Sum of positive eigenvalues. We know trace of a matrix is the sum of the eigenvalues of the given matrix. Suppose we know the characteristics polynomial of the matrix, is there any result which gives us the sum of the positive eigenvalues of the matrix?
Note that I need the sum of only the positive eigenvalues...not all eigenvalues.
 A: This is essentially just as hard as finding the individual eigenvalues...
...in particular, because if it were easy, you could use it to find the individual eigenvalues.
From the characteristic polynomial of a matrix $A$, it is easy to get the characteristic polynomial of $A-kI$, and the sum of all the positive eigenvalues of $A-kI$ is equal to the sum of all the eigenvalues greater than $k$ of $A$. In particular, if $k$ is between the largest eigenvalue and the second-largest, this sum will just be the largest eigenvalue of $A$.
We can find such a $k$ with binary search (note that in particular, the sum of the positive eigenvalues of $A$ is an upper bound on the largest eigenvalue of $A$, so we have a range to work with).
Once we figure out the largest eigenvalue of $A$, we could account for it and use a similar process to find the second-largest eigenvalue, third-largest, and so on.

Note that "just as hard as finding the individual eigenvalues" means that in particular, there can be no exact formula for finding this sum, when the matrix is $5\times 5$ or larger - in contrast to the sum of all eigenvalues, which is just the trace.
A: Let $C$ be a large positively-oriented square contour with the left side of the contour along the imaginary axis. Then
$$
            p(\lambda)=(\lambda-\lambda_1)^{r_1}(\lambda-\lambda_2)^{r_2}\cdots(\lambda-\lambda_n)^{r_n} \\
          \frac{p'(\lambda)}{p(\lambda)}=\frac{r_1}{\lambda-\lambda_1}+\frac{r_2}{\lambda-\lambda_2}+\cdots+\frac{r_n}{\lambda-\lambda_n} \\
        \frac{1}{2\pi i}\oint_{C}\lambda\frac{p'(\lambda)}{p(\lambda)}d\lambda = \sum_{\lambda_j > 0}r_j\lambda_j = \mbox{sum of roots in the right half-plane}
$$
If you know all the roots are real, then you have what you want. If not, then you can integrate close enough to the real axis, and you'll be summing only the roots on the real axis. They will be summed according to multiplicity.
