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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