General solution of ODE of a constant $3\times 3$ matrix Determine the general solution of the system $y'=Ay$ , where $A$ is a constant matrix, defined by
$$A = \begin{pmatrix}-5&-8&4\\2&3&-2\\6&14&-5\end{pmatrix}$$
After attempting to find the eigenvalues of the system, I end up with eigenvalues $\lambda=-1,-3,-3$, where $-3$ has a multiplicity of $2$. Then, finding the corresponding vector for $\lambda=-1$, 
$$
u=
\begin{pmatrix}3\\-1\\1\end{pmatrix},
$$
for $\lambda=-3$, 
$$v=
\begin{pmatrix}-2\\1\\1\end{pmatrix},
$$
and for the second $ \lambda =-3$ the vector 
$$w=\begin{pmatrix}-1-2t\\\frac{1}{2}+t\\t\end{pmatrix}.$$
I understand how to get the first two eigevectors $u$ and $v$, but how did they get the third eigenvector to be in terms of $t$? Is it because it has a multiplicity of $2$, so there is a second solution that can be described in terms of a variable? If so, what is the process to finding that third eigenvector? And does this process generalize for an eigenvalue that would have perhaps a multiplicity of $3$?
The general solution of the system is 
$$y(t) = C_{1}e^{-t}u + C_{2}e^{-3t}v + C_{3}e^{-3t}w$$
where $u$, $v$, and $w$ are defined above.
 A: To find the generalized eigenvector $w$ corresponding to the eigenvalue $\lambda=-3$, you have to solve the linear system of equations in $(A+3I)w=v$. 
This amounts to row reducing the augmented matrix: 
$$
\left( 
\begin{array}{rrr|r} 
-2 & -8 & 4 & -2 \\
2 & 6 & -2 & 1 \\ 
6 & 14 & -2 & 1 \\ 
\end{array}
\right) 
\stackrel{RREF}{\rightsquigarrow}  
\left( 
\begin{array}{rrr|r} 
1 & 0 & 2 & -1 \\
0 & 1 & -1 & \frac{1}{2} \\ 
0 & 0 & 0 & 0 \\ 
\end{array}
\right).
$$
Here, we put the augmented matrix into reduced row echelon form (RREF). 
So $w=(x_1,x_2,x_3)$ must satisfy 
$$
x_1 + 2 x_3 =-1 
\qquad 
\mbox{ and } 
\qquad  
x_2 -x_3 =\frac{1}{2}, 
$$
or 
$$
x_1 = -2 x_3 -1, 
\qquad  
x_2 = x_3 + \frac{1}{2}, 
\qquad 
\mbox{ and }
\qquad 
x_3 = x_3. 
$$
Since $x_3$ is a free variable, let's use $t$ as the free variable instead. 
So the generalized eigenvector $w$ must be of the form 
$$
\boxed{
w= 
t \begin{pmatrix}
-2 \\
1 \\ 
1 \\ 
\end{pmatrix}
+  
\begin{pmatrix}
-1 \\ 
\frac{1}{2} \\ 
0 \\ 
\end{pmatrix},
\qquad  
\mbox{ where }
t\in \mathbb{R}. 
}
$$
Note that if $w$ is a generalized eigenvector for the eigenvalue $\lambda=-3$, then $w$ must satisfy 
$$
(A+3I)w=v
$$ 
because $(A+3I)v=0$. So multiply $A+3I$ on the left to both sides of $(A+3I)w=v$
to obtain $(A+3I)^2 w = 0$, which is precisely what it means for a vector to be a generalized eigenvector. 
A: We have a defective matrix and have to resort to finding a generalized eigenvector.
Following Example 1, we can try
$$[A- \lambda I]v_2 = [A + 3I]v_2 = v_1$$
This gives us the RREF of 
$$\DeclareMathOperator{RREF}{RREF}\RREF
\left[\begin{array}{rrr|r}
-2 & -8 & 4 & -2 \\
2 & 6 & -2 & 1 \\
6 & 14 & -2 & 1
\end{array}\right]=
\left[\begin{array}{rrr|r}
1 & 0 & 2 & -1 \\
0 & 1 & -1 & \dfrac{1}{2}\\
0 & 0 & 0 & 0 
\end{array}\right] \implies v_2 = \begin{bmatrix}-1-2t\\\dfrac{1}{2}+t\\t\end{bmatrix}$$
You will learn about the Jordan Normal Form and Eigenvector Chaining.
