Eigenvalues of block matrix with zero diagonal I'm looking for the eigenvalues of a block matrix of the form
$$
M=\begin{pmatrix}
\boldsymbol{0} & C \\
C^\dagger & \boldsymbol{0}
\end{pmatrix}=
\begin{pmatrix}
\boldsymbol{0} & H(I-uu^\dagger) \\
(I-uu^\dagger)H & \boldsymbol{0}
\end{pmatrix}
$$
Where $H$ is a general positive semidefinite Hermitian matrix (dim: $2^n\times 2^n$) and $u$ is a general unit vector. If it makes things simpler, we can take the elements of $H$ and $u$ to be real. $M$'s clearly Hermitian so the eigenvalues are real. I can show 2 of $M$'s eigenvalues are zero, corresponding to
$$v=\begin{pmatrix}1\\0\end{pmatrix}\otimes u \quad\text{and}\quad v=\begin{pmatrix}0\\1\end{pmatrix}\otimes H^{-1}u$$
I can also show they need to come in pairs $\lambda, -\lambda$ since we have
$$
\left|\begin{pmatrix}
A & B \\
C & D
\end{pmatrix}\right| = \det (A)\det\left(D-CA^{-1}B\right)\to \left|\begin{pmatrix}
-\lambda I & C \\
C^\dagger & -\lambda I
\end{pmatrix}\right| = \det (-\lambda I)\det\left(-\lambda I+CC^\dagger/\lambda\right)
$$
which is clearly invariant under $\lambda\to -\lambda$.
What can be said about the remaining eigenvalues? An explicit formula would be awesome but bounds on their magnitudes or really anything would be helpful.
 A: By considering the Schur complement of the second diagonal sub-block of $\lambda I-M$, we have
\begin{aligned}
\det(xI-M)
=\det\pmatrix{xI&-C\\ -C^\ast&\lambda I}
&=\det(xI)\det\big(xI-C^\ast(\lambda I)^{-1}C\big)\\
&=\det(x^2I-C^\ast C).\\
\end{aligned}
Therefore the eigenvalues of $M$ are the zeroes of the polynomial
$$
\det(x^2I-C^\ast C)
=\prod_i\big(x^2-\lambda_i(C^\ast C)\big)
=\prod_i\big(x^2-\sigma_i(C)^2\big)
=\prod_i\big(x+\sigma_i(C)\big)\big(x-\sigma_i(C)\big).
$$
That is, they are the singular values of $C$ and their negatives.
Let $C$ be $N\times N$. By the definition of $C$, we have $C^\ast C=(I-uu^\ast)H^2(I-uu^\ast)$. Since $I-uu^\ast$ is an orthogonal projection onto a hyperplane, if we arrange the eigenvalues of $H^2$ and $C^\ast C$ in descending order, then by Cauchy's interlacing inequality (see Horn and Johnson, Matrix Analysis, 2/e, p.242, theorem 4.3.17),
$$
\lambda_1(H^2)\ge\lambda_1(C^\ast C)\ge\lambda_2(H^2)\ge\lambda_2(C^\ast C)\ge\cdots\ge\lambda_{N-1}(H^2)\ge\lambda_{N-1}(C^\ast C)\ge\lambda_{N}(H^2).
$$
It follows that
$$
\lambda_1(H)\ge\sigma_1(C)\ge\lambda_2(H)\ge\sigma_2(C)\ge\cdots\ge\lambda_{N-1}(H)\ge\sigma_{N-1}(C)\ge\lambda_N(H)\big(\ge\sigma_N(C)=0\big).
$$
