# 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.

• @user1551 interesting, is that just from $\det(-\lambda I+CC^\dagger/\lambda)=0$ or is there a deeper theorem you used? And can you be a little more explicit about what Cauchy's inequality implies? My linear algebra is a bit weak and I can't make sense of what you wrote, but it seems like what I want to know! Thanks! Commented Jul 2, 2022 at 1:24

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