- #1
kostoglotov
- 234
- 6
So, in a section on applying Eigenvectors to Differential Equations (what a jump in the learning curve), I've encountered
[tex]e^{At} \vec{u}(0) = \vec{u}(t)[/tex]
as a solution to certain differential equations, if we are considering the trial substitution [itex]y = e^{\lambda t}[/itex] and solving for constant coefficients. I understand the idea more or less, but the math gets quite involved, particularly with complex eigenvalues and eigenvectors.
I can see how (or at least I can follow and accept the explanation for) [itex]e^{At}[/itex] being an infinite series.
But how does [itex]e^{\Lambda t} = [/itex] a single matrix with [itex]e^{\lambda_i t}[/itex] on it's diagonal?
[tex]e^{At} \vec{u}(0) = \vec{u}(t)[/tex]
as a solution to certain differential equations, if we are considering the trial substitution [itex]y = e^{\lambda t}[/itex] and solving for constant coefficients. I understand the idea more or less, but the math gets quite involved, particularly with complex eigenvalues and eigenvectors.
I can see how (or at least I can follow and accept the explanation for) [itex]e^{At}[/itex] being an infinite series.
But how does [itex]e^{\Lambda t} = [/itex] a single matrix with [itex]e^{\lambda_i t}[/itex] on it's diagonal?