If I have an $n\times n$ hermitian matrix $A$ and I want to find all the eigenvalues of $A$, i.e $\{\lambda_{i}\}$, $i=1,...,n$ where $\lambda_{i+1}>\lambda_{i}$, if I only know the biggest eigenvalue (found using SDP), i.e $\lambda_{n}$, my question is:
How can I transform $\{\lambda_{i}\rightarrow \lambda'_{i} \}$ ($A \rightarrow A'$) in order to convert $\lambda_{n-1}$ in the 'new' biggest eigenvalue of $A'$, $\lambda'_{n}$, and then apply SDP to $A'$ finding the new biggest eigenvalue, i.e $\lambda'_{n}$, the second biggest eigenvalue of $A$, $\lambda_{n-1}$?
Set $A' = A - \lambda_n v_n v_n^*$ where $v_n$ is a normalized eigenvector for $\lambda_n$. The spectrum of $A'$ is that of $A$, except $\lambda_n$ is now replaced with 0.