Why is this 3D operator with degeneracies only giving me 2 eigenstates

[struggling_student]

The matrix representation of a certain operator in a certain basis is

$$\begin{bmatrix} 1 & 0 & 0 \\0 & 0 & -i \\ 0 & i & 0
\end{bmatrix} .$$

The eigenvalue problem leads to this equation

$$0=det\begin{bmatrix} 1-\lambda & 0 & 0 \\0 & -\lambda & -i \\ 0 & i & -\lambda
\end{bmatrix} =-(\lambda-1)^2(\lambda+1).$$

So the eigenvalues are $1$, $1$ and $-1$. There is a degenacy and this means that there are two distinct states with the same eigenvalue but I would like to find all the eigenstates, generically denoted as $\begin{bmatrix} a \\b \\ c \end{bmatrix}$.

First, the non-degenerate $\lambda=-1$ leads to the following equation

$$0=\begin{bmatrix} 2 & 0 & 0 \\0 & 1 & -i \\ 0 & i & 1 \end{bmatrix} \begin{bmatrix} a \\b \\ c \end{bmatrix},$$

$$\therefore 2a=0 \ and \ b=ic.$$

So the nondegenerate eigenstate is $\begin{bmatrix} 0 \\ic \\ c \end{bmatrix}=\begin{bmatrix} 0 \\i \\ 1 \end{bmatrix}$ because without the loss of generality we can set $c:=1$.

Now, for the degerate $\lambda=1$ the equation is

$$0=\begin{bmatrix} 0 & 0 & 0 \\0 & -1 & -i \\ 0 & i & -1 \end{bmatrix} \begin{bmatrix} a \\b \\ c \end{bmatrix},$$

$\therefore b=-ic$, any value for $a$ will do so the degerate state is $\begin{bmatrix} a \\-ic \\ c \end{bmatrix}$. How do I interpret this?

I expected to find two different vectors but instead I've got a two-parameter result. What is the degerate state? Is what I wrote not a valid operator? Am I mistaken in thinking that a degerate eigenvalue corresponds to exactly two distinct eigenstates?

 

[Keith_McClary]

If you put in two different values for $a$ you get two linearly independent vectors.

 

[Haborix]

Your result makes sense. The eigenvector calculation needs to give you all the vectors in the subspace (two dimensional here) with the degenerate eigenvalue. However, if you want to have a set of orthogonal vectors, then after you pick the first value of $a$ you determine the second vector by enforcing orthogonality.