<s>[INST] Basis for a eigenspace (multiple choice problem) The following (multiple choice) problem is from a test review.

For the given matrix $A$, find a basis for the corresponding eigenspace for the given eigenvalue.
$$A = \begin{bmatrix}1 & 6 & 6 \\ 6 & 1 & -6 \\ -6 & 6 & 13\end{bmatrix};\quad \lambda = 7.$$

The four given options are

A) $\left\{ \begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ -1 \end{bmatrix} \right\}$
B) $\left\{ \begin{bmatrix}0 \\ 1 \\ -1 \end{bmatrix} \right\}$
C) $\left\{ \begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix} \right\}$
D) $\left\{ \begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix} \right\}$

The correct answer is A.

My answer is not among the answer choices.  What am I doing wrong?
First, reduce $(A-\lambda I)$:
$$\begin{align}
A - \lambda I &=
\begin{bmatrix}1 & 6 & 6 \\ 6 & 1 & -6 \\ -6 & 6 & 13\end{bmatrix} - \begin{bmatrix}7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7\end{bmatrix} \\
&= \begin{bmatrix}-6 & 6 & 6 \\ 6 & -6 & -6 \\ -6 & 6 & 6\end{bmatrix} \\
&\sim \begin{bmatrix}-1 & 1 & 1 \\ 1 & -1 & -1 \\ -1 & 1 & 1\end{bmatrix} \\
&\sim \begin{bmatrix}1 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}
\end{align}$$
Parametric form solving for $(A-\lambda I) = \vec 0$:
$$\begin{align}
x_1 &= x_2 + x_3 \\
x_2 &= x_2 \\
x_3 &= x_3
\end{align}$$
has solutions $x_2(1,1,0) + x_3(1,0,1)$ for all $x_2 , x_3$.
So a basis for the null-space of $(A-\lambda I)$ consists of the vectors $\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$ and $\begin{bmatrix}1 \\0 \\ 1\end{bmatrix}$.
So, am I wrong or is the solution wrong?
 [/INST] As @GTonyJacobs points out in the comments, you are correct but the key is also correct. Vector spaces can have many different bases!
Note that you have proven that $A-7I$ has rank one so the rank-nullity theorem implies that the nullspace of $A-7I$ has dimension two. This allows us to immediately eliminate choices (B) and (C). 
Now, note that
\begin{align*}
A\begin{bmatrix}1\\0\\1\end{bmatrix}
&= \begin{bmatrix}7\\0\\7\end{bmatrix}
= 7\begin{bmatrix}1\\0\\1\end{bmatrix}
&
A\begin{bmatrix}0\\1\\-1\end{bmatrix}
&=\begin{bmatrix}0\\7\\-7\end{bmatrix}
=7\begin{bmatrix}0\\1\\-1\end{bmatrix}
\end{align*}
This proves that choice (A) is correct. 
Also, note that choice (D) is incorrect since
$$
A\begin{bmatrix}1\\0\\-1\end{bmatrix}
=\begin{bmatrix}-5\\12\\-19\end{bmatrix}
\neq7\begin{bmatrix}1\\0\\-1\end{bmatrix}
$$
</s>