Why does using orthogonal matrices as change of basis produce decoupled system of equation?

Question

Suppose I have the following set of differential equations

$\dot x_1 = -x_1 - 3x_2 \quad \dot x_2 = 2x_2$

This is the phase portrait of our system

But let's take a change of coordinate using $P = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}$ where each of the columns corresponds to an eigenvector

Then let $y = P^{-1}x$

Couple things:

1. The phase portrait has been "straightened"

2. The system of equation becomes decoupled (i.e. only diagonal elements exist)

What accounts for this amazing behavior? Why is it this particular change of coordinate matrix is able to yield a new set of equations with these properties?

Answer 1

While @Artem's answer is completely true, I wanted to give some intuition behind it. First observe that if $(\lambda, u)$ is an eigenpair of $A$, then $x(t) = e^{\lambda t} u$ solves the ODE $\dot{x} = Ax$. Then for simple eigenvalues, the general solution is $$x(t) = c_1 e^{\lambda_1 t} u_1 + c_2 e^{\lambda_2 t} u_2$$ This is why if you start on an eigenvector, e.g. $x(0) = u_1$, $c_1 = 1, c_2=0$, then you will stay on it forever. Of course, you can start from any $(c_1, c_2)$ pair, in fact if you plot the solutions for some of them, you obtain the phase portrait.

See $(c_1, c_2)$ is the coordinate values for the initial state (and trajectories) for the basis $\{ u_1, u_2 \}$. This basis may not be "pretty" as in your first plot. So we can change the basis of the states by defining $y = P^{-1} x$ where $P = [u_1 ~~ u_2]$ to "cancel out" the eigenvectors and obtain a "nice" basis like $\{e_1, e_2\}$ where $e_1^T = [1 ~~ 0]$ and $e_2^T = [0 ~~ 1]$. Also note that $e_1, e_2$ are eigenvectors of $P^{-1} A P$, which is the new system matrix for the states $y$.

Answer 2

To clarify the issue you need to prove the following linear algebra fact:

For any real 2 by 2 matrix $A$ there exists non-degenerate matrix $P$ such that $$ P^{-1}AP=J, $$ where $J$ can be in one of the following real Jordan's normal forms: $$ \begin{bmatrix} \lambda_1 & 0\\ 0 & \lambda_2 \end{bmatrix},\quad \begin{bmatrix} \lambda & 1\\ 0 & \lambda \end{bmatrix},\quad \begin{bmatrix} \alpha & \beta\\ -\beta & \alpha \end{bmatrix}. $$

After this observe what happens with your ODE $$ \dot x=Ax, $$ if you make the change $x=Py$.