have you tried computing β on basis elements of Mn(R), considered as elements of R^(n^2)? for example, in M2(R), we can identify
[0 0]
[1 0] with e3 in R^4.
you should get an n^2 x n^2 matrix, since dim(Mn(R)) = n^2.
Hi,
I've some troubles finding a matrix of a bilinear form:
Given n >= 1, n in IN and beta : M_{n}(\mathbf{R}) X M_{n}(\mathbf{R}) ----> IR the application defined by beta( A,B)=Tr({t}^A,B) \forall A,B \in M_{n}(\mathbf{R}
(IN= positive integers, IR= real numbers).
Find the matrix of beta in the standar basis of M_{n}(\mathbf{R} .
So I'm stuck here because I don't see what kind of n by n matrix would give me
A*(matrix)*B=\beta (A,B).
I am really sorry the Latex compiler doesn't seem to work .