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Oct 12, 2011 at 21:38 vote accept Wanderer
Sep 12, 2011 at 2:07 answer added David E Speyer timeline score: 10
Sep 12, 2011 at 1:52 comment added David E Speyer Although you can get $−1−t^2$; it's $\det \left( \begin{smallmatrix} 1 & t \\ t & -1 \end{smallmatrix} \right)$
Sep 12, 2011 at 1:43 comment added Alison Miller Do you care about the case of polynomials of degree $<n$?
Sep 12, 2011 at 1:21 comment added Robert Israel For $n=2$ and field $\mathbb R$, you can't get $1 + t^2$
Sep 12, 2011 at 1:14 answer added Andreas Thom timeline score: 4
Sep 12, 2011 at 0:54 answer added Alison Miller timeline score: 5
Sep 11, 2011 at 21:23 comment added Wanderer This has something to do with quadratic forms, of course.
Sep 11, 2011 at 21:09 comment added Wanderer Sorry, I forgot to mention the matrices should be symmetric.
Sep 11, 2011 at 21:08 history edited Wanderer CC BY-SA 3.0
added 12 characters in body; edited title
Sep 11, 2011 at 21:04 comment added David Loeffler en.wikipedia.org/wiki/Companion_matrix
Sep 11, 2011 at 21:00 history asked Wanderer CC BY-SA 3.0