Eigenvalues and eigenvectors

We look for solutions to (1.2) of the form  

$$X(t)=e^{\sigma t}\xi,$$ (5)

where $\xi$ is an $n\times 1$ constant vector. Upon substitution, $d/dt(e^{\sigma t}\xi)=Ae^{\sigma t}\xi$which yields $\sigma e^{\sigma t}\xi=Ae^{\sigma t}\xi$ or  

$$A\xi=\sigma\xi.$$ (6)

One solution of this is $\xi=0$. We seek non-zero solutions; these $\xi$ are the eigenvectors of A, and the corresponding $\sigma$ are the eigenvalues.

Example 2

$$A=\pmatrix{4&2\cr 3&3}.$$

To calculate the eigenpairs, we seek non-zero solutions $\xi$ to $(A-\sigma I)\xi=0$. Thus, $\det(A-\sigma I)=0$.

$$\left\vert\begin{array} {ll} 4-\sigma & 2\ 3 & 3-\sigma\end{array}\right\vert =(4-\sigma)(3-\sigma)-6$$

$$=\sigma^2-7\sigma+6=(\sigma-6)(\sigma-1).$$

For $\sigma=6$, $(A-\sigma I)\xi=0$ with

$$\xi=\pmatrix{\xi_1\cr \xi_2}$$

yields

$$\pmatrix{-2&2\cr 3&-3}\pmatrix{\xi_1\cr \xi_2}=\pmatrix{0\cr 0}.$$

Thus, $-\xi_1+\xi_2=0$ or $\xi_1=\xi_2$. Therefore,

$$\xi=\xi_1\pmatrix{1\cr 1}.$$

For $\sigma=1$,

$$\pmatrix{3&2\cr 3&2}\pmatrix{\xi_1\cr \xi_2}=\pmatrix{0\cr 0}.$$

Thus, $3\xi_1+2\xi_2=0$ or $\xi_2=-{3\over 2} \xi_1$. Therefore

$$\xi=\xi_1\pmatrix{1\cr -{3\over 2}}.$$

Recall: If eigenvalues are distinct, then their eigenvectors are linearly independent. If A is Hermitian ($A=\overline{A}^T=A^*$or if A is real symmetric), then the eigenvectors form an orthogonal basis.

The linearly independent solutions of the form (1.5) are

$$e^{6t}\pmatrix{1\cr 1},\quad e^t\pmatrix{1\cr -{3\over 2}},$$

$$\Omega=\pmatrix{ e^{6t} & e^t\cr e^{6t}& -{3\over 2}e^t }.$$

The general solution is $X(t)=\Omega C$ where $C=(c_1 \ c_2)^T$:

x₁(t)=c₁e^6t+c₂e^t,

$$x_2(t)=c_1e^{6t}-{3c_2\over 2}e^t.$$

Solution of $\dot X=AX+G$ by diagonalizing A.

Recall that A is diagonalizable if A has a basis of eigenvectors. In this case, $A\xi_i=\sigma_i\xi_i$, i=1,2, ..., n. Define

$$P=\pmatrix{\xi_1&\xi_2 & ...&\xi_n}.$$

Then

$$AP=PD, \quad D=\pmatrix{\sigma_1 &0 ..&...&... \cr 0&\sigma_... ... ...\cr \vdots&\vdots&\vdots&\vdots\cr ...& ...&...& \sigma_n}.$$

We have P^-1AP=D. Now set X(t)=PZ(t) in the differential equation: $P\dot Z=APZ+G$. Multiply on the left by P^-1. $P^{-1}P\dot Z=P^{-1}APZ+P^{-1}G$ or $\dot Z=DZ+P^{-1}G$.This procedure has decoupled the unknowns. Thus, in component form, $\dot z_1(t) =\sigma_1 z_1(t)+f_1$ which you can solve for z₁, and so on for z_i(t). Finally, set X(t)=PZ(t) to retrieve the solution.