Testing math stuff

Consider the ordinary differential equation

$\begin{displaymath} %%anchor: ODE \frac{dx}{dt} = \lambda x,\end{displaymath}$

(1)

where $ \lambda $ is a constant. This type of equation is called "separable," because you can algebraically get all the $ x $'s on one side and all the $ t $'s on the other:

$\begin{displaymath} %%anchor: ODE-separated \frac{dx}{x} = \lambda \, dt.\end{displaymath}$

(2)

Now, we can integrate both sides to get the general solution of (1):

$\begin{displaymath} \int \frac{dx}{x} = \lambda \int dt\end{displaymath}$

$\begin{displaymath} %%anchor: soln-log \ln |x| = \lambda t + C.\end{displaymath}$

(3)

Unfortunately, (3) isn't an explicit solution for $ x(t) $. But if we exponentiate both sides, we quickly find it:

$\begin{displaymath} |x| = e^{\lambda t + C} = e^C e^{\lambda t}.\end{displaymath}$

We can do better. Let $ A = \pm e^C $. Then, we have

$\begin{displaymath} %%anchor: soln-general x(t) = A e^{\lambda t}.\end{displaymath}$

(4)

Suppose $ x(t=0) = x_0 $. Then, from (4), we find $ A = x_0 $. Thus,

$\begin{displaymath} %%anchor: soln-x_0 x(t) = x_0 e^{\lambda t}.\end{displaymath}$

(5)

Theorem

If $ A $ is an $ n \times n $ matrix with the property that $ A^2 = A $, and $ \lambda $ is an eigenvalue of $ A $, then $ \lambda $ must be either 0 or 1.

Proof

By definition, if $ \lambda $ is an eigenvalue, then there is a non-trivial vector $ \vec{x} $ such that

$\begin{displaymath} %%anchor: eigenvalue A \vec{x} = \lambda \vec{x}.\end{displaymath}$

(6)

Left-multiplying (6) by $ A $:

$\begin{displaymath} A^2 \vec{x} = \lambda A \vec{x}.\end{displaymath}$

Using (6) and noting that $ A^2 = A $, we have

$\begin{displaymath} A \vec{x} = \lambda(\lambda \vec{x}) = \lambda^2 \vec{x}.\end{displaymath}$

Again using (6), we find

$\begin{displaymath} \lambda \vec{x} = \lambda^2 \vec{x},\end{displaymath}$

$\begin{displaymath} \lambda(1 - \lambda) \vec{x} = 0.\end{displaymath}$

Since $ \vec{x} $ is non-trivial, this can only be true if $ \lambda = 0 $ or $ \lambda = 1 $.

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