Finding eigenvectors for 3×3 system of differential equations

[VinnyCee]

Here is the problem:

$$X'\,=\,\left(\begin{array}{ccc}3 & -1 & -1\\1 & 1 & -1\\1 & -1 & 1\end{array}\right)\,X$$

Here is what i have so far:

$$det(A\,-\,rI)\,=\,0$$

$$det\left[\left(\begin{array}{ccc}3 & -1 & -1\\1 & 1 & -1\\1 & -1 & 1\end{array}\right)\,-\,\left(\begin{array}{ccc}r & 0 & 0\\0 & r & 0\\0 & 0 & r\end{array}\right)\right]\,=\,\left(\begin{array}{ccc}3\,-\,r & -1 & -1\\1 & 1\,-\,r & -1\\1 & -1 & 1\,-\,r\end{array}\right)$$

$$-(r\,-\,1)\,(r^2\,-\,4r\,+4)\,=\,0$$

$$r_1\,=\,1,\,\,\,\,r_2\,=\,2,\,\,\,\,r_3\,=\,2$$

$r\,=\,2$ is repeated once.

$$(A\,-\,r_1I)\,\xi\,=\,0$$

$$\left[\left(\begin{array}{ccc}3 & -1 & -1\\1 & 1 & -1\\1 & -1 & 1\end{array}\right)\,-\,\left(\begin{array}{ccc}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{array}\right)\right]\,\left(\begin{array}{ccc} \xi_1 \\ \xi_2 \\ \xi_3 \end{array}\right)=\,\left(\begin{array}{ccc}2 & -1 & -1\\1 & 0 & -1\\1 & -1 & 0\end{array}\right)\,\left(\begin{array}{ccc} \xi_1 \\ \xi_2 \\ \xi_3 \end{array}\right)\,=\,0$$

Now, multiplying out the right hand side of the equation above, and rref'ing, I get:

$$\left(\begin{array}{cccc} 1 & 0 & -1 & 0 \\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)$$

Which means that,

$$\xi_1\,-\,\xi_3\,=\,0$$
$$\xi_2\,-\,\xi_3\,=\,0$$

So then,

$$\xi_1\,=\,\xi_2\,=\,\xi_3?$$

Here is where I am stuck! What am I supposed to do now to solve this system?

 

[PhilG]

Just pick a number, like $$\xi_1 = \xi_2 = \xi_3 = 1$$ or something. That's just saying that any vector with all three components the same is an eigenvector corresponding to eigenvalue r = 1. Is that what you are asking?

 

[VinnyCee]

Ok, $\xi_1=\xi_2=\xi_3=1$.

What do I do with this information?

Do I just solve for the $r_2=r_3=2$ part and that is the answer?

I am still not understnding exactly what it is that I am supposed to be doing with the $\xi_n$ values.

 

[geosonel]

(note: unfortunately, there's a little difference between VinnyCee's notation and what follows. VinnyCee uses ξ₁, ξ₂, and ξ₃ to represent eigenvector components. In what follows, ξ₁, ξ₂, and ξ₃ are the actual eigenvectors themselves. sorry.)

from your results (& standard textbook): for single unique eigenvalue r₁=1 and its eigenvector ξ₁=<1, 1, 1>, a specific solution is:

$$X_{1}(t) \ = \ C_{1} \cdot \exp(r_1 t) \cdot \xi_{1} \ = \ C_1 \cdot \exp(t) \cdot \left ( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right )$$

then for repeated eigenvalue r₂₃=2 and its eigenvectors ξ₂=<2, 1, 1> and ξ₃=<1.33, -1, 2.33>, a specific solution is:

$$X_{23}(t) \ = \ C_{2} \cdot \exp(r_{23} t) \cdot \xi_{2} \ + \ C_{3} \cdot \exp(r_{23} t) \cdot ( \xi_{2} t \ + \ \xi_{3} ) \ = \ C_{2} \cdot \exp(2 t) \cdot \left ( \begin{array}{c} 2 \\ 1 \\ 1 \end{array} \right ) \ + \ C_3 \cdot \exp(2t) \cdot \left ( \left ( \begin{array}{c} 2 \\ 1 \\ 1 \end{array} \right ) \cdot t \ + \ \left ( \begin{array}{c} 1.33 \\ -1 \\ 2.33 \end{array} \right ) \right )$$

$$\mbox{General Solution} \ = \ \textbf{X(t)} \ = \ X_{1}(t) \ + \ X_{23}(t)$$

general solution is (X₁(t) + X₂₃(t)), and constants C₁, C₂, and C₃ obtained from initial cond.

 

[VinnyCee]

The general solution is...

Is this the correct general solution?:

$$X\,=\,\ C_1 \cdot e^t \cdot \left ( \begin{array}{c} 1 \\ 1 \\ 1 \end{array} \right ) \,+\,\ C_2 \cdot e^{2t} \cdot \left ( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right ) \,+\,\ C_3 \cdot e^{2t} \cdot \left [ \left( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right )\cdot t\,+\,\left ( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right ) \right]$$

This is using different eigenvectors than you calculated. I got:

$$\xi^{(2)}\,=\,\left ( \begin{array}{c} 1 \\ 0 \\ 1 \end{array} \right )$$ and $$\xi^{(2)}\,=\,\left ( \begin{array}{c} 1 \\ 1 \\ 0 \end{array} \right )$$.

I think that they are just multipules of your solutions.