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?
Best Answer
No. Repeated Eigen values don't necessarily have repeated Eigen vectors.
Counter Example: $$L =\begin{bmatrix}2&0\\0&2\end{bmatrix}$$ $$\lambda_{1,2}=2$$
$$v_1 =\begin{bmatrix}1\\0\end{bmatrix}$$
$$v_2 =\begin{bmatrix}0\\1\end{bmatrix}$$