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

*

*The phase portrait has been "straightened"


*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?
 A: 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 $x=Py$.
A: 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 $y$.
