Analytical solution to system of linear ODE's 
Given $x'=Ax$ with initial data $x(0)=\xi$, eigenvalues $\lambda_j=0,0,3$ and
  $$\begin{bmatrix}2&1&0\\0&2&4\\1&0&-1\end{bmatrix},$$ find the general
  solution to the system and a canonical Jordan form of the matrix $A$.
  Find all initial conditions such that solutions to the IVP will be
  bounded.

To do this, the book says I need to use the following:


Let the matrix $A$ have $s$ distinct eigenvalues
  $\lambda_1,\dots,\lambda_s$ with corresponding generalised eigenspaces
  $E(\lambda_j)$. Represent the initial data $x(0)=\xi$ for the solution
  $x(t)$ as a sum of its components from different generalised
  eigenspaces: $$\xi=\sum_{j=1}^{s}x^{0,j},\quad x^{0,j}\in
> E(\lambda_j).$$ Here $x^{0,j}\in E(\lambda_j)$ - are components of
  $\xi$ in the generalized eigenspaces
  $E(\lambda_j)=\ker(A-\lambda_j)^{m_j}$ of the matrix $A$. These
  subspaces intersect only in the origin and are invariant with respect
  to $A$ and $\exp(At)$. It implies that for the solution $x_z(t)$ with
  initial data $z\in E(\lambda_j)$, we have $x_z(t)\in E(\lambda_j)$ for
  all $t\in\Bbb R$.
Let $m_j$ be the algebraic multiplicity of the eigenvalue $\lambda_j$.
  We apply the formula $(11)$ to this representation and derive the an
  expression for solutions for arbitrary initial data as a finite sum
  (intead of series):
  $$\begin{equation}x(t)=e^{At}x_0=\sum_{j=1}^{s}\left(e^{\lambda_jt}\left[\sum_{k=0}^{m_j-1}(A-\lambda_jI)^{k}\frac{t^k}{k!}\right]x^{0,j}\right)\tag{13}\end{equation}.$$
  Series expressing $\exp(At)x^{0,j}$ terminates on each of the
  generalised eigenspaces $E(\lambda_j)$.

(Image that replaced text).

