Consider this matrix: $\begin{bmatrix}2-\lambda & -1 & 2\\-1&2-\lambda&2\\2&2&-1-\lambda\end{bmatrix}$ and the eigenvalues $-3$,$3$ and $3$.
The corresponding eigenvectors are $\begin{pmatrix}-1\\-1\\2\end{pmatrix}$, $\begin{pmatrix}1\\-1\\0\end{pmatrix}$ and $\begin{pmatrix}2\\0\\1\end{pmatrix}$.
The solution to the linear system can be $\begin{pmatrix}-a+2b\\a\\b\end{pmatrix}$, $\begin{pmatrix}a\\2b-a\\b\end{pmatrix}$ and $\begin{pmatrix}a\\b\\\frac{a+b}{2}\end{pmatrix}$.
My question is, why are there two eigenvectors for the eigenvalue $3$?
Since there are two variables $a$ and $b$, aren't there infinite linearly independent vectors? So why pick these two in particular? And why two and not one or three or more?
Best Answer
Whenever there is a double root in the characteristic equation giving a repeated eigenvalue as you have here, then there are infinitely many eigenvectors, correspondong to this eigenvalue, all of which are coplanar. You can therefore choose any two independent (non-parallel) vectors in this plane to be the eigenvectors.