Same eigenvalues, different eigenvectors I'm interested in the case of a specific matrix having different eigenvectors corresponding to two identical eigenvalues. The method I use for spectral decomposition returns different eigenvectors, even though the eigenvalue is the same. Is this possible, and if so, what this tells about the matrix?
 A: A trivial example: Consider the 2 by 2 identity matrix. It has only one eigenvalue, namely 1. However both $e_1=(1,0)$ and $e_2=(0,1)$ are eigenvectors of this matrix. 
A: Of course it's possible: 
$$
\begin{bmatrix}
2&0\\ 0&2
\end{bmatrix}
\,
\begin{bmatrix}
1\\ 0
\end{bmatrix}
=
2\;\begin{bmatrix}
1\\ 0
\end{bmatrix},
\ \ \ \ \ \ 
\begin{bmatrix}
2&0\\ 0&2
\end{bmatrix}
\,
\begin{bmatrix}
0\\ 1
\end{bmatrix}
=
2\;\begin{bmatrix}
0\\ 1
\end{bmatrix}.
$$
What it tells you about the matrix is that the geometric multiplicity of the eigenvalue is greater than $1$. 
A: Quick test for a $2 \times 2$ matrix where $a$ are (same) eigenvalues:
\begin{bmatrix} a & b \\ 0 & a\end{bmatrix}.
If $b = 0$, there are $2$ different eigenvectors for same eigenvalue $a$.
If $b \neq 0$, then there is only one eigenvector for eigenvalue $a$.
A: Every eigenvalue with multiplicity = n will be associated with n different (as in linearly independent) eigenvalues.
Multiplicity is how many "times" it shows up as an eigenvalue. It is like when you find only one solution to a second degree equation, which always has two roots. This solution has a multiplicity = 2.
