Block anti-diagonalize a square matrix?

Question

In Mathematica there is a very convenient function called JordanDecomposition[]. It takes a square matrix A as an argument and either diagonalizes it, or puts it into a block diagonal form A' and provides the transformation matrix s (such that A can be recovered by A=s.A'.Inverse[s]).

I was looking for a similar function which would anti-diagonalize the matrix A instead of diagonalizing it (see http://en.wikipedia.org/wiki/Anti-diagonal_matrix) and would also return a corresponding transformation matrix s. Is there such a function in Mathematica? Or maybe one can implement it? Thanks for any suggestion!

EDIT:

Some comments below suggest for me to better specify what I want to achieve. Let us look at an example:

A = DiagonalMatrix[{1, 2, 3}];
B = Table[Subscript[b, i, j], {i, 1, 3}, {j, 1, 3}];
Bi = Inverse[B] // Simplify;
vars = Flatten[B];
sol = FindInstance[{(Bi.A.B)[[1, 1]]==0,(Bi.A.B)[[2, 1]]==0,(Bi.A.B)[[3, 2]]==0,(Bi.A.B)[[3, 3]]==0,Det[B]==1},vars][[1]];

Note how the condition Det[B]==1 enforces the sought after transformation to be determinant preserving. The result I get is:

Bi.A.B /. sol // MatrixForm

$$ \left( \begin{array}{ccc} 0 & -\frac{5}{2} & -11 \\ 0 & 6 & 24 \\ 1 & 0 & 0 \end{array} \right) $$

As you can see, the transformation by an explicitly found B has put originally diagonal matrix A into a block anti-diagonal form. By inspection you can convince yourself that trying to get Bi.A.B to be even more anti-diagonal by introducing more constraints in FindInstance[...] will not yield any solution any more. This suggests that a block anti-diagonal form of the type found above is the best we can do for the matrix A. It is certainly true that if A was the unit matrix, it would already automatically be in its best block anti-diagonal form. However, having one such special example does not render the question invalid for other more general classes of matrices. Basically, I am looking for a solution that robustly automates the above procedure done "by hand" for matrices of arbitrary dimension and returns the transformation matrix B for a best-possible anti-diagonalisation of A.

Answer 1

The answer is that it's mathematically impossible to do what you ask in a general setting. As a counter-example, consider how you would bring the 3-dimensional unit matrix into anti-diagonal form. It would require a permutation with signature

Signature[{3, 2, 1}]

(* ==> -1 *)

so that the determinant of the anti-diagonal matrix has the opposite sign as the original matrix. But you're looking for a similarity transformation between the two. Since similarity transformations don't change the determinant, there exists no such transformation.

Answer 2

Reverse each row:

B = Reverse/@ A