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?

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    $\begingroup$ en.wikipedia.org/wiki/Companion_matrix $\endgroup$ – David Loeffler Sep 11 '11 at 21:04
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    $\begingroup$ Sorry, I forgot to mention the matrices should be symmetric. $\endgroup$ – Wanderer Sep 11 '11 at 21:09
  • $\begingroup$ This has something to do with quadratic forms, of course. $\endgroup$ – Wanderer Sep 11 '11 at 21:23
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    $\begingroup$ For $n=2$ and field $\mathbb R$, you can't get $1 + t^2$ $\endgroup$ – Robert Israel Sep 12 '11 at 1:21
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    $\begingroup$ Although you can get $−1−t^2$; it's $\det \left( \begin{smallmatrix} 1 & t \\ t & -1 \end{smallmatrix} \right)$ $\endgroup$ – David E Speyer Sep 12 '11 at 1:52

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.

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    $\begingroup$ It turns out that letting $\tau$ being the map that sends $b_0 + b_1 t + \dotsb + b_{n-1} t^{n-1}$ to its coefficient $b_{n-1}$ of $\tau^{n-1}$ is a useful choice. (In that it makes the pairing $(x,y)\mapsto \tau(xy)$ nondegenerate, even in the general case of arbitrary $f$.) $\endgroup$ – Alison Miller Sep 12 '11 at 3:17

The following is true: if the leading coefficient of $-P(t)$ is an $n$th power, then $P$ is achievable. Furthermore, replacing $A, B$ with $MAM^T$, $MBM^T$ for a matrix $M$ of determinant $d \in k$ shows that if $P$ is achievable, so is $d^2 P$. These two facts together imply that any polynomial of odd degree is achievable.

Also this seems related to Gross's recent talk on the arithmetic of pencils of quadrics which is online at http://techtv.mit.edu/collections/harris60/videos/13950-benedict-gross, but I'm not sure if there's a direct connection. Also, recent work of Melanie Wood may also be relevant: (in particular the parts of http://arxiv.org/PS_cache/arxiv/pdf/1008/1008.4781v1.pdf where she discusses symmetric tensors, and theorem 5.7, although you probably want to take your base ring $R$ to be a field).

The construction for matrices $A$ and $B$ in the case when $n$ is odd and $-P(t)$ is monic with no repeated roots is sketched in Bhargava and Gross's recent preprint on arithmetic invariant theory (http://www.math.harvard.edu/~gross/preprints/invariant.pdf, middle of page 8). Since it's buried mixed in with a lot of other stuff, I'll unpack it here. (It is essentially David Speyer's argument.)

Let $f(t) = (-1)^n P(t)$. By our assumption and rescaling, WLOG assume $f$ monic.

Let $K$ be the $k$-algebra $k[u]/f(u)$ (not necessarily etale, as David Speyer pointed out, since $f$ need not be separable), and let $\beta$ be the image of $u$ in $K$. We define trace and norm maps from $K$ down to $k$ in the standard way; namely, $\mathop{\mathrm{tr}}(x)$ and $\mathop{\mathrm{Norm}}(x)$ are the trace and norm respectively of the linear operator on $K$ given by multiplication by $x$.

We'll use the basis $1, \beta, \beta^2, \dotsc, \beta^{n-1}$ of $k$ over $K$. There's a nice nondegenerate pairing on $K$, which is the following: let $\langle a, b \rangle$ be $(-1)^{\binom{n}{2}}$ times the coefficient of $\beta^{n-1}$ in $ab$ (when written out in terms of our basis). It's easily checked that this pairing has determinant $1$ with respect to our basis (the entries above the antidiagonal are all $0$) so is nondegenerate.

(An earlier version of this answer used the trace pairing instead, but that runs into some issues, and this the pairing Bhargava and Gross actually use in their paper.)

Let $B$ be the matrix of the quadratic form $\langle, \rangle$ that we just defined. Let $A$ be the matrix of the quadratic form $(x, y) \mapsto \langle{\beta x, y}\rangle$.

Then $\det(A - t B)$ is the determinant of the pairing $(x, y) \mapsto \langle(\beta-t)x, y\rangle$ (I'm implicitly extending scalars to $k[t]$, or if $k$ is infinite, just think of this as a function of $t$). But that determinant is equal to $\mathop{\mathrm{Norm}}(\beta - t)$ times the determinant of the pairing $\langle, \rangle$ (with respect to our basis). But the latter determinant is just $1$, as stated above. Furthermore, by unpacking the definitions and using the definition of the characteristic polynomial, we get $\mathop{\mathrm{Norm}}(\beta - t) = (-1)^n f(t) = P(t)$, and so $\det(A - tB) = P(t)$ as desired. (Well, actually, not quite as desired, but $A - tB = A + t(-B)$, so we're good.)

Also note that nothing goes wrong if we choose $P$ to have repeated roots.

I still need to think about the general case.

  • $\begingroup$ I believe that Robert Israel's comment is correct, and that what I am currently actually able to show is the following: any such polynomial should be attainable provided that either $n$ is odd or $P(t)$ has negative leading coefficient. $\endgroup$ – Alison Miller Sep 12 '11 at 1:38
  • $\begingroup$ Actually, the previous comment is only true over $\mathbb{R}$. $\endgroup$ – Alison Miller Sep 12 '11 at 2:01
  • $\begingroup$ The correct version: if the leading coefficient of $-P(t)$ is an $n$th power in $k$, $P$ is attainable. $\endgroup$ – Alison Miller Sep 12 '11 at 2:36

This is (roughly) true over any ring even for polynomials in many variables. This result for $\mathbb R$ is from

J.W. Helton and S. McCullough and V. Vinnikov, Noncommutative convexity arises from linear matrix inequalities, J. Funct. Anal. 240 (2006), no. 1, 105–191.

A simple proof can be found in

R. Quarez, Symmetric Determinantal Representation of Polynomials, Preprint.

The precise result states

Theorem: Let $p(x_1,\dots,x_n)$ be a polynomial of degree $d$ in $n$ variables over a ring $R$ of characteristic different from $2$. Let $N := 2 {{n+⌊ d/2 ⌋}\choose{n}}$. Then, there is a symmetric $N × N$ affine linear pencil $A_0 + A_1 x_1 + \cdots A_n x_n$ with entries in $R$ such that $$p(x_1,\dots,x_n) = det(A_0 + A_1 x_1 + \cdots A_n x_n).$$

Note that in the case $n=1$, $N = 2 {{1+⌊ d/2 ⌋}\choose{1}} = 2 + 2⌊ d/2 ⌋$. So, the size of the matrix in this case is not precisely $d$, but $d+2$ is always enough.


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