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

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

• – glS
Sep 23, 2021 at 7:26

$$U_A$$ 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 $$-1$$.