No. But if the system can't be solved without that constraint then clearly it won't become solvable when adding it. It's not just fancy that caused me to comment it out -- with it in Mathematica claims ignorance. Charles Greathouse Analyst/Programmer Case Western Reserve University On Tue, Aug 5, 2014 at 4:54 PM, Dan Asimov <dasimov@earthlink.net> wrote:
Charles,
Just wondering, from your Mathematica comment (*==MatrixExp[A+B]*):
Is there some reason that the truth of
exp(A)exp(B) = exp(B)exp(A)
implies that these terms are also equal to exp(A+B) ???
--Dan
On Aug 5, 2014, at 7:48 AM, Charles Greathouse < charles.greathouse@case.edu> wrote:
Well, if you believe Mathematica, there are no such 2x2 real matrices:
A = {{a11, a12}, {a21, a22}}; B = {{b11, b12}, {b21, b22}}; Solve[MatrixExp[A].MatrixExp[B](*==MatrixExp[A+B]*)== MatrixExp[B].MatrixExp[A] && A.B != B.A, {a11, a12, a21, a22, b11, b12, b21, b22}, Reals] // Timing
{17.191310, {}}
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Mon, Aug 4, 2014 at 9:17 PM, Dan Asimov <dasimov@earthlink.net> wrote:
It's well known and easy to prove that if NxN matrices A and B commute, then
(*) exp(A)exp(B) = exp(A+B) = exp(B)exp(A)
.
But AB = BA is not a necessary condition for (*), as googling will readily reveal.
Does anyone know necessary and sufficient conditions on A and B for (*) to hold?
If not, how about a large class of A and B for which (*) holds despite AB and BA being unequal ?
I think such examples of complex matrices are easier to come by than of real ones, so I'm particularly interested in the case where A and B are real.
--Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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