Given a vector space of square matrices $M_n(R)$ over $R$. $B(M,N)=tr(MN)$, M, N elements of $M_n(R)$ . B is a bilinear form on $M_n(R)$.
What is the matrix of $M_n(R)$ in the basis $(E_{i,j})_{1\leq i, k\leq n}$ of $M_n(R)$ where $(E_{i,j})= 1, k=i, l=j \\ 0, else \\$
Im quite confused, since I read this $(E_{i,j})$ as the identity basis, so the matrix representation wouldnt change. Right? We then need to find whether it is symmetric, non-degenerate and the signature of B but i cant do any of this without understanding this. How should the definition of B as the trace of 2 elements of the subspace be brought into it? I just really dont know what more to say to succesfully expand on this. Please help!
I don't understand your reference to “identity matrix”. There is no identity matrix in the statement of your problem.
Note that$$B(E_{i,j},E_{k,l})=\begin{cases}1&\text{ if }k=j\text{ and }l=i\\0&\text{ otherwise.}\end{cases}$$So, matrix that you're after only has $0$'s and $1$'s. If, say, $n=2$ and your basis is $(E_{1,1},E_{1,2},E_{2,1},E_{2,2})$, then the matrix is$$\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}.$$The answer, of course, depends upon the order that you pick for the elements of your basis.