Derivative of the smallest eigenvalue Suppose $S_0, S$ are known real symmetric matrices and we have the function $f(x) := \lambda_{0}(S_0+x S)$ where $\lambda_0$ is the smallest eigenvalue.
By symmetry, the eigenvalues are real and I believe the smallest eigenvalue is a smooth function of $x$. How can I find the derivative $\frac{df(x)}{dx}$
 A: Yes, since the matrices $S_0$ and $S$ are symmetric, then the matrix $S_0+xS$ is diagonalizable, which means that all the eigenvalues are semisimple and, as a result, that they are smooth functions of the parameters.
If the minimum eigenvalue $\lambda_0(S_0)$ is simple, then the derivative is simply given by $v_0^TSv_0$ where $v_0$ is the eigenvector of $S_0$ associated with the eigenvalue $\lambda_0(S_0)$.
When the minimum eigenvalue $\lambda_0(S_0)$ is not simple, then we have that the derivative of all those eigenvalues to be the eigenvalues of $V_0^TSV_0$ where $V_0$ consists all the eigenvectors of $S_0$ associated with the minimum eigenvalue of $\lambda_0(S_0)$.
However, note that those results are only valid in a neighborhood of $x=0$. More generally, if you want to behavior of the eigenvalues around the point $x_0$, then you will need to substitute $S_0$ by $S_0+x_0S$ in the above expressions.
Reference: Seyranian, Mailybaev, "Multiparameter Stability Theory with Mechanical Applications"
