Finding solutions of system of differential equations with eigenvectors I was trying to solve this system of differential equations:

$$\frac{dx}{dt}=3x-y-z$$ 
  $$\frac{dy}{dt}=x+y-z$$
  $$\frac{dz}{dt}=x-y+z$$

I found the eigenvalues: $\lambda_1=1,\lambda_2=2$. The last one has multiplicity $2$. I found $\boldsymbol K_1 =\left(
\begin{array}{c}
1\\
1\\
1
\end{array}
\right)$ is an eigenvector, so a solution to the differential equation would be $\boldsymbol X_1=\boldsymbol K_1e^t.$ However when I want to find a solution with an eigenvector $\boldsymbol K_2 = \left(
\begin{array}{c}
k_1\\
k_2\\
k_3
\end{array}
\right)$ associated to $\lambda_2$ I have the equation $k_1-k_2-k_3=0$ three times. From this I know that two eigenvectors are $\left(
\begin{array}{c}
1\\
1\\
0
\end{array}
\right)$ and $\left(
\begin{array}{c}
1\\
0\\
1
\end{array}
\right)$. Now if I choose one of them, let's say the first one, then this will be a second eigenvector $\boldsymbol K_2$, such that $\boldsymbol X_2=\boldsymbol K_2e^{2t}$ solves the system. However if I try to find a third solution (which I already found: $\boldsymbol K_3e^{2t}$ from the third eigenvector; but I want to find it from here) that has the form $\boldsymbol X_3=\boldsymbol K_2te^{2t}+\boldsymbol P e^{2t}$, then I have to solve $(\boldsymbol A-2\boldsymbol I)\boldsymbol P = \boldsymbol K_2$. But this system doesn't have a solution:
$$(\boldsymbol A-2\boldsymbol I)\boldsymbol P = \boldsymbol K_2 \Longrightarrow \left(\begin{array}{ccc|c}1 & -1 & -1 & 1\\1 & -1 & -1 & 1\\1 & -1 & -1 & 0\\ \end{array}\right)???$$
Is it possible to obtain a third solution from the second one? I'm trying to construct this solution which should end up with an expression that involves the third eigenvector.
I appreciate any thoughts on this. Thanks.
 A: We have to look at the matrix $${\bf A} = \begin{bmatrix} 3 & -1 & -1\\ 1 & 1 & -1 \\ 1 & -1 & 1\end{bmatrix},$$ and the system ${\bf X}' = {\bf A}{\bf X}$.
I didn't checked, but I'll trust you that the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 2$, they're too nice to be wrong. Let's go find eigenvectors:


*

*$\lambda_1 = 1$: Solve $({\bf A}-{\bf I}){\bf K}_1 = {\bf 0}$, that
is: $$\begin{bmatrix} 2 & -1 & -1\\ 1 & 0 & -1 \\ 1 & -1 &
   0\end{bmatrix}\begin{bmatrix} a \\ b \\ c\end{bmatrix} =
   \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}$$ The first line is the sum
of the other, so it is useless. From the second equation we get $a=c$
and from the third, $a = b$. So far, so good: $${\bf K}_1 =
   \begin{bmatrix} a \\ b \\ c\end{bmatrix} = \begin{bmatrix} a \\ a \\
   a\end{bmatrix} = a\begin{bmatrix} 1 \\ 1 \\ 1\end{bmatrix},$$ so we
pick the vector $${\bf k}_1 = \begin{bmatrix} 1 \\ 1 \\
   1\end{bmatrix},$$ and get one row for the fundamental matrix of the
system: $$e^{\lambda_1t}{\bf k}_1 = e^{1t}\begin{bmatrix}1 \\ 1 \\ 1 \end{bmatrix} =
   \begin{bmatrix}e^t \\ e^t \\ e^t\end{bmatrix}.$$

*$\lambda_2 = 2$: Solve $({\bf A}-2{\bf I}){\bf K}_2 = {\bf 0}$, that 
is: $$\begin{bmatrix} 1 & -1 & -1\\ 1 & -1 & -1 \\ 1 & -1 &   
   -1\end{bmatrix}\begin{bmatrix} a \\ b \\ c\end{bmatrix} =    \begin{bmatrix} 0 \\ 0 \\ 0\end{bmatrix}.$$ Better impossible, we
have the relation $a-b-c = 0$, that is, $a = b+c$. Then: $${\bf K}_2
   =    \begin{bmatrix} a \\ b \\ c\end{bmatrix} = \begin{bmatrix} b+c \\ b \\    c\end{bmatrix} = b\begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix}
   + c\begin{bmatrix} 1 \\ 0 \\ 1\end{bmatrix}.$$ Pick the vectors: $${\bf k}_2^{(1)} = \begin{bmatrix} 1 \\ 1 \\ 0\end{bmatrix} \quad
   \mbox{and} \quad{\bf k}_2^{(2)} = \begin{bmatrix} 1 \\ 0 \\
   1\end{bmatrix}.$$ This gives us the more independent rows of the
fundamental matrix: $$e^{\lambda_2t}{\bf k}_2^{(1)} =
   e^{2t}\begin{bmatrix}1 \\ 1 \\ 0 \end{bmatrix} =   
   \begin{bmatrix}e^{2t} \\ e^{2t} \\ 0\end{bmatrix} \quad \mbox{and}
   \quad e^{\lambda_2t}{\bf k}_2^{(2)} = e^{2t}\begin{bmatrix}1 \\ 0 \\
   1 \end{bmatrix} =    \begin{bmatrix}e^{2t} \\ 0 \\
   e^{2t}\end{bmatrix}.$$


So, we get as fundamental matrix for the system, the matrix: 
$${\cal A} = \begin{bmatrix} e^t & e^{2t} & e^{2t}\\ e^t & e^{2t} & 0 \\ e^t & 0 & e^{2t}\end{bmatrix},$$ and hence, given $c_1,c_2,c_3 \in \Bbb R$, we get the general solution to the system: $${\bf X} = {\cal A}{\bf C} \implies \begin{bmatrix}x(t) \\ y(t) \\ z(t)\end{bmatrix} = \begin{bmatrix} e^t & e^{2t} & e^{2t}\\ e^t & e^{2t} & 0 \\ e^t & 0 & e^{2t}\end{bmatrix}\begin{bmatrix}c_1 \\ c_2 \\ c_3\end{bmatrix} = \begin{bmatrix}c_1e^t+(c_2+c_3)e^{2t} \\ c_1e^t+c_2e^{2t} \\ c_1e^t+c_3e^{2t}\end{bmatrix}.$$
