Let $H=M \oplus M^{\perp}$ be a Hilbert space and an operator $A$ on $H$ be described with respect to the given decomposition as:
$A= \left( \begin{array}{ccc} |z|^{2} & -\bar{z}\langle Bu,v\rangle \\ -z\langle v, Bu\rangle & |\langle Bu,v \rangle|^{2}+\|(\alpha_{z}I-C)u\|^{2} \\\end{array} \right) $.
Then $A$ is self-adjoint. The question is to show that the minimum eigenvalue $\lambda_{min}$ of $A$ is strictly smaller than $|z|^{2}$, i.e. $\lambda_{min} < |z|^{2}$. How can this be proved?
The only idea I have got so far is as follows:
The eigen spectrum $\sigma_{e}(A)$ is contained in the numerical range $w(A)=\{\langle Ax, x\rangle: \|x\|=1\}$. Choosing a unit vector $\xi \in M$, and considering the unit vector $x=\xi \oplus 0$, $|z|^{2}=\langle Ax, x \rangle$. Can something be deduced from here?
This is just a thought which might be helpful but is a bit long to be a comment.
Consider a $N\times N$ matrix $A$ whose $(1,1)$ entry is $|z|^2$ . Let $e_1=[1,0,\dots,0]^T$ denote first column of $N\times N$ Identity matrix. Then note that, $$\lambda_{min}(A) = \min_{x^Tx=1}~x^TAx\leq e_1^TAe_1=|z|^2$$