Why are all the eigenvalues of a "Hermitian block-encoding" equal to $\pm1$?

Question

I was looking at the paper : https://arxiv.org/abs/2002.11649 and the eigenvalue discussion is not clear to me.

Block-encoding is a general technique to encode a nonunitary matrix on a quantum computer. Let $A \in \mathbb{C}^{N \times N}$ be an $n$-qubit Hermitian matrix. If we can find an $(m+$ $n)$-qubit unitary matrix $U \in \mathbb{C}^{M N \times M N}$ such that $$ U_{A}=\left(\begin{array}{cc} A & \cdot \\ \cdot & \cdot \end{array}\right) $$ holds, i.e., $A$ is the upper-left matrix block of $U_{A}$, then we may get access to $A$ via the unitary matrix $U_{A}$. In particular, $$ A=\left(\left\langle 0^{m}\right| \otimes I_{n}\right) U_{A}\left(\left|0^{m}\right\rangle \otimes I_{n}\right) $$

If $U_{A}$ is Hermitian, it is called a Hermitian block-encoding. In particular, all the eigenvalues of a Hermitian block-encoding $U_{A}$ are $\pm 1$.

It is not clear why all the eigenvalues are $\pm 1$? any guidance? Thanks

Answer 1

$U_A$ is defined to be a unitary matrix in the paper and your question. Consider the eigenvalue $\lambda$ of the general unitary matrix $U$ given by $U|\lambda\rangle=\lambda|\lambda\rangle$.

$$U|\lambda\rangle=\lambda|\lambda\rangle$$ $$\langle\lambda| U^{\dagger}=\lambda^*\langle\lambda|$$ $$\langle\lambda| U^{\dagger}U|\lambda\rangle=\lambda\lambda^*\langle\lambda|\lambda\rangle$$ $$\rightarrow |\lambda|^2=1$$

In the paper it also says "If $U_A$ is Hermitian then it is called Hermitian Block Encoding" on the first line of page three. Now obviously we know that the eigenvalues of a Hermitian Matrix are real.

Therefore we know the eigenvalues $\lambda$ of the matrix $U_A$ are real and have magnitude 1. There are only 2 such numbers; $+1$ and $-1$.