# Systems of linear differential equations - eigenvectors

Solve the following system of equations

$\begin{cases} x_1^{'}(t)=x_1(t)+3x_2(t) \\ x_2^{'}(t)=3x_1(t)-2x_2(t)-x_3(t) \\ x_3^{'}=-x_2(t)+x_3(t)\end{cases}$.

First, I create the column vectors $X$ and $X^{'}$. Then the matrix $$A= \begin{bmatrix} 1 & 3 & 0 \\ 3 & -2 & -1 \\ 0 & -1 & 1 \\ \end{bmatrix}$$

Now, I find the eigenvalues, $-4,3,1$ and their corresponding eigenvectors $(-3,5,1)^T (-3,-2,1)^T (1,0,3)^T$.

I'm just not sure how to take it from here and solve the system of differential equations. I want a diagonal matrix $D$ so that I can read the solutions easy, but I'm not sure how to do it.

EDIT

Building on @Francisco 's answer, I'd have that:

$$X=c_1 (-3,5,1)^T e^{-4t} + c_2 (-3,-2,1)^T e^{3t} + c_3 (1,0,3)^T e^{t}$$. But I believe this could be written in a simpler form.

• Have you tried writing the matrix in a basis of its eigenvectors? (i.e in the form $S M S^{-1}$) where M is a diagonal matrix – Pol van Hoften Feb 28 '14 at 17:07
• To write $X^{'}=S^{-1}DSX$? – jacob Feb 28 '14 at 17:17
• You have $X'(t)=A X(t)$. Then $X=\vec v e^{\lambda t}$ (where $\vec v$ is an eigenvector and $\lambda$ is an eigenvalue) is a solution to the system. The general solution is given by $X=c_1 \vec v_1 e^{\lambda_1 t} +c_2 \vec v_2 e^{\lambda_2 t} + c_3 \vec v_3 e^{\lambda_3 t}$ where $c_1$ $c_2$ and $c_3$ are arbitrary constants. – Francisco Feb 28 '14 at 17:25
• What would be a faster approach? – jacob Feb 28 '14 at 17:56
• @Francisco I updated my question. Also, is there a wiki page for the formula you mentioned? – jacob Feb 28 '14 at 19:28

The matrix $A$ that you have is symmetric. So it has an orthonormal basis of eigenvectors. The eigenvectors you have found are mutually orthogonal (which they must be because they correspond to different eigenvalues.) So, if you normalize your eigenvectors and make those normalized vectors the columms of a matrix $U$, then $U^{T}U=I$ is automatic, and $U^{T}AU=D$ is diagonal. Explicitly, $$U = \left[ \begin{matrix} -\frac{3}{\sqrt{35}} & -\frac{3}{\sqrt{14}} & \frac{1}{\sqrt{10}} \\ \frac{5}{\sqrt{35}} & -\frac{2}{\sqrt{14}} & 0 \\ \frac{1}{\sqrt{35}} & \frac{1}{\sqrt{14}} & \frac{3}{\sqrt{10}} \end{matrix}\right]$$ The inverse of $U$ is the transpose $U^{T}$ of $U$. And, $$U^{T}AU = \left[\begin{matrix}-4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1\end{matrix}\right]=D.$$ Equivalently, $$A = U\left[\begin{matrix}-4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1\end{matrix}\right]U^{T}=U DU^{T}.$$ The general solution is expressed in terms of $a=x_{1}(0)$, $b=x_{2}(0)$, $c=x_{3}(0)$ as $$\left[\begin{matrix}x_{1}\\x_{2}\\x_{3}\end{matrix}\right] = e^{tA}\left[\begin{matrix}a\\b\\c\end{matrix}\right] = Ue^{tD}U^{T}\left[\begin{matrix}a \\ b \\ c\end{matrix}\right] = U\left[\begin{matrix} e^{-4t} & 0 & 0 \\ 0 & e^{3t} & 0 \\ 0 & 0 & e^{t}\end{matrix}\right]U^{T} \left[\begin{matrix}a \\ b \\ c\end{matrix}\right]$$
• This would be the same as $$\begin{cases} x_1= -3c_1 e^{-4t} - 3c_2e^{3t} +c_3e^{t} \\ x_2= 5c_1 e^{-4t} -2c_2e^{3t} \\ x_3= c_1 e^{-4t} + c_2e^{3t} +3c_3e^{t}. \end{cases}$$ yes? – jacob Feb 28 '14 at 20:55
• The forms are equivalent, except that the constants in mine are $x_{1}(0)$, $x_{2}(0)$, $x_{3}(0)$, while your constants are not that. Yours are $x_{1}(0)=-3c_{1}-3c_{2}+c_{3}$, etc. Mine will be the same if you set $a=-3c_{1}-3c_{2}+c_{3}$, $b=\dots$, etc. – Disintegrating By Parts Feb 28 '14 at 21:10