To analyze the problem:

We are given that A is a Hermitian matrix such that $A^2 = I$, where $I$ is the identity matrix. The goal is to determine which property of $A$ must be true.

Solution and Explanation:

1. Hermitian matrix properties:

   • A Hermitian matrix satisfies $A = A^\dagger$, where $A^\dagger$ is the conjugate transpose of $A$.
   • The eigenvalues of a Hermitian matrix are always real.

2. Given condition $A^2 = I$:

   • This implies $A \cdot A = I$, which means $A$ is its own inverse: $A^{-1} = A$.
   • Therefore, the eigenvalues of $A$ must satisfy $\lambda^2 = 1$, which means $\lambda = \pm 1$.

3. Analyzing the options:

   • Option 1: $A$ is singular
     This is incorrect because $A^2 = I$ implies $\det(A) \neq 0$, so $A$ is nonsingular.

   • Option 2: The eigenvalues of $A$ are all $\pm 1$
     This is correct. From $A^2 = I$, we derived that $\lambda^2 = 1$, so all eigenvalues must be $\pm 1$.

   • Option 3: $A$ is a diagonal matrix
     This is not necessarily true. While $A$ can be diagonalized (because it is Hermitian), it is not required to already be diagonal.

   • Option 4: The trace of $A$ is zero
     This is not necessarily true. The trace is the sum of eigenvalues, and since the eigenvalues are $\pm 1$, the trace could be zero, positive, or negative depending on the matrix.

Correct Answer:

The eigenvalues of $A$ are all $\pm 1$.

Would you like me to go into further details or discuss related properties?
Here are 5 related questions to expand your understanding:

1. Why do Hermitian matrices have real eigenvalues?
2. How does $A^2 = I$ imply that $A$ is invertible?
3. What other types of matrices satisfy $A^2 = I$ besides Hermitian ones?
4. How does the trace of $A$ relate to its eigenvalues?
5. What is the significance of diagonalizability for Hermitian matrices?

Tip: Remember that Hermitian matrices are always diagonalizable with real eigenvalues, making them particularly important in quantum mechanics and other physical applications!