Generalized eigenvectors and differential equations

[drawar]

Let $A$ be an 3x3 matrix such that $A\mathbf{v_1}=\mathbf{v_1}+\mathbf{v_2}, A\mathbf{v_2}=\mathbf{v_2}+\mathbf{v_3}, A\mathbf{v_3}=\mathbf{v_3}$ where $\mathbf{v_3} \neq \mathbb{0}$. Let $B=S^{-1}AS$ where $S$ is another 3x3 matrix.
(i) Find the general solution of $\dot{\mathbf{x}}=B\mathbf{x}$.
(ii) Show that 1 is the only eigenvalue of $B$.

It's clear that $\mathbf{v_3},\mathbf{v_2}$ and $\mathbf{v_1}$ form a chain of generalized eigenvectors associated with $\lambda=1$ and hence are linearly independent. From this I can find the general solution of $\dot{\mathbf{x}}=A\mathbf{x}=SBS^{-1}\mathbf{x}$ but how can I proceed from here to find the general solution of $\dot{\mathbf{x}}=B\mathbf{x}$?
Any help is much appreciated, thank you!

 

[AlephZero]

You have $A = S B S^{-1}$.

If $x$ is a solution of $\dot x = Bx$, let $x = S^{-1} y$.

Then $\dot y = Ay$.

 

[drawar]

That's brilliant, thanks!