# Why do we say that the roots of the characteristic polynomial of an ODE, are the eigenvalues?

Suppose I have a differential equation, $$\ddot{x} - \dot{x}-6x = 0$$

I've read two ways to obtain the solution,

$$i)$$ Assume $$x = e^{\lambda t}$$, then I obtain the characteristic equation, $$\lambda^2 - \lambda -6 =0$$Now I obtain the values for $$\lambda$$, namely $$\lambda_1 = 3$$ and $$\lambda_2=-2$$. Now the solution is $$x(t) = c_1 e^{3t} +c_2e^{-2t}$$

$$ii)$$ Form a system of equations by introducing dummy variables, $$\dot{x} = v\\ \dot{v} = 6x+v$$

$$\begin{bmatrix} \dot{x}\\ \dot{y} \end{bmatrix}=\begin{bmatrix} 0 & 1\\ 6 & 1 \end{bmatrix}\begin{bmatrix} x\\ v \end{bmatrix}$$ $$\dot{X}=AX$$ Find the eigen values and vectors of $$A$$, $$\lambda_1 = 3$$ and $$\xi_1 =[1\,\,\,3]$$, $$\lambda_2=-2$$ and $$\xi_2=[1\,\,\,-0.5]$$

$$\begin{bmatrix} x(t)\\ v(t) \end{bmatrix}=\begin{bmatrix} 1 & 1\\ 3 & -0.5 \end{bmatrix}\begin{bmatrix} e^{3t} & 0\\ 0 & e^{-2t} \end{bmatrix}\begin{bmatrix} 1 & 1\\ 3 & -0.5 \end{bmatrix}^{-1}\begin{bmatrix} x(0)\\ y(0) \end{bmatrix}$$ $$X(t)=TDT^{-1}X(0)$$ Finally, it also yields $$x(t) = c_1 e^{3t} +c_2e^{-2t}$$

My Question is:

Why do we get the characteristic equation while solving for the eigen values? What is the relation between these two methods?

• It is in general true and the aim of the construction that, given some polynomial $p$, the characteristic polynomial of its companion matrix is again the original polynomial $p$. Commented Jan 1, 2020 at 13:52
• @LutzLehmann Ahh, so the reason is, we tailored the companion matrix to give us the characteristic polynomial? Commented Jan 1, 2020 at 14:28
• There are more organic interpretations, for instance that the companion matrix is the matrix for the multiplication operator $x$ modulo $p(x)$ in the monomial basis, which corresponds here to the construction of the first order system via the derivatives array. Commented Jan 1, 2020 at 14:40

Considering the equivalent system

$$\cases{ \dot x_1 = x_2\\ \dot x_2 = x_2 + 6x_1 },\ \ \ \text{or}\ \ \ \dot X = A X$$

supposing we can factorize $$A = Q^{-1}\Lambda Q$$ we have then

$$\dot X = Q^{-1}\Lambda Q X$$

and now introducing the change of variables $$Y = Q X$$ we follow with

$$\dot Y = \Lambda Y$$

so concluding, if we can do $$A = Q^{-1}\Lambda Q$$ then there exist a reference coordinate system such that the solutions are associated to the $$A$$ eigenvalues.

NOTE

Here $$\Lambda$$ is the diagonal eigenvalues matrix and $$Q$$ the eigenvectors matrix, both associated to $$A$$ matrix.

$$Q = \left( \begin{array}{cc} 1 & -1 \\ 3 & 2 \\ \end{array} \right),\ \ \ \Lambda = \left( \begin{array}{cc} 3 & 0 \\ 0 & -2 \\ \end{array} \right)$$

The solution for $$\dot Y = \Lambda Y$$ is

$$\cases{ y_1 = c_1 e^{3t}\\ y_2 = c_2 e^{-2t} }$$

and to recover the initial coordinates we should do

$$X = Q^{-1}Y = \left( \begin{array}{c} \frac{2}{5} e^{3 t} c_1+\frac{1}{5} e^{-2 t} c_2 \\ \frac{1}{5} e^{-2 t} c_2-\frac{3}{5} e^{3 t} c_1 \\ \end{array} \right)$$

• Thank you very much, but can you please explain the last sentence a bit more. Why is that true, an example maybe, please. Commented Jan 1, 2020 at 12:27
• Detailed as required. Commented Jan 1, 2020 at 13:41