In revision for an upcoming exam, I've come across the following question: Let the bilinear form (A,B) be defined as tr(AB) on the space of 2x2 real matrices. Find an orthogonal basis for the form.
I know that when working with vectors, with (x,y) = (xt)Ay that I find the matrix by taking the bilinear form of basis elements, but I'm not sure how to do that here when working with matrices.
I also know I'm supposed to get a number from this form (cause its the trace) so I started considering the matrices as 4-d vectors then, although I'm really not sure if I'm on the right track or not. (Not too sure where to go from here) One thought that turned me against it was if I just write the four entries of the vector as the 1st row of the matrix, then 2nd row, I'm not sure how that will get me tr(AB) as opposed to the trace of A times the transpose of B. I do know how to use Gram-Schmidt, but I presume that comes afterwards?
Help would be greatly appreciated, as this question has really exposed my lack of knowledge on this topic. Thanks
This is the sort of problem that is easily done by trial and error. You need to find four matrices $M_1, ..., M_4$ such that $Tr({M_i}^2) \neq 0$ and $Tr(M_i M_j)=0$ for $i \neq j$. We would like to find matrices with lots of zeroes to make the orthogonality condition easy to satisfy.
So the matrices $\begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix}$ are great candidates. Now look at $\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$.
You can check that these four matrices solve the problem.