Solving a system of linear differential equations with repeated eigen values I have this problem where to solve the system,
$$x'=4x+y-z$$
$$y'=2x+5y-2z$$
$$z'=x+y+2z$$
using a linear algebraic solution. I have found the eigen values of the 
$$\begin{bmatrix}
        4 & 1 & -1 \\
        2 & 5 & -2 \\
        1 & 1 & 2 \\
        \end{bmatrix}$$
as 3,3 and 5. When evaluating the corresponding eigen vectors for 3, the following occurs.
$$(A-3I)x=0$$
$$\begin{bmatrix}
        1 & 1 & -1 \\
        2 & 2 & -2 \\
        1 & 1 & -1 \\
        \end{bmatrix}
\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} = 0$$
We can say ok and $x_3=x_1+x_2$ and then a set of eigen vectors which are not multiples of each other are formed. As the next step, I have to find $\rho$ such that $(A-3I)\rho = \eta_{\lambda =3}$. There I'm getting nowhere because of the ambiguity of $\eta_{\lambda=3}$. I am new to this eigen things. Am I doing something terribly wrong or what?
 A: You have correctly calculated your eigenvalues. To avoid ambiguous notation, denote your eigenvalues as follows:
$\lambda_1=5$
$\lambda_2=3$
$\lambda_3=3$.
Using your eigenvalues, use the fact that:
$A\mathbf{x}=\lambda_1 \mathbf{x}$, $A\mathbf{x}=\lambda_2 \mathbf{x}$ and $A\mathbf{x}=\lambda_3 \mathbf{x}$, where $\mathbf{x}=\begin{pmatrix}x\\y\\z \end{pmatrix}$ and $A=\begin{pmatrix} 4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2 \\ \end{pmatrix}$ to evaluate your eigenvectors.
Hence for $\lambda_1$: $\begin{pmatrix} 4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2 \\ \end{pmatrix}\begin{pmatrix}x_1\\y_1\\z_1 \end{pmatrix}=5 \begin{pmatrix}x_1\\y_1\\z_1 \end{pmatrix}$.
For $\lambda_2$: $\begin{pmatrix} 4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2 \\ \end{pmatrix}\begin{pmatrix}x_2\\y_2\\z_2 \end{pmatrix}=3 \begin{pmatrix}x_2\\y_2\\z_2 \end{pmatrix}$.
For $\lambda_3$: $\begin{pmatrix} 4 & 1 & -1 \\ 2 & 5 & -2 \\ 1 & 1 & 2 \\ \end{pmatrix}\begin{pmatrix}x_3\\y_3\\z_3 \end{pmatrix}=3 \begin{pmatrix}x_3\\y_3\\z_3 \end{pmatrix}$.
From this, we deduce for $\lambda_1$ the following system:
$4x_1+y_1-z_1=5x_1$
$2x_1+5y_1-2z_1=5y_1$
$x_1+y_1+2z_1=5z_1$
Note that these simultaneous equations are redundant – they are all essentially the same. Hence, we deduce from one of these equations that $y_1=x_1+z_1$.
We repeat the same process for $\lambda_2$ (no need for $\lambda_3$ since both have same eigenvalues) and obtain $z_2=x_2+y_2$, as you mentioned earlier. The next step to find the eigenvectors is to find the smallest possible nonzero integer values for $x_n, y_n, z_n$ for each of your corresponding eigenvalues such that it satisfies your equation. These are your eigenvectors.
For $\lambda_1$, we obtain $v_1=\begin{pmatrix}1\\2\\1 \end{pmatrix}$. For $\lambda_2$ we obtain two possible solutions $v_2=\begin{pmatrix}1\\0\\1 \end{pmatrix}$ and $v_3=\begin{pmatrix}-1\\1\\0 \end{pmatrix}$. The eigenvectors will be the same for $\lambda_3$.
Therefore, from the eigenvalues and eigenvectors, we may substitute and obtain the general solutions:
$\begin{pmatrix}x(t)\\y(t)\\z(t) \end{pmatrix}=c_1\begin{pmatrix}1\\2\\1 \end{pmatrix}e^{5t}+c_2\begin{pmatrix}1\\0\\1 \end{pmatrix}e^{3t}+c_3\begin{pmatrix}-1\\1\\0 \end{pmatrix}e^{3t}$.
Please do not hesitate to ask if you have any doubts or questions.
