Generalized eigenvectors and differential equations

Click For Summary
The discussion centers on finding the general solution of the differential equation \(\dot{\mathbf{x}}=B\mathbf{x}\) given a specific 3x3 matrix \(A\) and its relationship to matrix \(B\). It is established that the vectors \(\mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3}\) form a chain of generalized eigenvectors associated with the eigenvalue \(\lambda=1\), indicating their linear independence. The transformation \(x = S^{-1} y\) is suggested to relate solutions of \(\dot{x} = Bx\) to \(\dot{y} = Ay\). This approach simplifies finding the general solution for the system. The discussion concludes with an acknowledgment of the effectiveness of this method in solving the problem.
drawar
Messages
130
Reaction score
0
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!
 
Physics news on Phys.org
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##.
 
That's brilliant, thanks!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
656
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K