Eigenvalue with algebraic multiplicity > 1 If I have some $\lambda_1$ as one of my eigenvalues, and it has algebraic multiplicity of $n$ with $n > 1$. When I try to get some eigenvectors associated with this $\lambda_1$, assuming that the geometric multiplicity for $\lambda_1$ is going to be $2$, how should I compute these two eigenvectors? It's easy when we have one eigenvalue associating to just one eigenvector, but how about this case that I've mentioned?
Can someone please show me an example of that computation to find these eigenvectors?!
Thanks!
 A: Let $A$ be a $3 \times 3$ matrix defined as follows
\begin{equation}
A=
  \begin{bmatrix}
    0 & 1 & 1\\
    1 & 0 & 1\\
    1 & 1 & 0
  \end{bmatrix}
\end{equation}
After some straight forward calculations, the characteristic polynomial is computed as $P(\lambda) = det(A - \lambda I ) = - \lambda^3 + 3\lambda + 2 = - (\lambda - 2)(\lambda + 1)^2$. So $\lambda_1 = -1$ is of multiplicity $2$ and $\lambda_2 = 2$ is of multiplicity $1$. Your question is how do we compute the eigenvectors associated with $\lambda_1$. 
Let $x_1$ and $x_2$ be the two eigenvectors associated with $\lambda_1$, then 
\begin{equation}
(A - \lambda_1 I )x_k = 0 \qquad k =1,2 
\end{equation}
which gives
\begin{equation}
 \begin{bmatrix}
    1 & 1 & 1\\
    1 & 1 & 1\\
    1 & 1 & 1
  \end{bmatrix}x_k = 0 \qquad k =1,2 
\end{equation}
Let $x_k = [a ,\ b ,\ c]^T$. So we get $a + b + c = 0$ which is an equation of 3 unknowns. Therefore any $3 \times 1$ vector satisfying $a + b +c =0$ is an eigenvector of $A$. Also note that this space of eigenvectors is of dimension $2$ (due to the repeated eigenvalue). What you can do is pick any $a$ and $b$, then compute $c$ accordingly. So for $a = b = 1$, you get $c = -2$ and hence
\begin{equation}
x_1 =
 \begin{bmatrix}
    1 \\
    1 \\
    -2 
  \end{bmatrix}
\end{equation}
Now, choose $x_2$ such that it is linearly independent of $x_1$, meaning that $x_2 \neq \alpha x_1$ for any $\alpha\in \mathbb{R}$. So for $x_2$ you could pick $a = -2$ and $b = 1$ to get $c = 1$ hence
\begin{equation}
x_2 =
 \begin{bmatrix}
    -2 \\
    1 \\
    1 
  \end{bmatrix}
\end{equation}
A: You have to solve, as usual, $(A-\lambda_1I)v=0$. If the geometric multiplicity is $2$, it means you have a linear system of codimension 2 (rank $= \dim A-2$), which you solve as any linear system, knowing in advance the set of solutions will depend on two parameters.
