Let $k$ be a field. Can each degree $n$ polynomial $P(t) \in k[t]$ be written as the determinant of the matrix $A + tB$, where $A$ and $B$ are two *symmetric* $(n \times n)$-matrices with entries in $k$? 


Over an algebraically closed field this is pretty obvious, but is this also true for non-closed fields? 



Which polynomials are determinants of a symmetric matrix with linear entries?
Here is a simple argument showing that you can get any polynomial up to a constant factor. As Rob Israel and my comments above show, you might not be able to get rid of that constant.


If $\det(A\_1 t + B\_1) = f\_1(t)$ and $\det(A\_2 t + B\_2) = f\_2(t)$, then $\det \left( \begin{smallmatrix} A\_1 t+B\_1 & 0 \\ 0 & A\_2t + B\_2 \end{smallmatrix} \right) = f\_1(t) f\_2(t)$. Thus, we are immediately reduced to the case that $f$ is irreducible.


We also may ignore the case that $f(t) = t$, as that is easy to achieve.


Let $k$ be our ground field, and let $L = k(t)/f(t)$. Since $f$ is irreducible, this is a field. Choose a nonzero $k$-linear map $\tau: L \to k$. (If our extension is separable, it would be elegant to choose the trace, but this choice doesn't actually matter.) Let $u \in L$ be such that $\tau(u) \neq 0$. 


Define two $k$-linear symmetric bilinear forms on $L$: $\langle x,y \rangle\_1 = \tau(xy)$ and $\langle x,y \rangle\_2 = \tau(xyt)$. These are symmetric because multiplication is commutative. The first is non-degenerate because, for any non-zero $x \in L$, we have $\langle x, u x^{-1} \rangle\_1 \neq 0$. A similar argument applies to $\langle \ , \ \rangle\_2$, using that $t \neq 0$. (This is why we had to exclude $f(t)=t$.)


Observe that 
$$\langle x,y \rangle\_2 = \langle x, T y \rangle\_1$$
where $T$ is the linear map $L \to L$ given by multiplication by $t$.
Choose a $k$-basis for $L$. Let $A$ and $B$ be the matrices of $\langle \ , \ \rangle\_1$ and $\langle \ , \ \rangle\_2$ in this basis. Then the above equation shows that $A = BT$.


Since our bilinear forms are symmetric, $A$ and $B$ are symmetric matrices. Since our bilinear forms are nondegenerate, they are invertible. Since $A = BT$, we have $\det(A \lambda+B) = \det(B) \det(\lambda T + \mathrm{Id})$. The characteristic polynomial of $T$ is $f(t)$, so we are done.


**ADDED:** It looks like Allison Miller has just about the same argument, and also has some references to where this idea can be found in the literature.




