Question 31

Let $A$ be an $n \times n$ real symmetric matrix of rank 2. Because $A$ is real symmetric, all its eigenvalues are real. Because the rank of $A$ is 2, exactly two eigenvalues are non-zero, say $\lambda_1$ and $\lambda_2$, while the remaining $n - 2$ eigenvalues are zero.

The sum of the squares of all elements of $A$ equals the square of the Frobenius norm $\|A\|_F^2$:

Substituting the eigenvalues:

Now examine Statement (I): "One eigenvalue must be in $[-5, 5]$".
Suppose for contradiction that both non-zero eigenvalues have magnitude strictly greater than 5, i.e., $|\lambda_1| > 5$ and $|\lambda_2| > 5$. Then $\lambda_1^2 > 25$ and $\lambda_2^2 > 25$, which implies:

This contradicts $\lambda_1^2 + \lambda_2^2 = 50$. Therefore, at least one of $\{\lambda_1, \lambda_2\}$ must satisfy $|\lambda| \le 5$, meaning it lies in $[-5, 5]$. (The $n - 2$ zero eigenvalues also trivially lie in $[-5, 5]$). Hence, Statement (I) is necessarily correct.

Now examine Statement (II): "The eigenvalue with the largest magnitude must be strictly greater than 5".
Consider the case where $\lambda_1 = 5$ and $\lambda_2 = 5$ (or $\lambda_1 = 5, \lambda_2 = -5$). Then:

The largest magnitude among all eigenvalues is $\max(|\lambda_1|, |\lambda_2|) = 5$, which is not strictly greater than 5. Hence, Statement (II) is not necessarily true.

Therefore, only Statement (I) is necessarily correct.