1. Jun 22, 2005

### 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?

Last edited: Jun 22, 2005
2. Jun 23, 2005

### 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?

3. Jun 23, 2005

### 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.

4. Jun 23, 2005

### 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.

Last edited: Jun 23, 2005
5. Jun 24, 2005

### 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.