Let me try: An eigenvector corresponding to $\lambda = 0$ can be taken as $v_1=\begin{bmatrix}1&-2&1\end{bmatrix}^T$. For $\lambda=3$ we can take eigenvector $v_2=\begin{bmatrix}4&4&1\end{bmatrix}^T$. Since the eigenvalue of zero is of algebraic multiplicity $2$ we need to find an extra generalized eigenvector $v_1^{(1)}$ such that we can express $\xi$ as a linear combination of these. We solve the equation $(A-\lambda I)v_1^{(1)}=v_1$. Solving this we get that $v_1^{(1)}=\begin{bmatrix}1&-1&0\end{bmatrix}^T$. Since we are in $\mathbb{R}^3$ we know that there can not be more than three generalized eigenvectors. In my case, $s=2$ since the multiplicity of the zero eigenvalue is $2$. Thus expression $(13)$ can be written as
\begin{align}
x(t)&=\sum_{j=1}^2\left(e^{\lambda_jt}x^{0,j}\left[\sum_{k=0}^{1}(A-\lambda_jI)\frac{t^k}{k!}\right]\right)\\
&= \sum_{j=1}^2\left(e^{\lambda_jt}x^{0,j}\left[1+(A-\lambda_jI)\right]\right)
\end{align}
The problem I have here is that I don't understand how to express the $x^{0,j}$. 
 A: First, let’s deal with the errors in the way you applied (13) to this problem. You need to be more careful about reordering terms: matrix multiplication is in general not commutative. You’ve switched the order of $x^{0,j}$, which is a $3\times1$ vector, and a $3\times3$ matrix, so the product in your expression doesn’t make sense.  
Also, it looks like you’ve somehow combined the contributions of the two eigenvalues into one term. The inner sum in particular is different for different multiplicities. Taking first the eigenvalue $3$ with algebraic multiplicity $1$, the inner sum for this eigenvalue is simply $I$. Its contribution to the solution is therefore just $e^{3t}x^{0,3}$. (I’ve changed the notation a bit—instead of using an index $j$, I’m labeling the components of $\xi$ with the eigenvalues.) The other eigenvalue $0$ has algebraic multiplicity $2$, so the inner sum is $I+tA$, and its contribution to the solution is $(I+tA)x^{0,0}$, therefore the solution to this initial value problem is $$e^{3t}x^{0,3}+(I+tA)x^{0,0}.$$ 
To find the components $x^{0,j}$ of $\xi$, you basically need to apply the same change of basis to $\xi$ that you would to $A$ to convert it into its Jordan normal form. So, let $$P=\begin{bmatrix}1&1&4\\-2&-1&4\\1&0&1\end{bmatrix}$$ so that $$P^{-1}AP=\begin{bmatrix}0&1&0\\0&0&0\\0&0&3\end{bmatrix}.$$ The elements of $P^{-1}\xi$ are the coordinates of $\xi$ relative to this basis, and the $x^{0,j}$ are constructed via corresponding linear combinations of the three generalized eigenvectors. Now, since the product of a matrix and vector can be interpreted as a linear combination of the columns of the matrix with coefficients given by the elements of the vector, we can write $$x^{0,0}=P\begin{bmatrix}1&0&0\\0&1&0\\0&0&0\end{bmatrix}P^{-1}\xi = P_0\xi\\ x^{0,3}=P\begin{bmatrix}0&0&0\\0&0&0\\0&0&1\end{bmatrix}P^{-1}\xi = P_3\xi.$$ The matrices $P_0$ and $P_3$ are projectors onto the respective generalized eigenspaces with the property that $P_0P_3=P_3P_0=0$.  

Addendum: If $\xi = C_1v_1+C_2v_1^{(1)}+C_3v_2$ then 
formula (13) gives $$x(t) = (I+tA)\left(C_1v_1+C_2v_1^{(1)}\right)+C_3e^{3t}v_2$$ as the solution to this IVP, but we can simplify this. Recall that $v_1$ is an eigenvector with eigenvalue $0$, so $Av_1=0$. Expanding the first term of $x(t)$, we therefore have $$\begin{align} x(T) &= C_1(I+tA)v_1+C_2(I+tA)v_1^{(1)}+C_3e^{3t}v_2 \\ &= C_1v_1+C_2(I+tA)v_1^{(1)}+C_3e^{3t}v_2. \end{align}$$
A: Since the matrix is not diagonalizable, one needs to solve this equation as $$\frac{d X}{dt}=M X \Rightarrow X =\exp[M t] C.$$ Here $C=X(0)=\xi.$ The exponential of matrix can be found by Mathematica using the comand $MarixExp[M t]$.
We get solution as
$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}=
9^{-1}\begin{pmatrix} 
 (5 + 4 e^{3 t} + 6 t) & (-4 + 4 e^{3 t} - 3 t) & 
  4(-1 + e^{3 t} - 3 t) \\ 4 (-1 + e^{3 t} - 3 t) & 
   (5 + 4 e^{3 t} + 6 t) &
  4(-1 + e^{3 t} + 6 t) \\  (-1 + e^{3 t} + 6 t) & 
   (-1 + e^{3 t} - 3 t) &  (8 + e^{3 t} - 12 t)
\end{pmatrix}
\begin{pmatrix} \xi_1 \\  \xi_2 \\ \xi_3 \end{pmatrix}$$ 
If $\xi_1=\xi_2=\xi_3=\xi$, then
$$\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}==\frac{\xi}{3} \begin{pmatrix}
 (-1 + 4 e^{3 t} - 3 t) \\  (-1 + 4 e^{3 t} + 6 t) \\ 
   (2 + e^{3 t} - 3 t) \end{pmatrix}.$$
