Assessing the geometric multiplicity of an eigenvalue. Suppose $\lambda_1 = 1$ is an eigenvalue of the hypothetical matrix $\mathbf A$
$$
\mathbf A =
        \begin{bmatrix}
        a_{11} & a_{12} & a_{13} \\
        a_{11} -1 & a_{12} +1 & a_{13} \\
        a_{11} -1 & a_{12} & a_{13} +1 \\
        \end{bmatrix}
$$
Let $\mathbf I_3$ represent the identity matrix for $\mathbb R^3$, and let $\mathbf 0_3$ represent the zero vector of $\mathbb R^3$.  To find the eigenspace corresponding to $\lambda _1$ it is necessary solve the following linear system of homogeneous equations.
$$
\left(
        \begin{bmatrix}
        a_{11} & a_{12} & a_{13} \\
        a_{11} -1 & a_{12} +1 & a_{13} \\
        a_{11} -1 & a_{12} & a_{13} +1 \\
        \end{bmatrix}
- \lambda_1 \mathbf I _3
\right)
        \begin{bmatrix}
        x_1 \\
        x_2 \\
        x_3 \\
        \end{bmatrix}
= \mathbf 0_3
$$
Elementary row operations are used to obtain the following augmented matrix.
$$ \left[
    \begin{array}{ccc|c}
      a_{11} & a_{12} & a_{13} & 0\\
      0&0&0&0 \\
      0&0&0&0
    \end{array}
\right] \\ 
$$
Therefore
$$
x_3 = s \quad ; \quad x_2 = t \quad ; \quad a_{11} x_1 = -a_{12}t -a_{13}s 
$$
The solution of this system of equations is 
$$
s
        \begin{bmatrix}
        {-a_{13} \over a_{11}} \\
        0 \\
        1 \\
        \end{bmatrix}
+t
        \begin{bmatrix}
        {-a_{12} \over a_{11}} \\
        1 \\
        0 \\
        \end{bmatrix}
$$
Conclusion: The geometric multiplicity of $\lambda_1 $ is $2$. The eigenspace of $\lambda_1 $ is a two-dimensional subspace of $\mathbb R ^2$ with basis
$$
\left\{
        \begin{bmatrix}
        {-a_{13} \over a_{11}} \\
        0 \\
        1 \\
        \end{bmatrix}
,
        \begin{bmatrix}
        {-a_{12} \over a_{11}} \\
        1 \\
        0 \\
        \end{bmatrix}\right\}
$$
Here is the question:  Is this analysis correct?

Not exactly. By row reductions, the correct row echelon form should be
$$
\left[\begin{array}{ccc|c}
a_{11}-1 & a_{12} & a_{13} & 0\\
0&0&0&0 \\
0&0&0&0
\end{array}\right]
$$
instead. Since you don't know the exact values of the $a_{ij}$s, you cannot use as denominators in divisions. All you can say is that the eigenspace of $A$ corresponding to the eigenvalue $\lambda=1$ is the set of all vectors orthogonal to $(a_{11}-1,\ a_{12},\ a_{13})^T$.
Since $(a_{11}-1,\ a_{12},\ a_{13})^T$ may be the zero vector (when $A=I$), the geometric multiplicity of $\lambda=1$ can also be $3$ rather than $2$.
