Link between two forms of solution for a linear dynamical system

Question

I have a linear dynamical system defined by $\dot{\mathbf{x}}(t) = \mathbf{A} \mathbf{x}(t)$. Assuming that $\mathbf{A}$ is diagonalizable and can be decomposed as $\mathbf{A} = \Phi \Lambda \Phi^{-1}$, the solution to the system of equations is:

$\mathbf{x}(t) = \Phi \exp(\Lambda t) \Phi^{-1} \mathbf{x}(0)$.

I have seen that this solution can also be expressed as $\mathbf{x}(t) = \exp(\mathbf{A}t) \mathbf{x}(0)$, meaning that:

$\exp(\mathbf{A}t) = \Phi \exp(\Lambda t) \Phi^{-1}$

This equality has already been used in the answer to a similar question, but I do not understand how one side can be obtained from the other. How would you demonstrate it?

Answer 1

For real numbers, certainly you have seen the Taylor series.

$e^{at} = \sum_{0}^{\infty} \frac {(at)^n}{n!}$

Now we replace the real $a$ with the matrix $A.$

And if $A$ is diagonalizable $A^2 = \Phi \Lambda \Phi^{-1} \Phi \Lambda \Phi^{-1} = \Phi \Lambda^2 \Phi^{-1}.$ With similar reasoning, you can conclude that $(At)^n = \Phi(\Lambda t)^n \Phi^{-1}.$

Applying it to the Taylor series above, $e^{At} = \Phi e^{\Lambda t} \Phi^{-1}.$

Answer 2

This folows from the fact that $Pexp(A)P^{-1}=exp(PAP^{-1})$. To see this, remark that recursively that $(PAP^{-1})^n=PA^nP^{-1}$, it is true for $n=0$, suppose it is true for $n$ $(PAP^{-1})^{n+1}=(PAP^{-1})^n(PAP^{-1}=PA^nP^{-1}PAP^{-1}=PA^{n+1}P^{-1}.$

You deduce that $\Phi (exp(\Lambda(t))\Phi^{-1}=exp(\Phi\Lambda(t)\Phi^{-1})=exp(At)$.