Finding eigenvalues and eigenvectors and then determining their geometric and algebraic multiplcities I have the following matrix:
$A = \begin{bmatrix} 
    1 && 7 && -2 \\ 
    0 && 3 && -1 \\
    0 && 0 && 2 \end{bmatrix}$
and I am trying to find the eigenvalues and eigenvectors followed by their respective geometric multiplicity and algebraic multiplicity. 
What I have so far:
$\det(A - \lambda I) = \det\begin{bmatrix} 
    1-\lambda && 7 && -2 \\ 
    0 && 3-\lambda && -1 \\
    0 && 0 && 2-\lambda \end{bmatrix}$
I see that it is an upper triangular matrix so determinant is just the diagonal. Which gives me 
$(1-\lambda)(3-\lambda)(2-\lambda)$ which gives me $\lambda = 1,3,2$. I also notice that all three have the algebraic multiplicity of 1 (their exponents were 1).
Following that I move on to the geometric multiplicity:
$ A - 3I = \begin{bmatrix} 
    -2 && 7 && -2 \\ 
    0 && 0 && -1 \\
    0 && 0 && -1 \end{bmatrix}$ which has RRE of $\begin{bmatrix} 
    1 && -\frac{7}{2} && 0 \\ 
    0 && 0 && 1 \\
    0 && 0 && 0 \end{bmatrix}$ which yields 
$\begin{bmatrix} 
    x \\ 
    y \\
    z \end{bmatrix} = \begin{bmatrix} 
    \frac{7}{2}\\ 
    1 \\
    0\end{bmatrix} s$ which has a geometric multiplicity of 1.
$ A - 2I = \begin{bmatrix} 
    -1 && 7 && -2 \\ 
    0 && 1 && -1 \\
    0 && 0 && 0 \end{bmatrix}$ which has RRE of $\begin{bmatrix} 
    1 && 0 && -5 \\ 
    0 && 1 && -1 \\
    0 && 0 && 0 \end{bmatrix}$ which yields 
$\begin{bmatrix} 
    x \\ 
    y \\
    z \end{bmatrix} = \begin{bmatrix} 
    5\\ 
    1 \\
    1\end{bmatrix} s$ which has a geometric multiplicity of 1.
Finally, 
$ A - I = \begin{bmatrix} 
    0 && 7 && -2 \\ 
    0 && 2 && -1 \\
    0 && 0 && 1 \end{bmatrix}$ which has RRE of $\begin{bmatrix} 
    0 && 1 && 0 \\ 
    0 && 0 && 1 \\
    0 && 0 && 0 \end{bmatrix}$
This is where I am stuck, I'm not sure what is the resulting  $\begin{bmatrix} 
    x \\ 
    y \\
    z \end{bmatrix}$ that column of 0's is confusing me.
Any help would be appreciated.
Edit: Would I just say that column of zeros is a free variable? Thus giving me $\begin{bmatrix} 
    1 \\ 
    0 \\
    0 \end{bmatrix}$ ? or is that wrong?

$$\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}\iff \begin{bmatrix}y\\z\\0\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix}$$
meaning that $y=z=0$, while leaving $x$ to be whatever you want, results in a kernel vector.
