I am investigating permutation matrices. These are matrices made by
beginning with the identity matrix I and then permuting its columns,
until a new matrix M results. M has the effect of "scrambling" or
permuting a vector x, and this is reversible: given y = Mx, M has an
inverse.
Let M^u = M * M * M, u times. Then it is an interesting fact that for
all 4x4 permutation matrices (of which there are 4!=24 possible
permutation matrices), M^(12+x) = M^x for non-negative integers x. Can
any one shed some light on this phenomenon? Thus M^5=M^17=M^29 etc. It
is also interesting that M^12=I, e.g. whatever this magic number 12
is, M to that power yields the identity matrix.
Also I am interested in finding the set of permutation matrices M such
that M ^ u = I for some non-negative integer u. I developed a program
to do it, and it gives me M for u = 6 and u = 12 for instance. It
works all right -- all possible answers have been multiplied by
themselves to yield I after the requisite number of multiplications --
although it is not exhaustive since we consider only permutation
matrices. I found that M is a set -- there are for instance several
possible permutation matrices M such that M ^ 2 = I (10 to be exact
for a 4x4 matrix system).
Can anyone please help me figure out the meaning of M ^ (3 / 5) = I ?
That is, I can compute M ^ 3 = I by brute force; it's a set of
permutation matrices that when taken to the power of 3, yields I. But
how to I proceed for that -SET- of matrices to a new set of matrices M
^ (3 / 5) ? I tried computing the cube of each element in the set and
found the result is wrong (how do I know? I tried to compute I ^ (2 /
12) = M by first computing I ^ (1/12) and then finding the square of
each element in that set. The results are in fact right -- taken to
the 6th power they yield I, but they are not exhaustive -- if I
compute I^(1/6) directly I find additional elements in the set not
found by taking I ^ (1/12) elements squared).
I also tried taking all permutation matrix M to the power of 3 and
then computing that result to the power of (1/5), a set that, if it
contiains only the identity matrix, was taken to be the M such that M
^ (3/5) = I. It got the same, wrong answer.
I do not want to introduce negative numbers or imaginary numbers, or
anything other than 0's and 1's in the permutation matrix.
Your help greatly appreciated . . .
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