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I would like to transform matrix $\mathbf A = \begin{pmatrix} a&b&i&j\\ c&d&k&l \\ e&f&m&n \\ g&h&o&p \end{pmatrix}$ into matrix $\mathbf B = \begin{pmatrix} -p&o&-h&g\\ -n&m&-f&e \\ -l&k&-d&c \\ -j&i&-b&a \end{pmatrix}$.

If possible I would like to generalize this to $\{n\times n,\ n>4\}$ size matrices. Is there a simple way to do this in Mathematica?

Thanks in advance.

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  • $\begingroup$ Please do provide at least A as a Mathematica expression. $\endgroup$
    – Yves Klett
    Feb 17, 2013 at 15:50

4 Answers 4

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Surely a better solution exists! Assuming m your matrix.

m = RandomReal[1, {1000, 1000}];
pat = Array[(-1)^# &, First@Dimensions[m]];
B1 = (pat #) &  /@ Reverse[m, {1, 2}]; // AbsoluteTiming

{0.124800, Null}

Though Table is intuitive but will be pretty slow for big lists.

u = Length[m];
B = Table[
m[[u - r, u - s]]*(-1)^(s + 1), {r, 0, u - 1}, {s, 0,u - 1}]; // AbsoluteTiming

{6.567611, Null}

Testing!

B2 === B

True

Now if you want to go even faster with Mathematica use Compile to external language C. However this is a solution only if your matrix has Number entries. However you will not get much more speed up as Map and Reverse are already pretty optimized in Mathematica.

fun = Compile[{{x, _Real, 2}}, Module[{pat},
pat = Array[(-1)^# &, First@Dimensions[x]];
(pat #) &  /@ (Reverse[Reverse /@ x])],
CompilationTarget -> "C"];
B3 = fun[m]; // AbsoluteTiming

{0.062400, Null}

Testing again!

B === B2 === B3

True

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  • $\begingroup$ Ok! This works fine, and you sure are convincing! Thanks a lot! $\endgroup$
    – Meclassic
    Feb 17, 2013 at 14:38
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How do you like this?

A = {{a, b, i, j}, {c, d, k, l}, {e, f, m, n}, {g, h, o, p}};
B = Reverse[A,{1,2}].DiagonalMatrix[{-1, 1, -1, 1}]
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  • $\begingroup$ Thanks for your contribution! How can we generalize your proposition for large matrices? $\endgroup$
    – Meclassic
    Feb 17, 2013 at 14:42
  • 4
    $\begingroup$ @jrojasqu change the list inside the diagonal matrix to (-1)^Range@Length@A $\endgroup$
    – rm -rf
    Feb 17, 2013 at 14:59
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Ok,

I feel sorry for myself by now... This was easier than I thought... So here's my proposition for a solution :

Given matrix $\mathbf A$ written in mathematica as :

A = {{a, b, i, j}, {c, d, k, l}, {e, f, m, n}, {g, h, o, p}}

To obtain matrix $\mathbf B$ I did this :

u = Length[A];
B = Table[A[[u - r, u - s]]*(-1)^(s + 1), {r, 0, u-1}, {s, 0, u-1}]

There it is, I hope this proves useful to someone.

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  • $\begingroup$ u = Length@A instead $\endgroup$
    – Öskå
    Feb 17, 2013 at 12:38
  • $\begingroup$ @Öskå You're absolutely right, this code can be highly optimized, it's just that I posted it as soon as it came to me... $\endgroup$
    – Meclassic
    Feb 17, 2013 at 12:42
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 ClearAll[trnsfrmF1, trnsfrmF2];
 trnsfrmF1 = MapAt[-# &, Reverse /@ Reverse@#, {;; , ;; ;; 2}] &;
 trnsfrmF2 = Module[{temp = Reverse /@ Reverse@#},
 temp[[;; , ;; ;; 2]] = (-1) temp[[;; , ;; ;; 2]]; temp] &;

Example usage:

 aa = ArrayReshape[CharacterRange["a", "p"], {4, 4}];
 Grid[{{"aa", "trnsfrmF1[aa]", "trnsfrmF2[aa]"},
  MatrixForm /@ {aa, trnsfrmF1[aa], trnsfrmF2[aa]}}, Dividers -> All]

enter image description here

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