I am not sure if a square matrix A is symetric ( ) when we have the relation . Has anyone an idea how to prove this?
Thanks Ulrich
I think you are after something like this
Symmetric Matrices
Yes, but I think it is not possible to transform the first equation into the second because the matrices are supposed to not commute. Maybe it should be done by supposing that the first equation is also valid for
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...
Hi all,
I had some time recently and also found, as a "by-product", the solution to this problem. In fact the matrices
and
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.