Let $A\in\mathbb{R}^{m \times m}$ and $C\in\mathbb{R}^{n\times m}$. The pair $(A,C)$ is observable if $Cx \ne 0$ for every right eigenvector $x$ of $A$. Therefore, if the pair $(A,C)$ is NOT observable, then
\begin{align} \exists \lambda \in \mathbb{C}, x \in \mathbb{C}^m, x\ne 0 \: \text{ such that } Ax = \lambda x \text{ and } Cx = 0. \tag{1} \end{align}
In the above case that ($1$) holds, is there any concept or result that study the multiplicity of the eigenvalue $\lambda$ through which the observability is violated? For instance, what is the difference between the case that $\lambda$ in ($1$) is a simple eigenvalue and the case where $\lambda$ is not simple? Is there a way to identify the multiplicity of $\lambda$ for example through observability matrix?
Thank you for your comments and thoughts.
One answer to question 1: the notion of "detectability" gives us a bit of a lens into the nature of the eigenvalues of $A$ for which (1) holds. In particular, a system is said to be detectable if all such eigenvalues satisfy $\operatorname{Re}(\lambda) < 0$ (or $|\lambda|<1$ for discrete systems).
When the system is detectable, we find that the system's state can be recovered from the output up to an exponentially decreasing error. When the eigenvalue for which (1) holds is on the opposite side of the threshold (i.e. $\operatorname{Re}(\lambda) > 1$ or $|\lambda|>1$), we find that the error in observation grows exponentially. The simplicity of the eigenvalue hasn't played a role so far, but it matters for the edge case where (1) holds for an eigenvalue on the "boundary", i.e. one for which $\operatorname{Re}(\lambda) = 0$ (or $|\lambda| = 1$ in the discrete case), and (1) does not hold for any other eigenvalues outside the open left half plane.
If the boundary eigenvalue is simple, then the error in observation will be constant. If the boundary eigenvalue fails to be simple, then the error in observation grows polynomially.
Regarding question 2: yes, there is a way to identify the multiplicity of $\lambda$ using the observability matrix. In particular, if $x$ is in the kernel of the observability matrix, which is to say that $$ \mathcal O(A,C) x = \pmatrix{C\\CA\\ \vdots \\ CA^{m-1}}x = 0, $$ then we have $CA^k x = 0$ for all integers $k \geq 0$, which is to say that $\ker(C)$ contains the invariant subspace generated by $x$.
With that in mind: let $\{x_1,\dots,x_{k_1}\}$ be a basis for $\ker \mathcal O(A,C)$. Extend this to a basis $\{x_1,\dots,x_{k_1}, y_1,\dots,y_{k_2}\}$ of $\ker C$. Extend this to a further basis $\{x_1,\dots,x_{k_1},y_1,\dots,y_{k_2},z_1,\dots,z_{k_3}\}$ of $\Bbb R^m$. Let $S$ denote the matrix whose columns are the elements of this basis, i.e. $x_1,\dots,x_{k_1},y_1,\dots,y_{k_2},z_1,\dots,z_{k_3}$. If we apply a change of basis to the state space to obtain the new matrices $\bar C = CS$ and $\bar A = S^{-1}AS$, then we find that the matrices $\bar C$ and $\bar A$ will have the form $$ \bar C = \pmatrix{0_{n \times k_1}& 0_{n \times k_2} & C_0}, \quad \bar A = \pmatrix{A_{11} & A_{12} & A_{13}\\0 & A_{22} & A_{23}\\ 0 & A_{32} & A_{33}}, $$ where $C_0$ is invertible of size $k_3$ and $A_{11}$ has size $k_1$.
In terms of these matrices, the eigenvalues for which (1) holds are precisely the eigenvalues of $A_{11}$.
The decomposition I give can be thought of as an alternative version of the Kalman decomposition. In terms of the Kalman decomposition, we are concerned with the eigenvalues of $A_{r \bar o}$ and possibly some of those from $A_{\overline{ro}}$.