TITLE: The relation between trace and determinant of a matrix
QUESTION [65 upvotes]: Let $M$ be a symmetric $n \times n$ matrix. 
Is there any equality or inequality that relates the trace and determinant of $M$?

REPLY [19 votes]: Due to OP's fairly general formulation there's diverse bunch of answers by now. In addition to these, I'd like to mention some concrete relations expressing the determinant in terms of traces. They hold without the symmetry hypothesis, just assume dealing with a general complex matrix.
Despite being "quite different beasts", both $\det(M)$ and $\operatorname{tr}(M)$ of an $n\times n$ matrix $M$ are given by some $n$-variable symmetric polynomial, evaluated at the eigenvalues of $M$.

For any fixed matrix size $n$ there's a polynomial in $n$ variables, of total degree $n$, which yields $\det(M)$ when evaluated at
  $\left(\operatorname{tr}(M^n),\operatorname{tr}(M^{n-1}),\ldots,\operatorname{tr}(M)\right)$:
  $$
\det M\; =\;
\begin{cases}
\operatorname{tr}M & n=1\\[1.67ex]
\frac{1}{2}\big((\operatorname{tr}M)^2
 - \operatorname{tr}\left(M^2\right)\big) & n=2\\[1.67ex]
\frac{1}{6}\big((\operatorname{tr}M)^3
 - 3\operatorname{tr}\left(M^2\right)(\operatorname{tr}M)
 + 2\operatorname{tr}\left(M^3\right)\big) & n= 3\\[1.67ex]
 \quad\ldots & n\ge4
\end{cases}
$$

Case $\mathbf{n=1}$ is clear.
$\mathbf{n=2}$ is straightforward when applying Cayley-Hamilton to $M$
$$
M^2\:-\:(\operatorname{tr}M)\:M\:+\:(\det M)\pmatrix{1&0\\ 0&1}\;=\;\pmatrix{0&0\\0&0}
$$
and taking the trace.
Referring to Paul's answer/example featuring
$M=\left(\begin{smallmatrix}1&k\\ k&1\end{smallmatrix}\right)$,
thus $M^2=\left(\begin{smallmatrix}1+k^2&2k\\ 2k&1+k^2\end{smallmatrix}\right)$,
one obtains
$$\det M\;=\;\frac{1}{2}\big(4-2(1+k^2)\big) = 1-k^2$$
for the sake of illustration, presumably not the quickest way towards $\det(M)$
(the parameter has been renamed to $k$ since $n$ denotes the matrix size).
Cases $\mathbf{n\ge3}$ get more expensive $\ldots$ and are available:
A suitable entry point is the corresponding subsection in the Wikipedia entry on determinants.

If you clicked the preceding link then you may scroll down just a bit to get into a det-tr-inequality for a positive-definite matrix, also worthwhile as answer to the OP. Or skip directly down to it ;-)