu}-e^{t\lambda}}{\mu-\lambda}\,,$$ where $\mu,\lambda$ are the eigenvalues of $A$. Thus we only have to prove that $g$ is log-concave, that is $$gg''-g'^2=-e^{t(\mu+\lambda)}\le0.$$ Notice that the assumption is implicitely used in that it implies $g>0$.
Edit. The formula above seems to be a particular case of a more general one. Suppose $A$ is $n\times n$. With Cayley-Hamilton, we have $$e^{tA}=f(t)I_n+g(t)A+\cdots+h(t)A^{n-1}.$$ Let us form the Hankel matrix $M_h(t)=\left(h^{(i+j-2)}(t)\right)_{1\le i,j\le n}$. Then $\det M_h(t)=(-1)^{n+1}e^{t{\rm Tr}\,A}$.
Remark that a smooth function $h$ satisfies a linear ODE of order $n-1$ with some constant coefficients if, and only if, $\det M_h\equiv0$.