# Thread: Systems of first order linear equations

1. ## Systems of first order linear equations

$x' = \left[ \begin{array}{cccc} -\frac{5}{2} & 1 & 1 \\ 1 & -\frac{5}{2}& 1 \\ 1 & 1 & -\frac{5}{2} \end{array} \right]x$

I have found that the eigenvalues are $\lambda_{1,2}=-\frac{7}{2}$ and $\lambda_{3} = -\frac{1}{2}$

However finding the eigenvectors is where I get a little confused. Looking first at the $-\frac{7}{2}$ value we get:

$\left[ \begin{array}{cccc} 1 & 1 & 1 \\ 1 & 1& 1 \\ 1 & 1 & 1 \end{array} \right] \begin{bmatrix} \epsilon_{1} \\ \epsilon_{2} \\ \epsilon_{3} \end{bmatrix}=0$

So for the eigenvector do I just choose any values for e2 and e3. For instance would the vector: $\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ work?

Further since this is a repeated roots case don't I need to find another matrix $\eta$ such that:
$(A+\frac{7}{2}I)\eta = \epsilon$

However, trying to solve this I get the equation:
$\left[ \begin{array}{cccc} 1 & 1 & 1 \\ 1 & 1& 1 \\ 1 & 1 & 1 \end{array} \right] \begin{bmatrix} \eta_{1} \\ \eta_{2} \\ \eta_{3} \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}$ which I think has no solutions...

help?

2. For

$
\underbrace{\left[ \begin{array}{cccc} 1 & 1 & 1 \\ 1 & 1& 1 \\ 1 & 1 & 1 \end{array}\right]}_{A} \begin{bmatrix} \epsilon_{1} \\ \epsilon_{2} \\ \epsilon_{3} \end{bmatrix}=0
$

Finding the RREF(A) is simply:

$\begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}$

and so

$\epsilon_1 + \epsilon_2 + \epsilon_3 = 0$

$\Rightarrow \epsilon_1 = -\epsilon_2 - \epsilon_3,\quad \epsilon_2,\epsilon_3 \in \mathbb{R}$.

So the solution set $S = \text{Ker}(A) = \text{span}\lbrace \begin{bmatrix} -1\\1\\0\end{bmatrix}, \begin{bmatrix} -1\\0\\1\end{bmatrix} \rbrace$ and so those are your two general e-vectors for $\lambda_1 = \lambda_2$.

(This is a linear algebra question and really belongs in that forum)

3. Originally Posted by scorpion007
For

$
\underbrace{\left[ \begin{array}{cccc} 1 & 1 & 1 \\ 1 & 1& 1 \\ 1 & 1 & 1 \end{array}\right]}_{A} \begin{bmatrix} \epsilon_{1} \\ \epsilon_{2} \\ \epsilon_{3} \end{bmatrix}=0
$

Finding the RREF(A) is simply:

$\begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}$

and so

$\epsilon_1 + \epsilon_2 + \epsilon_3 = 0$

$\Rightarrow \epsilon_1 = -\epsilon_2 - \epsilon_3,\quad \epsilon_2,\epsilon_3 \in \mathbb{R}$.

So the solution set $S = \text{Ker}(A) = \text{span}\lbrace \begin{bmatrix} -1\\1\\0\end{bmatrix}, \begin{bmatrix} -1\\0\\1\end{bmatrix} \rbrace$ and so those are your two general e-vectors for $\lambda_1 = \lambda_2$.

(This is a linear algebra question and really belongs in that forum)
But pertains to differential equations so I would say it's in the right forum.

4. Originally Posted by scorpion007
For

$
\underbrace{\left[ \begin{array}{cccc} 1 & 1 & 1 \\ 1 & 1& 1 \\ 1 & 1 & 1 \end{array}\right]}_{A} \begin{bmatrix} \epsilon_{1} \\ \epsilon_{2} \\ \epsilon_{3} \end{bmatrix}=0
$

Finding the RREF(A) is simply:

$\begin{bmatrix}1&1&1\\0&0&0\\0&0&0\end{bmatrix}$

and so

$\epsilon_1 + \epsilon_2 + \epsilon_3 = 0$

$\Rightarrow \epsilon_1 = -\epsilon_2 - \epsilon_3,\quad \epsilon_2,\epsilon_3 \in \mathbb{R}$.

So the solution set $S = \text{Ker}(A) = \text{span}\lbrace \begin{bmatrix} -1\\1\\0\end{bmatrix}, \begin{bmatrix} -1\\0\\1\end{bmatrix} \rbrace$ and so those are your two general e-vectors for $\lambda_1 = \lambda_2$.

(This is a linear algebra question and really belongs in that forum)
Thanks for the reply. I am still stuck on the second part of my question, perhaps you could help me there. This really is a differential equations... finding the eigenvectors are just part of the process, not the main process.

5. Hi Tau. I'm not entirely sure about this but since Scorpion found you two linearly-independent eigenvectors for the repeated root, the solution is simply:

$X=c_1\left(\begin{array}{c}1 \\ 1\\ 1\end{array}\right) e^{-1/2 t}+c_2\left(\begin{array}{c}-1 \\ 0\\ 1\end{array}\right) e^{-7/2 t}+c_3\left(\begin{array}{c}-1 \\ 1\\ 0\end{array}\right) e^{-7/2 t}$

And that solution does solve the system although I'm a little rusty with the linear algebra part.

6. Thanks for advice shawsend. I can't explain why, but I know that is not right... that is how you would do it if there weren't repeated roots, but since there are repeated roots you must do something special.

7. Hi Tau. I'm stickin' to my guns. Sides, it's satisfies the system. My understanding is you only need to seek solutions of the form $tf(t)$ if you cannot find linearly independent eigenvectors. But we have two. Tell you what, it's Sunday. If it's due tomorrow and it was mine, this is what I'm turning in.

8. Actually now that I think about it you may be right. I think you just made me understand it with your last post... Im glad I doubted you hah. Im gonna mull it over a little more... but right now Im on your side. Thanks.