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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

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    $\begingroup$ I am deleting my answer, it was misguided. You cannot do Gram-Schmidt because this bilinear form is not an inner product. I am not sure how you attempted it even, because for $e_3$ when you try to compute ${\rm tr}(e_3^2)$ you get $0$, and you can't divide by $0$ as GS would have you do. $\endgroup$ – anon Dec 21 '13 at 18:41
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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.

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  • $\begingroup$ I think it's worth pointing out that even though this "guessing" approach feels a bit like cheating, in the long run, this will be better than Gram-Schmidt. For instance, if I you had to do this problem for $10 \times 10$ matrices, Gram-Schmidt would be an absolute nightmare but you'd still be able to guess a basis without too much trouble. $\endgroup$ – Alexander Dec 21 '13 at 18:29
  • $\begingroup$ Thanks, I'd found these matrices after trying to G-S it, without actually getting it through G-S, just by looking at it as you said. I'd just wanted to get the GS right myself. Nonetheless, I got them anyway, so thanks! $\endgroup$ – Np92 Dec 21 '13 at 18:41
  • $\begingroup$ You're welcome! Never underestimate the power of guessing the answer to a problem. $\endgroup$ – Alexander Dec 21 '13 at 18:45
  • $\begingroup$ Also, if you are happy with my answer, you should accept it as this will increase the likelihood of a lot of attention to your future questions. $\endgroup$ – Alexander Dec 21 '13 at 19:12
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    $\begingroup$ @Justanothermathstudent Analogously to the first 2 matrices, you want $n \times n$ matrices that are nonzero in exactly 1 diagonal entry. The analogue of the third matrix is symmetric matrices that are 0 on the diagonal and have exactly 2 nonzero entries. The analogue of the 4th matrix is skew-symmetric matrices with exactly two nonzero entries. Then you just need to check that these actually work! $\endgroup$ – Alexander Oct 20 '18 at 20:52

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