Hermitian matrix?

Question

Note that this challenge requires no handling or understanding of complex numbers.

Given a non-empty square matrix where every element is a two-element (Re,Im) integer list, determine (giving any truthy/falsy values or any two consistent values) whether this represents a Hermitian matrix.

Note that the input is a 3D array of integers; not a 2D array of complex numbers. If your language cannot take a 3D array directly, you may take a flat list (and the n×n or n×n×2 shape if that helps).

A matrix is Hermitian if it equals its own conjugate transpose. In other words, if you flip it across its top-left to bottom-right diagonal and negate the second element of all the two-element leaf-lists, it is identical to the input matrix. Note that the order of flipping and negating is irrelevant, so you may negate first, and flip afterwards.

Walk-though example

This example uses JSON with superfluous white-space to ease reading:

[[ [2, 0] , [2, 1] , [4, 0] ],
 [ [2,-1] , [3, 0] , [0, 1] ],
 [ [4, 0] , [0,-1] , [1, 0] ]]

Transpose (flip across NW—SE diagonal):

[[ [2, 0] , [2,-1] , [4, 0] ],
 [ [2, 1] , [3, 0] , [0,-1] ],
 [ [4, 0] , [0, 1] , [1, 0] ]]

Negate second elements of leaf-lists:

[[ [2, 0] , [2, 1] , [4, 0] ],
 [ [2,-1] , [3, 0] , [0, 1] ],
 [ [4, 0] , [0,-1] , [1, 0] ]]

As this is identical to the input, the matrix is Hermitian.

Test cases

Hermitian

[[[2,0],[2,1],[4,0]],[[2,-1],[3,0],[0,1]],[[4,0],[0,-1],[1,0]]]

[[[1,0],[2,0]],[[2,0],[1,0]]]

[[[1,0],[2,-3]],[[2,3],[1,0]]]

[[[42,0]]]

Non-Hermitian

[[[2,0],[2,1],[4,0]],[[2,-1],[3,0],[0,1]],[[4,0],[0,-1],[1,-1]]]

[[[0,1],[0,2]],[[0,2],[0,1]]]

[[[1,0],[2,3]],[[2,3],[1,0]]]

[[[3,2]]]

Answer 1

R, 71 48 47 bytes

function(A)all(Conj(t(B<-A[,,1]+A[,,2]*1i))==B)

Takes a 3D array of real numbers, make a 2D array of imaginary numbers, transpose, conjugate and compare.

Thanks to @Giuseppe for reducing the byte count by an astounding 23 bytes, and to @Vlo for the final 1!

Try it online!

Example:

> A <- array(c(2,2,4,2,3,0,4,0,1,0,-1,0,1,0,-1,0,1,0),dim=c(3,3,2))
> A
, , 1

     [,1] [,2] [,3]
[1,]    2    2    4
[2,]    2    3    0
[3,]    4    0    1

, , 2

     [,1] [,2] [,3]
[1,]    0    1    0
[2,]   -1    0    1
[3,]    0   -1    0

> f <- function(A)all(Conj(t(B<-A[,,1]+A[,,2]*1i))==B)
> f(A)
[1] TRUE

Answer 2

Octave, 39 34 31 bytes

@(x)(y=x(:,:,1)+j*x(:,:,2))==y'

Try it online!

Saved 3 bytes thanks to Luis Mendo who informed me about the clarifications in the challenge text.

Explanation:

In MATLAB and Octave, ' is the conjugate complex transpose, not the "regular" transpose.

We create a variable y inline that's the first layer of the 3D matrix plus the second layer multiplied with the complex unit j, i.e. a complex matrix where the real term is the first "layer", and the imaginary is the second "layer". We then check if it equals itself complex conjugate transposed.

This will output a matrix containing only 1 if true, and a matrix containing at least one 0 if false. These are considered true and false in Octave (Proof).

