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**.

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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?