Question

Let \( A \in \mathbb{R}^{n \times n} \) be such that \( A^3 = A \). Which one of the following statements is ALWAYS correct?

• A. \( A \) is invertible
• B. Determinant of \( A \) is 0
• C. The sum of the diagonal elements of \( A \) is 1
• D. \( A \) and \( A^2 \) have the same rank

Hint:

If \( A^3 = A \), then \( A^2 = A \), and both \( A \) and \( A^2 \) have the same rank. This property holds true for idempotent matrices.

Answer 1

The Correct Option is D

We are given that \( A^3 = A \), which means that \( A \) is an idempotent matrix. An idempotent matrix is one where \( A^2 = A \), but in this case, the matrix also satisfies \( A^3 = A \), which implies \( A^2 = A \) as well.

Let’s analyze each option:

Option (A): \( A \) is invertible. For \( A \) to be invertible, \( \text{det}(A) \neq 0 \). However, from the equation \( A^3 = A \), we know that the eigenvalues of \( A \) must satisfy \( \lambda^3 = \lambda \), which gives \( \lambda = 0 \) or \( \lambda = 1 \). Therefore, \( A \) could have eigenvalue 0, making \( \text{det}(A) = 0 \), meaning that \( A \) is not necessarily invertible. This statement is not always true.

Option (B): The determinant of \( A \) is 0. As mentioned earlier, since \( A \) can have eigenvalue 0, the determinant could indeed be 0, but this is not guaranteed. \( A \) could also have only eigenvalue 1 (in which case \( \text{det}(A) = 1 \)). Hence, this statement is not always true.

Option (C): The sum of the diagonal elements of \( A \) is 1. The sum of the diagonal elements of a matrix is the trace of the matrix, which is the sum of its eigenvalues. Since \( A^3 = A \), the eigenvalues of \( A \) can only be 0 or 1. Therefore, the trace (sum of eigenvalues) is the number of 1’s among the eigenvalues, but this is not guaranteed to be 1. For example, if all eigenvalues are 0 or if there are multiple eigenvalues equal to 1, the trace could be different. Therefore, this statement is not always true.

Option (D): \( A \) and \( A^2 \) have the same rank. This is the correct statement. If \( A^3 = A \), then \( A^2 = A \). This implies that the rank of \( A^2 \) is equal to the rank of \( A \), because the non-zero eigenvalues of both \( A \) and \( A^2 \) must be the same. Therefore, this statement is always true.

Thus, the correct answer is (D) \( A \) and \( A^2 \) have the same rank.