Which of the following is(are) NOT the square of a 3 × 3 matrix with real entries?<br><br>(A) $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$ (B) $\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$ (C) $\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ (D) $\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$
$\begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$
$\begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$
Step-by-Step Solution
Key Concept: A matrix A is the square of a real matrix B (i.e., A = B^2) if and only if its determinant is non-negative (if it has real eigenvalues) or more generally, if it can be diagonalized with non-negative eigenvalues or has Jordan blocks corresponding to negative eigenvalues that can be paired. Specifically, for a diagonal matrix, the square of a real matrix must have non-negative diagonal entries if the eigenvalues are real.
Let $A = B^2$. If $A$ is a diagonal matrix with diagonal entries $\lambda_1, \lambda_2, \lambda_3$, then these are the eigenvalues of $A$. Since $A = B^2$, the eigenvalues of $A$ must be the squares of the eigenvalues of $B$. If $B$ is a real matrix, its eigenvalues are either real or occur in conjugate pairs. For a diagonal matrix $A$ to be a square of a real matrix, its negative eigenvalues must come in pairs (to allow for imaginary eigenvalues of $B$). In (A), the eigenvalues are 1, 1, -1. There is only one negative eigenvalue, so it cannot be the square of a real matrix. In (B), the eigenvalues are -1, -1, -1. There are three negative eigenvalues, which cannot be paired, so it cannot be the square of a real matrix. (C) and (D) have non-negative eigenvalues or pairs of negative eigenvalues, so they can be squares of real matrices.
Correct Answer: 1, 2