I think you are after something like this
and then arrive at some contradiction. Or equivalently puting
, from which follows
and by comparision
Does this imply that B is the identity matrix? If yes, why? It's some time ago when I took courses in linear algebra...
I had some time recently and also found, as a "by-product", the solution to this problem. In fact the matrices
do not commute but are nevertheless special because where (this is also true for the 3-dimensional matrices I mentionned before).
From this follows that with B not necessarily being the identity matrix. So if one imposes that B and P do not commute, then implies that A is symmetric.
I also asked earlier how it is possible to show from that B is the identity matrix. This can be done as follows:
Given P and Q two arbitray matrices, we want to determine another matrix B such that is verified. From some theorem (sorry forgot the name) there is a matrix E such that Q can be diagonalised. In other words, we get where E is the matrix formed by the eigenvectors of Q, let's say . D is the diagonal matrix with the eigenvalues of Q. This also yields
So EB is the matrix formed with the eigenvectors of P, that is , which yields .
This is why, if we have like above, B must be the identity matrix.