what would be the eigen vector for this value? So I have a $3\times 3$ matrix
$$A=\begin{pmatrix}
2&1&1\\
1&2&1\\
1&1&2
\end{pmatrix}.$$
My instructions are to find the eigenvalues and eigenvectors of the matrix. For each eigenvalue, find as many linearly independent eigenvectors as possible that correspond to it. I was able to find three eigenvalues, which are $1$, $1$, and $4$: In order to find the eigenvectors I am supposed to plug in each of those eigenvalues into the characteristic equation of the coefficient matrix. When I plug  eigenvalue $1$ into the coefficient matrix, I end up with the matrix
$$\begin{pmatrix}
1&1&1\\
1&1&1\\
1&1&1
\end{pmatrix}.$$
The next step is to multiply that matrix by the
$$\begin{pmatrix}a\\b\\c\end{pmatrix}$$
matrix and set that equal to the zero matrix. I end up with a single equation:
$$a+b+c=0.$$
My question is what would be the eigenvector for this value? Typically in other problems similar to this one, I end up with 3 different equations in $a$, $b$, and $c$, and am able to solve for those variables and with that I'd have my eigenvector, but this problem has me stunned.
 A: Eigenvectors solve the equation $(A-\lambda I)x = 0$.
The quantity $A-\lambda I$ is a matrix. Therefore, any solution to $(A-\lambda I)x = 0$ lies in the null space of $A-\lambda I$.
The dimension of the null space of a square matrix is the size of the matrix minus is rank (this is the rank-nullity theorem, in a nutshell).
If you look at the matrix above, you have $\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix}$. What can you say about the rank of this matrix? Its rank is one. Since it is a $3\times 3$ matrix, the dimension of its null space is $3-1=2$.
Since the eigenvectors form a basis to the null space of the matrix, and the null space has dimension 2, there are two eigenvectors associated with the repeated eigenvalue.
To compute these two eigevectors, you must simply find the basis for the null space of $\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{pmatrix}$.
A: @eigenvalue 1, you correctly created the new matrix, which I'm gonna call $[B]$: 
$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1\end{bmatrix}$
The eigenvector is whatever will produce the zero vector $\overrightarrow{0}$when you multiply it with the $(A−λI)$. So, it's just a matter of solving:
$(A−λI)\overrightarrow{X}=\overrightarrow{0}$
which is actually in this case (since you already did the $A-λI$ part):
$[B]\overrightarrow{X}=\overrightarrow{0}$
You do your row reductions and manipulations to end up with 
$\begin{bmatrix} 1 & 1 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} = \begin{bmatrix} 0 \\0 \\  0\end{bmatrix}$
Then, each component is:
$x_1= -x_2 - x_3$
$x_2= s$
$x_3= t$
Basically, you should be realizing that this matrix produces infinite solutions along two vectors. The solutions take the "shape" of:
$\overrightarrow{X}= s \begin{bmatrix} -1 \\1 \\  0\end{bmatrix}+
t\begin{bmatrix} -1 \\0 \\  1\end{bmatrix}$
So $\begin{bmatrix} -1 \\1 \\  0\end{bmatrix}, \begin{bmatrix} -1 \\0 \\  1\end{bmatrix}$are two of your eigenvectors. Of course, this is again only @eigenvalue1, you'll have to do the same thing for eigenvaue 4
