Eigenvector multiplication

Question

I don't understand how multiplying eigenvetors by an expression like $e^{-2t}$ works, and results in this graph. Can someone explain this to me?

Answer 1

It works like multiplying a vector by a scalar ($e^{2t}$ is a scalar-valued function of $t$).

That is:

$$ \begin{pmatrix}x_1(t) \\ x_2(t)\end{pmatrix} = e^{-2t}\begin{pmatrix}1 \\ - 1\end{pmatrix} = \begin{pmatrix}e^{-2t} \\ - e^{-2t}\end{pmatrix}. $$

The plots are of the first coordinate vs. time, and of the second coordinate vs. time. This solution involves $3$ variables (one input variable, $t$, and two output variables $x_1$ and $x_2$), and so your options for plotting the solutions on a $2$D piece of paper are limited.

Perhaps the best way to visualize this solution would be a movie of a particle moving though $2$-D space: each frame corresponds to a specific value of $t$, and in that frame, the particle is at the position given by the solution.

Answer 2

We are given the system:

$$ x' = \begin{bmatrix}1 & 3\\3 & 1\end{bmatrix} x$$

We are not yet given any initial conditions, so we find the eigenvalues/eigenvectors and have the general solution:

$$X(t) = \begin{bmatrix}x_1(t)\\x_2(t)\end{bmatrix} = c_1 e^{-2t}\begin{bmatrix}1 \\-1\end{bmatrix} + c_2 e^{4t}\begin{bmatrix}1\\1\end{bmatrix}$$

Next, they give us three different sets of initial conditions (look at the bottom of the figure).

For the first $(x_1(0), x_2(0)) = (2, -2)$, we get $c_1 = 2, c_2 = 0$

This gives us the solution:

$$X(t) = \begin{bmatrix}x_1(t)\\x_2(t)\end{bmatrix} = 2 e^{-2t}\begin{bmatrix}1 \\-1\end{bmatrix} = \begin{bmatrix}2e^{-2t}\\-2e^{-2t}\end{bmatrix}$$

So, for the first set of initial conditions, $(x_1(0), x_2(0)) = (2, -2)$, we get:

$$x_1(t)= 2e^{-2t}, x_2(t) = - 2 e^{2t}$$

For the second set of initial conditions, $(x_1(0), x_2(0)) = (0.1, 0.1)$, we get:

$$x_1(t)= \dfrac{e^{4 t}}{10}, x_2(t) = \frac{e^{4 t}}{10}$$

For the third set of initial conditions, , we get:

$$x_1(t) = e^{-2 t} \left(1.9975\, +0.0025 e^{6 t}\right), x_2(t) = e^{-2 t} \left(0.0025 e^{6 t}-1.9975\right)$$

Now, they are merely plotting the three different $x_1(t)$ results to compare them. Here they are superimposed on the same plot (they are identical to the book and you can plot each one to verify that; note that blue is the first solution, orange the second and green the third).