How to solve this differential equation system? The following system is given:
$$
\dot{x} = y + z \\
\dot{y} = x + z \\
\dot{z} = x + y
$$
The first thing I did was to find out the eigenvalues. I found out, that -1 is a doubled and 2 a single eigenvalue, so
$$
\lambda_{1,2} = -1,\ \ \lambda_3 = 2
$$
In the excercises ago, the ideas were to determine $ y=e^{\lambda x} \underline{u} $. so I tried to do the following:
$$
\begin{pmatrix}
0-\lambda & 1 & 1 \\
1 & 0-\lambda & 1 \\
1 & 1 & 0-\lambda
\end{pmatrix}
$$
Is this step right? I tried to find a scheme as in the excercises ago and in the line $ \dot{x} = y +z$ I don't have an x but one y and one z.
When inserting  $ \lambda_1 = -1 $ I have
$$
A-\lambda E = \underline{0} \rightarrow
\begin{pmatrix}
1 & 1 & 1 \\
1 & 1 & 1 \\
1 & 1 & 1
\end{pmatrix} = \underline{0}
$$
which means that $ x_i + y_i + z_i = 0\ for\ i ={1,2,3} $. Here is the point on which I don't know how to go on. One solution is the trivial one, so $x=y=z=0$. Can I use this solution?
I think that I have to use something like $$
y = C_1 *
\begin{pmatrix}
u_1*e^{\lambda_1 x} \\
u_2*e^{\lambda_1 x}\\
u_3*e^{\lambda_1 x}
\end{pmatrix} + C_2 \begin{pmatrix} ... \end{pmatrix} + C_3 \begin{pmatrix} ... \end{pmatrix}$$ 
in the case $ \lambda_1 = 1 $, but how to I get my u here exactly?
 A: The equations can be written as $\dot{p} = Ap$, with $p \in \mathbb{R}^3$ and
$A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{bmatrix}$.
Note that $A= v v^T -I$, where $v=(1,1,1)^T$, so it has one eigenvalue at 2 corresponding to the eigenvector $v$, and two at -1 corresponding to the eigenspace $\{v\}^\bot$. (Note that $A$ is symmetric hence has an orthonormal basis of eigenvectors.)
Hence if you write $p = \alpha v + w \in \mathbb{R}^3$, where $w \bot v$, you will have
$Ap = \alpha 2 v -w$, and, in general,
$A^k p = \alpha 2^k v + (-1)^k w$, from which we see that
$e^{At} p = \alpha e^{2t} v+e^{-t} w$.
The projection of a point $p \in \mathbb{R}^3$ onto $v$ is straightforward to compute.
The point here is that you don't need to explicitly find eigenvectors for the eigenspace $\{v\}^\bot$.
A: Your differential equation is of the form $\vec u'=A\vec u$, where $\vec u=\begin{bmatrix} x\\ y\\ z\end{bmatrix}$ and $A=\begin{bmatrix} 0 & 1 & 1\\ 1& 0 & 1\\ 1 & 1 & 0\end{bmatrix}$.
If there is a choice of eigenvectors of $A$ that form a basis of $\mathbb R^{3\color{grey}{\times 1}}$, then, assuming the eigenvectors are $v_1, v_2, v_3$ with corresponding eigenvalues $\lambda _1, \lambda _2, \lambda _3$, then a basis of solutions of $\vec u'=A\vec u$ is $\left(t\mapsto e^{\lambda _1 t}v_1, t\mapsto e^{\lambda _2 t}v_2, t\mapsto e^{\lambda _3 t}v_3\right)$.
I know beforehand that the hypothesis of the if above is verified, so this is just a linear algebra problem now.
You found $\lambda _1=-1, \lambda _2=-1$ and $\lambda _3=2$.
Now you need to find a basis of the eigenspaces of each eigenvalue.
$\bbox[5px,border:2px solid #0000FF]{\text{Eigenvalue }-1}$
You wish to find a basis of the vector space $\left\{\begin{bmatrix} a\\ b\\ c\end{bmatrix}\in \mathbb R^{n\times 1}\colon (A+I)\begin{bmatrix} a\\ b\\ c\end{bmatrix}=0_{n\times 1}\right\}$.
The condition $(A+I)\begin{bmatrix} a\\ b\\ c\end{bmatrix}=0_{n\times 1}$ is equivalent to $a+b+c=0$ and to $a=-b-c$, so the set above is equal to $\left\{\begin{bmatrix} -b-c\\ b\\ c\end{bmatrix}\colon b,c\in \mathbb R\right\}$ which is equal to $\left\{b\begin{bmatrix} -1\\ 1\\ 0\end{bmatrix}+c\begin{bmatrix} -1\\ 0\\ 1\end{bmatrix}\colon b,c\in \mathbb R\right\}$ and we can now immediately identify the basis $\left\langle \begin{bmatrix} -1\\ 1\\ 0\end{bmatrix},\begin{bmatrix} -1\\ 0\\ 1\end{bmatrix} \right\rangle$. So you can take $v_1=\begin{bmatrix} -1\\ 1\\ 0\end{bmatrix}$ and $v_2=\begin{bmatrix} -1\\ 0\\ 1\end{bmatrix}$.
$\bbox[5px,border:2px solid #0000FF]{\text{Eigenvalue }2}$
Do it yourself.
