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 need a basis of the generalized eigenspace. Therefore you need to make sure that the vectors are linearly independent.

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.)

What is the significance of this observation?

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy