Edit: since we seem a bit deadlocked at this point, let me weaken the question. It's fairly easy to see that the set of 8-tuples of reals which can be the eigenvalues of a matrix of the desired form is closed. We know from jjcale and Caleb Eckhardt that its complement is nonempty. Is its complement dense? That is, would a generic 8-tuple not be the eigenvalues of such a matrix?




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First, here is a baby version of the question, that I already know the answer to. Consider complex Hermitian $4\times 4$ matrices of the form $$\left[\begin{matrix}a I\_2&A\cr A^\*&b I\_2\end{matrix}\right]$$ where $A \in M\_2(\mathbb{C})$ and $a,b \in \mathbb{R}$ are arbitrary. Can any four real numbers $\lambda\_1 \leq \lambda\_2 \leq \lambda\_3\leq \lambda\_4$ be the eigenvalues of such a matrix, or is there some restriction? Answer: there is a restriction, we must have $\lambda\_1 + \lambda\_4 = \lambda\_2 + \lambda\_3$.


The real question is: what are the possible eigenvalues of Hermitian $8\times 8$ matrices of the form $$\left[\begin{array}{c|c}aI\_4&A\cr \hline A^\*&\begin{matrix}bI\_2& B\cr B^\*&cI\_2\end{matrix}\end{array}\right]$$ with $a,b,c\in\mathbb{R}$, $A \in M\_4(\mathbb{C})$, and $B \in M\_2(\mathbb{C})$? Can any eight real numbers be the eigenvalues of such a matrix? (I suspect not. If they could, that would tell you that any Hermitian $8\times 8$ matrix is unitarily equivalent to one of this form.)



What are the possible eigenvalues of these matrices?
Let $V$ be the real vector space of the $8\times 8$ matrices of the form given in the question.


Where is an open set in $\mathbb{R}^8$ of possible $8$-tuples of eigenvalues of matrices in $V$.


**Proof :**


Choose $M\_1 \in V$ such that $M\_1$ has no degenerated eigenvalues.


Let $v\_1,...,v\_8$ be an orthonormal base of eigenvectors of $M\_1$ .


Then first order perturbation theory tells us that it suffices to show that the map


$V \rightarrow \mathbb{R}^8$, $M \mapsto (v\_1^\* M v\_1,...,v\_8^\* M v\_8)$ has rank $8$.


So I need $7$ additional matrices $M\_2,...,M\_8$ in $V$ such that the matrix
$X = (v\_i^\* M\_j v\_i)\_{ij}$ is nonsingular . 


Let $f(a,b,c,A,B)$ be the corresponding matrix.


Choose $$ A\_1 = \left[\begin{matrix}4&2&3&4\cr5&6&7&8\cr9&10&11&12\cr13&14&15&16\end{matrix}\right] $$ .


Then choose


$M\_1 = f(2.7,1,-1,A\_1,diag(2,1))$,


$M\_2 = f(1,0,0,0,0)$,


$M\_3 = f(0,1,0,0,0)$,


$M\_4 = f(0,0,1,0,0)$,


$M\_5 = f(0,0,0,diag(1,1,1,2),0)$,


$M\_6 = f(0,0,0,diag(1,0,0,0),diag(0,1))$,


$M\_7 = f(0,0,0,diag(0,1,0,0),diag(1,2))$,


$M\_8 = f(0,0,0,diag(0,0,1,0),diag(3,2))$.


This gives $det X = 21.661...$ .


