Consider a non diagonalizable square matrix $A \in \mathbb{R}^{N \times N}$, and consider its distinct eigenvalues counted only once $\lambda_1, \lambda_2, \ldots, \lambda_L \in \mathbb{C}$, with $1 \leq L < N$ (the non diagonalizability of $A$ implies that $L$ is strictly lower than $N$).
For each of the previous eigenvalue, we can find at least one eigenvector. The eigenvectors are $v_1, v_2, \ldots, v_L \in \mathbb{C}^{N}$.
Now consider the matrix $$B = A - \eta I,$$
with $\eta \in \mathbb{R}$.
If we multiply $B$ by an eigenvector $v_j$ of $A$, we get the following:
$$Bv_j = Av_j - \eta v_j = (\lambda_j - \eta) v_j.$$
This means that $\lambda_1-\eta, \lambda_2-\eta, \ldots, \lambda_L-\eta \in \mathbb{C}$ are also eigevalues of $B$, and $v_1, v_2, \ldots, v_L \in \mathbb{C}^{N}$ are the corresponding eigenvectors.
The question is:
Can I say that $\lambda_1-\eta, \lambda_2-\eta, \ldots, \lambda_L-\eta \in \mathbb{C}$ are the unique eigenvalues of $B$? Can there be other eigenvalues of $B$?
Motivation of the question
If $\lambda_1-\eta, \lambda_2-\eta, \ldots, \lambda_L-\eta \in \mathbb{C}$ are the unique eigenvalues of $B$, then I can choose $\eta$ such that the real part of each eigenvalue of $B$ is negative, i.e.
$$ \eta > \mathcal{Re}(\lambda_q),$$
such that $$\mathcal{Re}(\lambda_q) = \max_{i=1, \ldots, L} \mathcal{Re}(\lambda_i) .$$
If there are more eigenvalues than the listed, then I can't be sure of choice of $\eta$.
The answer is yes, as can be seen immediately using the characterisation that $\lambda$ is an eigenvalue of a matrix $M$ if and only if $\det(M-\lambda \operatorname{Id}_N)=0$.
In fact, if the list of distinct eigenvalues of a matrix $M\in \mathcal{M}_{N\times N}(\mathbb{C})$ is $\lambda_1, \ldots,\lambda_k $ then the list of eigenvalues of $(M-\lambda \operatorname{Id}_N)$ is $\lambda_1-\lambda, \ldots,\lambda_k -\lambda$, and this is true for literally any matrix, diagonalisable or not.
Now to answer your question in the comments, if $B$ was diagonalisable, you would get $$P(A-\eta \operatorname{Id})P^{-1}=D$$ where $D$ is some diagonal matrix, therefore $$PAP^{-1}=D+\eta \operatorname{Id}$$ would be diagonal as well which is impossible. Note that we used the fact that $\operatorname{Id}$ commutes with any matrix (here $P$).