Let $A$ be $n \times n$ real valued square symmetric matrix of rank 2 with $\sum_{i=1}^n \sum_{j=1}^n A_{ij}^2 = 50$. Consider the following statements. (I) One eigenvalue must be in $[-5, 5]$ (II) The eigenvalue with the largest magnitude must be strictly greater than 5 Which of the above statements about eigenvalues of $A$ is/are necessarily CORRECT?

Since $A$ is a real symmetric matrix, all its eigenvalues are real.

Given $\text{rank}(A) = 2$, $A$ has exactly 2 non-zero eigenvalues, say $\lambda_1$ and $\lambda_2$, while the remaining $n-2$ eigenvalues are $0$.

The Frobenius norm squared of $A$ satisfies:

• Analysis of Statement (I): The zero eigenvalues (for $n > 2$) lie in $[-5, 5]$. Furthermore, since $\lambda_1^2 + \lambda_2^2 = 50$, the smaller squared magnitude satisfies $\min(\lambda_1^2, \lambda_2^2) \le 25$, meaning at least one non-zero eigenvalue satisfies $|\lambda| \le 5$, i.e., $\lambda \in [-5, 5]$. Thus, Statement (I) is necessarily correct.

• Analysis of Statement (II): The maximum squared magnitude satisfies $\max(\lambda_1^2, \lambda_2^2) \ge 25$, which implies $|\lambda_{\max}| \ge 5$. However, if $\lambda_1 = 5$ and $\lambda_2 = 5$ (or $\lambda_2 = -5$), we have $\lambda_1^2 + \lambda_2^2 = 25 + 25 = 50$, where the largest magnitude is exactly $5$, not strictly greater than $5$. Thus, Statement (II) is not necessarily correct.

Therefore, only statement (I) is necessarily correct, so the correct option is (B).