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.

  • $\begingroup$ 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? $\endgroup$ – Viktor Glombik Dec 21 '18 at 22:32
  • $\begingroup$ 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. $\endgroup$ – alphanzo Dec 21 '18 at 22:37

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