On my awaking, there was a whiff of sulfur in the air, and a greenish
haze... and somehow in my mind the idea that that last program won't
work, because of the possibility of negative zero. I'll stay relegated
to imp status.
Henry Rich
On 1/15/2013 6:20 PM, Henry Rich wrote:
Nah, that's not beyond impish. The devilish solution is to take the
bitwise OR of the matrix with its conjugate transpose (but that's easier
in assembler language than in J:
(23 b.&.(a.&i.)&.(2&(3!:5))&.+. +@|:))
). And you need to be sure that the zeros on the lower diagonal and
below are true zeros!
Henry Rich
On 1/15/2013 6:03 PM, km wrote:
Oh, boy! (v1`v2) } y <--> (v1 y) } (v2 y)
Brief and devilish, take care for your soul, Henry!
--Kip
Sent from my iPad
On Jan 15, 2013, at 3:39 PM, Henry Rich <henryhr...@nc.rr.com> wrote:
hft =: 0&=`(,: +@|:)}
Henry Rich
On 1/15/2013 5:25 AM, km wrote:
This is an easy one. A Hermitian matrix matches its conjugate
transpose. Write a verb hft that creates a Hermitian matrix from a
triangular one that has a real diagonal.
ishermitian =: -: +@|:
]A =: 2 2 $ 1 2j3 0 4
1 2j3
0 4
]B =: hft A
1 2j3
2j_3 4
ishermitian A
0
ishermitian B
1
Kip Murray
Sent from my iPad
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