Somehow, this makes sense, yes. We don't know how to divide by anything
but a matrix, so the right rank is 2, but we know how to multiply two
things that are more than matrices, so it's ok to have a left right _.
It took me some time to understand why you were mentioning A%.A, as
matrix multiplication is not commutative. Turns out, by reading the
documentation, that x%.y is actually (%.y) +/ .* x, not x +/ .* %.y as I
implicitly assumed. Btw, why is that? Why does dividing on the right by
y actually multiplies on the left by the inverse? Isn't that weird?
So, in the end, I have the impressions that it all boils down to the
rank of u .v, and how . works in general.
Adrien Mathieu
On 16/05/2022 21:50, Raul Miller wrote:
First off, this is a good question.
That said, it's matrix divide because we are "dividing by" a matrix.
But maybe it's easier to illustrate this than explain it:
'A B C'=:ABC=: p:i.3 3 3
ABC%.A
3.92308 4 2.25641
2 1.84615 1.58974
0.923077 _0.410256 _0.974359
5.23077 5 6.23077
6 3.46154 3.23077
2.23077 3.23077 0.923077
_4.30769 _4 _3.64103
_3 _0.615385 0.025641
1.69231 2.02564 4.4359
A+/ .*ABC%.A
2 3 5
7 11 13
17 19 23
29 31 37
41 43 47
53 59 61
67 71 73
79 83 89
97 101 103
Compare this with:
A%.A
1 0 0
0 1 0
0 0 1
In other words, %. would not work for higher ranked arrays if its left
rank was 2.
I hope this makes sense,
----------------------------------------------------------------------
For information about J forums see http://www.jsoftware.com/forums.htm
