Eigenvalues in terms of trace and determinant for matrices larger than 2 X 2 The eigenvalues of a $2\times2$ matrix can be expressed in terms of the trace and determinant. 
$\lambda_\pm = \frac{1}{2}\left(\textrm{tr} \pm \sqrt{\textrm{tr}^2-4\det}\right)$
Is there a similar formula for higher dimensional matrices?
Approach
The trace and determinant of a matrix are equal to the trace and determinant of the matrix in Jordan normal form. For a matrix in Jordan canonical form, $\textrm{tr } =\sum \lambda$ and $\det =\prod \lambda $.
Substituting these latter two identities into the first results in an identity, which is encouraging. I'm not sure how to check this assumption for larger matrices. I'm not sure how generate more than two eigenvalues from the first formula. For the $3\times3$ case, the first formula seems to break down. 

There is if you generalize in the correct manner.  The characteristic equation $\lambda^n+\sum\limits_{i=0}^{n-1}c_i\lambda^i=0$ can be expressed with coefficients in terms of the trace and the determinant of the matrix, but as $n$ grows, this gets extremely laborious. Please see this Wikipedia article.
Of particular interest are $c_{n-1}=-\DeclareMathOperator{\tr}{tr} \tr(M)$ and $c_0=(-1)^n\det(M)$.