Answer 3

Python 2, 50 bytes

lambda m:[[[a,-b]for a,b in l]for l in zip(*m)]==m

Try it online!

Answer 4

APL (Dyalog Unicode), 22 15 9 7 bytes

⍉≡⊢∘-\¨

Try it online!

Tacit prefix function.

Thanks to Adám for 7 bytes on the Dfn, and to both Adám and ErikTheOutgolfer for putting up with my stupidity helping me find the tacit version.

Thanks to ngn for 2 bytes on the tacit version.

How?

⍉≡⊢∘-\¨ ⍝ Anonymous tacit function.
      ¨ ⍝ Apply to each element of the argument:
     \  ⍝ Cumulative reduction, using
  ⊢∘-   ⍝ Ignore the first element, then negate the second
 ≡      ⍝ And match
⍉       ⍝ To the argument's transposition.

Answer 5

Wolfram Language (Mathematica), 45 34 33 26 21 18 bytes

• Saved eleven bytes thanks to JungHwan Min.
• Saved a byte thanks to Martin Ender.
• Saved seven bytes thanks to alephalpha.
• Saved five bytes thanks to alephalpha.
• Saved three bytes.

#==#&[#.{1,I}]&

Try it online!

Answer 6

Java 8, 137 136 134 126 119 116 bytes

m->{int r=1,l=m.length,i=l*l,j,k;for(;i-->0;)r=m[j=i/l][k=i%l][0]!=m[k][j][0]|m[j][k][1]!=-m[k][j][1]?0:r;return r;}

-3 bytes thanks to @ceilingcat.

Returns 1 if Hermitian, 0 otherwise.

Explanation:

Try it online.

m->{                 // Method with 3D integer-array as parameter and boolean return-type
  int r=1,           //  Flag-integer `r`, starting at 1
      l=m.length,    //  The size of the 3D input array
      i=l*l,j,k;     //  Index-integers
  for(;i-->0;)       //  Loop over the rows and columns
    r=m[j=i/l][k=i%l][0]!=m[k][j][0]
                     //   If the first numbers diagonally aren't equal,
      |m[j][k][1]!=-m[k][j][1]?
                     //   or the second numbers aren't negatives of each other:
       0             //    Set the flag `r` to 0
      :              //   Else:
       r;            //    Leave the flag `r` the same
  return r;}         //  Return the flag `r`

Answer 7

J, 14 bytes

[:(+-:|:)j./"1

Try it online!

Explanation

[:(+-:|:)j./"1  Input: 3d array
         j./"1  Reduce by complex combine at rank 1
[:              Cap, operate on the 2d array of complex values
   +              Conjugate
      |:          Transpose
    -:            Match?

Answer 8

Haskell, 50 bytes

-7 bytes thanks to H.PWiz.

(==)<*>map(map((0-)<$>)).foldr(zipWith(:))e
e=[]:e

Try it online!

Answer 9

Jelly,  6  5 bytes

Z×Ø+⁼

A monadic link returning 1 for a Hermitian input and 0 otherwise.

Try it online!

How?

Z×Ø+⁼ - Link: list of lists of lists, M
Z     - transpose
  Ø+  - literal = [1,-1]
 ×    - multiply (vectorises)
    ⁼ - equal to M?

Answer 10

05AB1E, 9 bytes

øεεX®‚*]Q

Try it online!

Explanation

ø           # transpose
 ε          # apply to each 2-d array
  ε         # apply to each pair
   X®‚*     # multiply by [1,-1]
       ]    # end apply(s)
        Q   # compare to input for equality

Answer 11

Ruby, 46 bytes

->m{m.transpose.map{|l|l.map{|a,b|[a,-b]}}==m}

Try it online!

