Bounds on eigenvalue perturbation of block matrix

Question

Consider some square $(n+m)\times(n+m)$ block matrix of the form $$M=\left[\matrix{A & B\\ \epsilon C & D}\right]$$ where $\epsilon$ is some perturbation parameter and $A,D$ are square matrices of dimensions $n\times n$ and $m\times m$ respectively. I am interested in finding some bounds on how much the eigenvalues of $M$ differ from the eigenvalues of $A$ and $D$ (for small $\epsilon>0$) i.e. $$|\lambda_i(A)-\lambda_i(M)|\leq f(\epsilon,B,C)\quad \forall i=1,2,\cdots,n$$ and $$|\lambda_i(D)-\lambda_i(M)|\leq g(\epsilon,B,C)\quad \forall i=n+1,n+2,\cdots,n+m$$

Answer 1

$f,g$ depend also on $A,D$. Indeed, the eigenvalues $(\lambda_i)$ of $M$ may be non-derivable when they are multiple. In particular, an upper bound depends on the values of the $\lambda_i-\lambda_j,i,j$.

Then I think about a bound as

$|\Delta(\lambda)|\leq \dfrac{||B||||C||\epsilon}{|discrim(\chi)|^{\alpha}}$

where $discrim(\chi)$ is the discriminant of $\chi$, the characteristic polynomial of $M$ and $\alpha >0$.

On the other hand, when $D-\lambda I$ is invertible and $m=n$, $\det(M-\lambda I)=\det((A-\lambda I)(D-\lambda I)-B(D-\lambda I)^{-1}C(D-\lambda I))$.

Possibly $cond(D)$ or $cond(A)$ may also appear in the bound.

EDIT. I come back quickly. Let $M=\begin{pmatrix}A&B\\C&D\end{pmatrix}$ where $A,B,D$ are fixed and $C$ is a $C^1$ matrix function of $t\in\mathbb{R}$ and $f(\lambda)=\det(M-\lambda I)$.

Since $\lambda\in spectrum(M)$ is simple, it is a $C^1$ function of the $(c_{i,j})$, $f'(\lambda)=tr((M'-\lambda'I)adj(M-\lambda I))=0$ and

$\lambda'=\dfrac{tr(M'adj(M-\lambda I))}{tr(adj(M-\lambda I))}=\dfrac{tr(C'adj(M-\lambda I))}{tr(adj(M-\lambda I))}$.

Now $|\lambda'|\leq \dfrac{||C'||||adj(M-\lambda I)||}{|tr(adj(M-\lambda I))|}$ where $||.||$ denotes the Frobenius norm.

Since $rank(adj(M-\lambda I))=1$, $|tr(adj(M-\lambda I)|=\rho(adj(M-\lambda I))$ and finally

$|\lambda'|\leq \dfrac{||C'||||adj(M-\lambda I)||}{\rho(adj(M-\lambda I))}$.

Note that $\rho(adj(M-\lambda I))$ strongly depends on $|\lambda-\mu|$ where $\mu$ is the closest element of $\lambda$ in $spectrum(M)\setminus \{\lambda\}$.