Finding three independent solutions for the system I'm stuck on this assignment. Not sure how to begin.
Let $\begin {bmatrix}2&1&1\\ 0&2&3\\ 0&0&2\end {bmatrix} = A$ as part of the system $Ax=x'$. Find three independent solutions for the system.
The system has eigenvalue 2 with multiplicity 3.You need to develop a way to extend methods for dealing with repeated eigenvalues from multiplicity 2 to multiplicity 3. Also, you will need to justify this method, following a similar derivation to that of eigenvalues for a two equation system with a multiplicity of 2. Be sure to verify that the solutions are linearly independent.
 A: For the first eigenvalue, we can write: $[A-2I]v_1 = 0$.
This leads to:
$$\begin {bmatrix}0&1&1\\ 0&0&3\\ 0&0&0\end {bmatrix}v_1=\begin {bmatrix}0\\0\\0 \end {bmatrix}$$ 
This leads to eigenvector of: $v_1 = (1,0,0)$.
For the next, repeated, eigenvalue, we can form the generalized eigenvector: $[A - 2I]v_2 = v_1$.
This leads to:
$$\begin {bmatrix}0&1&1\\ 0&0&3\\ 0&0&0\end {bmatrix} v_2=\begin {bmatrix}1\\0\\0 \end {bmatrix}$$ 
This results in an eigenvector of: $v_2 = (0,1,0)$.
We can try repeating this process again, so, for the next, repeated, eigenvalue, we can form the generalized eigenvector: $[A - 2I]v_3 = v_2$.
$$\begin {bmatrix}0&1&1\\ 0&0&3\\ 0&0&0\end {bmatrix} v_3=\begin {bmatrix}0\\1\\0 \end {bmatrix}$$
This results in an eigenvector of: $\displaystyle v_3 = (0,-\frac{1}{3}, \frac{1}{3})$.
Note this was called the bottom-up method, but there are other ways to approach it, like the adjoint method or top-down method and others. For example, sometimes the bottom-up method does not work and you have to resort to the top-down.
Now, you need to show that these $v_i$ are linearly independent, but I suppose you know how to do that.
