If an eigenvalue is repeated, is the eigenvector also repeated? I am not sure if I am doing this correctly, but... I have a Laplaican matrix as follows:
$$L =\begin{bmatrix}2&-1&-1\\-1&2&-1\\-1&-1&2\end{bmatrix}$$
And I would like to find the eigenvalues and eigenvectors. I end up with three eigenvalues ($e_1=0, e_2=3, e_3=3$).
I know that to find their corresponding eigenvectors, I need to solve for $(L-eI)v = 0$ (where $e$ is an eigenvalue and $v$ is an eigenvector).
For the first eigenvector, I end up with a vector of $[1,1,1]$.
I am asking about the second/third eigenvector. Here is my work. I know I need to solve for:
$$(L-eI)v = 0$$
Since $e=3$ (for both second and third eigenvector), then $L-e$ is:
$$\begin{bmatrix}-1&-1&-1\\-1&-1&-1\\-1&-1&-1\end{bmatrix}$$
This leaves me solving for a system of equations where $(L-eI)v = 0$. Originally, I came up with two eigenvectors for $v_2$ and $v_3$: $[1, 1, -2]$.
But, can these two eigenvectors be the same? Is there some rule about the three eigenvectors needing to be perpendicular to one another? And if so, how would I apply it in this case?
 A: No. Repeated Eigen values don't necessarily have repeated Eigen vectors. 
Counter Example:
$$L =\begin{bmatrix}2&0\\0&2\end{bmatrix}$$
$$\lambda_{1,2}=2$$
$$v_1 =\begin{bmatrix}1\\0\end{bmatrix}$$
$$v_2 =\begin{bmatrix}0\\1\end{bmatrix}$$
A: Actually, you have $2$ different eigenvectors for which $(L-3I) v = 0$. One is $[1,0,-1]$ and one is $[1,-1,0]$, for example.
A: Since the matrix is symmetric, it is diagonalizable, which means that the eigenspace relative to any eigenvalue has the same dimension as the multiplicity of the eigenvector.
It doesn't make sense to speak about a “repeated eigenvector”; you can find a basis of the eigenspace, which is the null space of the matrix $L-\lambda I$ (where $\lambda$ is the eigenvalue under consideration.
For $\lambda-3$ you have
$$
L-3I=
\begin{bmatrix}
-1 & -1 & -1 \\
-1 & -1 & -1 \\
-1 & -1 & -1
\end{bmatrix}
\xrightarrow{\text{Gaussian elimination}}
\begin{bmatrix}
1 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{bmatrix}
$$
which means that the eigenvectors satisfy $x_1=-x_2-x_3$, so a basis of the eigenspace is
$$
\left\{
\begin{bmatrix}
-1\\1\\0
\end{bmatrix}\,,
\begin{bmatrix}
-1\\0\\1
\end{bmatrix}
\right\}
$$
