Eigenvalues of a matrix given its derivative.

Suppose I have a large matrix $$A(x,t)$$ depending on two free parameters (so each input of $$A$$ is a continuous function of $$x$$ and $$t$$). Furthermore, $$B(x,t)=\frac d{dt}A(x,t)\\ C(x,t)=\frac d{dx}B(x,t)$$ If I know all the eigenvalues and eigenvectors of $$A$$, and I know all the eigenvalues and eigenvectors of $$C$$, then can I use any of that information to find the complete eigenvalues and eigenvectors of $$B$$?

I know that if such a relationship exists that it is non-trivial, since, for example, if $$A=\sum \lambda_i |v_i\rangle\langle v_i|$$ then $$B$$ is given by $$B=\sum (\frac d{dt}\lambda_i) |v_i\rangle\langle v_i|+ \lambda_i |\frac d{dt}v_i\rangle\langle v_i|+ \lambda_i |v_i\rangle\langle \frac d{dt}v_i|$$ which shows that $$B$$ is not diagonalized the same as $$A$$, but surely a relationship must exist between them.

• Something seems to be wrong with your notation (or I'm not familiar with it): What is $$\lambda_i | v_i \rangle \langle v_i|$$ supposed to mean? – Viktor Glombik Dec 21 '18 at 22:32
• Very sorry, it is common notation in physics and engineering. $\lambda_i$ is the ith eigenvalue, and $|v_i\rangle$ (resp. $\langle v_i|$) is the ith column (resp. row) eigenvector. It is a convenient way of represented the Jordon decomposition for a diagonalizable matrix. – alphanzo Dec 21 '18 at 22:37