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.
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 $$ \lambda_1(H)\ge\sigma_1(C)\ge\lambda_2(H)\ge\sigma_2(C)\ge\cdots\ge\lambda_{N-1}(H)\ge\sigma_{N-1}(C)\ge\lambda_N(H)\big(\ge\sigma_N(C)=0\big). $$