12
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Given the following matrix:

m = Array[Subscript[a, #, #2] &, {4, 4}]

enter image description here

how can I find the skew diagonal or anti-diagonal or back diagonal of the matrix (shown in red)

enter image description here

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    $\begingroup$ What is a "back diagonal"? Do you mean the "skew diagonal" or "anti diagonal"? $\endgroup$ – rm -rf Apr 13 '13 at 19:35
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    $\begingroup$ You can try to apply Reverse to the matrix before Diagonal. The elements will be in reverse order so you might want to apply Reverse again to the result... $\endgroup$ – Spawn1701D Apr 13 '13 at 19:38
  • $\begingroup$ Consider MapThread[Part, {Array[C, {5, 5}], -Range[5]}]. $\endgroup$ – J. M.'s technical difficulties Apr 13 '13 at 19:54
9
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More options:

a = Range@12 ~Partition~ 4;

a // MatrixForm

$\left( \begin{array}{cccc} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \end{array} \right)$

Diagonal[Reverse @ a]
{9, 6, 3}
Diagonal[a ~Reverse~ 2, 1]
{3, 6, 9}
Diagonal[a ~Reverse~ 2]
{4, 7, 10}
Diagonal[Reverse @ a, 1]
{10, 7, 4}
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11
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Diagonal[Reverse /@ Array[f, {4, 4}]]

{f[1, 4], f[2, 3], f[3, 2], f[4, 1]}

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    $\begingroup$ ...and if you want the off-antidiagonals, just use the second argument of Diagonal[]. $\endgroup$ – J. M.'s technical difficulties Apr 14 '13 at 1:39
6
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I believe you can take diagonal off a rectangular matrix too. "Back 'diagonal'".

SeedRandom[1];
m = RandomInteger[6, {3, 4}]
(* {{6, 4, 2, 4}, {0, 1, 6, 0}, {0, 2, 0, 6}} *)

Table[m[[i, -i]], {i, Min@Dimensions@m}]
(* {4, 6, 2} *)
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    $\begingroup$ This is two orders of magnitude faster than what I proposed. Were I to answer this question today I would use this method. A very belated +1. $\endgroup$ – Mr.Wizard Aug 1 '18 at 10:18
5
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Here's a flexible version:

antidiag[m_] := Diagonal[m[[-1 ;; 1 ;; -1, 1 ;; -1 ;; 1]]]
antidiag[m_, offset_] := 
 Diagonal[m[[offset ;; -1, offset ;; -1]][[-1 ;; 1 ;; -1, 
   1 ;; -1 ;; 1]]]

with no second argument, it gives what you want.

MatrixForm[m = Array[Subscript[a, #1, #2] &, {5, 5}]]
antidiag[m]
antidiag[m, 2]

enter image description here

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  • $\begingroup$ This is probably the "correct" way to approach it, though perhaps less transparent than Reverse. $\endgroup$ – Mr.Wizard Apr 13 '13 at 23:01
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    $\begingroup$ @Mr.Wizard I was not aware that you were an advocate of "transparency" :) thanks. $\endgroup$ – acl Apr 13 '13 at 23:04
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    $\begingroup$ acl: transparent~ less~ than $\endgroup$ – Dr. belisarius Apr 14 '13 at 1:12
  • $\begingroup$ @acl Touché! (Most of my coding is quite transparent... to myself.) $\endgroup$ – Mr.Wizard Apr 14 '13 at 3:28
4
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If by "back diagonal", you mean the diagonal from the NE corner to the SW corner of the matrix, then you can obtain it the following way:

Clear@AntiDiagonal
AntiDiagonal[m_?MatrixQ] /; Equal @@ Dimensions@m := 
    Composition[Reverse, Diagonal, Reverse][m]
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    $\begingroup$ If by "NE corner to the SW corner", you mean the top right to the bottom left corner, then you ... ;-) $\endgroup$ – Sjoerd C. de Vries Apr 13 '13 at 20:43

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