Boolean matrix canonisation is equivalent to bipartite graph canonisation is equivalent to graph canonisation

As someone on stack exchange pointed out my problem is actually ptime equivalent to the graph canonisation problem for bipartite graphs under the isomorphim equivalence.
There is a funny proof for this.

A bipartite graph is a graph where the nodes can be partitioned into two disjoint sets A and B, so
that there are no edges between nodes that belong to the same set.

Example with adjacency matrix.

The nodes can be partitioned into the blue set and the red set and there are no edges beween node of the same set, which corresponds to two 0-blocks in the adjacency matrix of the graph.

Write \(G\cong H\) if the two graphs \(G,H\) are isomorphic.


The canonisation problem (call it BGC) for bipartite graphs is to find an efficiently computable function N on bipartite graphs (or adjacency matrice of them) so that for any bipartite graphs G,H:
1. \(N(G)\cong G\)
2. \(N(G)=N(H) \Leftrightarrow G\cong H\)

My problem (call it BMC) is to find an efficiently computable function N on boolean matrices, so that
for any boolean matrices A,B:
1. \(N(A)\sim A\)
2. \(N(A)=N(B) \Leftrightarrow A\sim B\)

where \(\sim\) is defined by
\(A\sim B\) iff there exists permutation matrices \(P,Q\) so that \(B=PAQ\)


Turns out the two problems are equivalent. Surprise, surprise!
How so?


We can transform any boolean matrix A into the adjacency matrix Bip(A) of a bipartite graph as follows:

Notice that since A is duplicated and mirrored in Bip(A) any row in A is transformed into a row and a column of Bip(A) with the same index and analogously every column of A is transformed into a row and column of Bip(A) with the same index.
(Row i of A is mapped onto row n+i and column n+i of Bip(A). Column j of A is mapped onto column j and row j of Bip(A))
Thus permuting rows of A corresponds to permuting rows and columns with the same index in Bip(A) and analogously for columns of A.
In particular there are permutation matrices P,Q with B=PAQ iff there is a single permutation matrix Bip(P) so that Bip(B)=Bip(P)^t Bip(A) Bip(P).
That means
\(A\sim B \Leftrightarrow Bip(A)\cong Bip(B)\)
 and if there is a canonisation function N for boolean matrices under independent row and column perms, then there is a canonisation function Bip(N) for adjacency matrices of bipartite graphs.
The whole thing also works in the other direction, so that BGC and BMC are in fact polynomial time reducible to each other.
Since it in particular works for adjacency matrices of Graphs, the graph isomorphism problem is also polynomially equivalent to the bipartite graph isomorphism problem.

Thus my problem is just as hard as the graph isomorphism problem. :-D
But somehow i find it intuitvely easier to permute rows and columsn independently from each other and i find equivalence of boolean matrices easier than graph isomorphism.
Graphs are so confusing and their visual representation contains more information than necessary. Boolean matrices only contain the relevant information (0 or 1) and nothing more.