Why is this the eigenvector? For the  eigenvector how are they getting \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} when you have \begin{bmatrix} 0 & -1 & -1 \\ 0 & -1 & -3 \\ 0 & 0 & -2 \end{bmatrix} multiplied by\begin{bmatrix} v_1 \\ v_2 \\ v_3 \end{bmatrix} which turns to three equations...:
$(0) * v_1 -v_2 -v_3 = 0$ 
$(0) * v_1 - v_2 -3v_3 = 0$
$(0) * v_1 + (0) * v_2 -2v_3 = 0$ 
so $v_1$ turns to $0$ when you multiply by $0$ so how are they getting $1$ in the $v_1$ position.. What am i missing here?
 A: We have
$$\begin{bmatrix} 0 & -1 & -1 \\ 0 & -1 & -3 \\ 0 & 0 & -2 \end{bmatrix}
  \cdot \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = 0\cdot \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$$
So the column vector is indeed an eigenvector of the matrix, with eigenvalue zero.
I'm not sure what you mean by "the smallest eigenvector" since there are others, with eigenvalues $-1$ and $-2$.
As for your analysis, @Hayden is correct in saying that you can choose $v_1$ arbitrarily. If you continue solving those equations you will get $v_2=0$ and $v_3=0$. Therefore, all the eigenvectors with eigenvalue zero are
$$\begin{bmatrix} v_1 \\ 0 \\ 0 \end{bmatrix}$$
where $v_1\ne 0$. If you set $v_1=0$ the equalities hold, but by definition an eigenvector must be non-zero.
A: For the eigenvalue $\lambda=0$, that means that for the corresponding eigenvector, $x$, you have $Ax = \lambda x$ which in this case corresponds to $Ax=0$.
To solve for $x$, you can set up and row reduce the augmented matrix:
$\left[\begin{array}{ccc|c}
0 & -1 & -1 & 0\\
0 & -1 & -3 & 0\\
0 & 0 & -2 & 0\end{array}\right]$
By row reducing, we arrive at:
$\left[\begin{array}{ccc|c}
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0\end{array}\right]$
Interpreting this result, this means that for $Ax=0$ where $x=\begin{bmatrix} v_1\\v_2\\v_3\end{bmatrix}$ you must have:
$\begin{cases} v_2=0\\ v_3=0\end{cases}$, while $v_1$ can be anything.
Thus, the eigenvector is anything of the form $\begin{bmatrix}v_1\\0\\0\end{bmatrix}$.  One such representative is $\begin{bmatrix}1\\0\\0\end{bmatrix}$.

In general, for $Ax=\lambda x$, you will manipulate the expression to get it into the form $(A-\lambda I)x = 0$ and row reduce to find the eigenvector or eigenvectors associated with that eigenvalue.