Port of my Python answer

Answer 12

Perl 5, -a0 48 bytes

Old counting: 50 bytes (+2 for a0). Not bad for a language that has no builtin transpose (I'm not jealous at all, no sirree)

Give the input matrix on STDIN with , between the real and imaginary part, so e.g.:

2,0 2,1 4,0
2,-1 3,0 0,1
4,0 0,-1 1,0

Will print 1 for hermitian, nothing otherwise

#!/usr/bin/perl -a0
say@F~~[map/(\S+,)(\S+)/gc?$1.-$2:(),(/.+/g)x@F]

Try it online!

Answer 13

Husk, 7 bytes

=¹mmṀ_T

Try it online!

How?

Note that † should work instead of mm , but there's an annoying bug that prevents me from using it :(

=¹mmṀ_T – Full program. Takes input from Command line args, as a list of lists of tuples.
  m   T – For each list in the input's tranpose...
   mṀ_  – ... Negate the last value of each tuple they contain.
=¹      – Check whether this is the same as the input.

Answer 14

JavaScript (ES6), 53 bytes

Saved 2 bytes thanks to @Neil

Returns false for Hermitian or true for non-Hermitian.

m=>m.some((r,y)=>r.some(([a,b],x)=>m[x][y]!=a+[,-b]))

Try it online!

Answer 15

C (gcc), 107 103 100 bytes

• Saved four bytes thanks to Steadybox; golfed A[0] to *A twice.
• Saved three bytes thanks to ceilingcat.

j,k,r;f(A,s)int***A;{for(r=0,j=s;j--;)for(k=s;k--;)r|=*A[j][k]-*A[k][j]|A[j][k][1]+A[k][j][1];A=!r;}

Try it online!

Answer 16

Actually, 13 bytes

┬⌠⌠Çá╫k⌡M⌡Mß=

Try it online!

How it works?

This submission actually makes use of complex numbers. If taking input as a matrix of complex entries was allowed, then it would be 8 bytes.

┬⌠⌠Çá╫k⌡M⌡Mß=  –> Full program.
┬              –> Transpose.
 ⌠       ⌡M    –> For each list in the input's transpose do the following:
  ⌠    ⌡M         –> For each two-element list of the form [a, b]...
   Ç              –> Turn it into a complex number (a+bi).
    á             –> Find its complex conjugate: Push (a-bi).
     ╫k           –> Push [Re(N), Im(N)], so [a, -b].
           ß=  –> Check whether the result equals the input.

Answer 17

Pyth, 9 bytes

qCmm,hk_e

Explanation:

qCmm,hk_ekdQQ  Autofill variables
    ,hk_ek     [a,-b]...
  mm      dQ    ...for each [a,b] in the input (m...Q)'s rows (m...d).
 C             Transpose.
q           Q  Is this result equal to the original?

Test suite.

Answer 18

C,  111   110  108 bytes

Thanks to @Jonathan Frech for saving a byte and thanks to @ceilingcat for saving two bytes!

i,j,r;f(A,n)int*A;{for(r=i=0;i<n*2;i+=2)for(j=n*2;j;r|=A[i*n+j]-A[j*n+i]|A[i*n-~j]+A[j*n-~i])j-=2;return!r;}

Try it online!

C (gcc),  106  104 bytes

i,j,r;f(A,n)int*A;{for(r=i=0;i<n*2;i+=2)for(j=n*2;j;r|=A[i*n+j]-A[j*n+i]|A[i*n-~j]+A[j*n-~i])j-=2;A=!r;}

Try it online!

Answer 19

Actually, 13 bytes

;┬⌠⌠d±@q⌡M⌡M=

Try it online!

Explanation:

;┬⌠⌠d±@q⌡M⌡M=
;              make a copy
 ┬             transpose copy
  ⌠⌠d±@q⌡M⌡M   for each row:
   ⌠d±@q⌡M       for each cell in row:
    d              remove last element from list
     ±             swap sign
      @q           insert at end of list
            =  compare equality with original

Answer 20

Stax, 9 bytes

ü♦╧Γ♥├eï∞

Run and debug it