# Multiple eigenvalue solutions problem

In a problem regarding multiple eigenvalue solutions (defective eigenvalues, complete eigenvalues, the like) I have a 4x4 matrix with one complete eigenvalue, and another incomplete eigenvalue with a defect of 1 and a multiplicity of 3.

The associated eigenvector to the latter is defined by the system:

$2a + c = 0$

$2b + d = 0$

Thus, depending on our choices for $a, b$ and $c, d$ respectively, we obtain two eigenvectors:

$u_1 = [1, 0, -2, 0]^T$

$u_2 = [0, 1, 0, -2]^T$

Because this eigenvalue has defect $1$, we are tasked with finding a generalized eigenvector of rank $2$.

After solving for $(A + 2I)^2 v_2 = 0$ {the eigenvalue of concern is -2}, we have $v_2 = [0, 0, 1, -1]^T$ is such a vector and satisfies $(A + 2I)v_2 = v_1$ {therefore $v_2$ is an associated eigenvector}.

Thus {$v_1$, $v_2$} is the length 2 chain we need. Now here is my question (I applaud those who made it this far): The book reads, "The eigenvector $v_1$ is neither of the two eigenvectors found previously, but we observe that $v_1 = u_1 - u_2$."

What is the significance of this observation?

The book continues: "For a length $1$ chain {$w_1$} to complete the picture, we can choose any combination of $u_1$ and $u_2$ that is independent of $v_1$; for instance,

$w_1 = u_1 + u_2$"

How am I able to do this? What does this mean? What leads us to this conclusion? I know that the associated eigenvectors of different eigenvalues have to be linearly independent, but do the associated eigenvectors of a single eigenvalue have to be linearly independent?

Please do excuse my tedious rambling, as I am unable to post a picture of my problem (10 reputation apparently). I also realize this is more than one question, however, I feel the longevity and detail of the problem warrants more than a single question!

You have $v_1,v_2$ and now you need to complete them to a basis of the generalized eigenspace by adding a vector from the eigenspace that is independent of the two. (It actually suffices that it is independent of $v_1$ but I find it easier to think about it that way.)
This is just said to make clear that $v_1$ is not independent of $u_1,u_2$. So it is in the eigenspace but (necessarily) a linear combination of the two, as the dimension of the eigenspace is two.