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?
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.)
Best Answer
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 $det X = 21.661...$ .