Which of the following statements is/are TRUE for an idempotent matrix $A$ (where $A^2 = A$) of order $n$?
Options:
(A) The only possible eigenvalues of $A$ are $0$ and $1$
(B) The matrix $I - A$ is also an idempotent matrix
(C) The trace of matrix $A$ is equal to the rank of matrix $A$
(D) If $A$ is invertible, then $A$ must be the identity matrix $I$
Answer Key:
A, B, C, D
Step-by-Step Solution & Explanation
Let $A$ be an idempotent matrix ($A^2 = A$):
1. If $\lambda$ is an eigenvalue with eigenvector $x \ne 0$, then $Ax = \lambda x \implies A^2 x = \lambda^2 x$. Since $A^2 = A$, $\lambda^2 = \lambda \implies \lambda(\lambda - 1) = 0$, so $\lambda \in \{0, 1\}$ (A is true).
2. $(I - A)^2 = I - 2A + A^2 = I - 2A + A = I - A$, so $I - A$ is idempotent (B is true).
3. Any idempotent matrix is diagonalizable, and since its eigenvalues are 0 or 1, $\operatorname{trace}(A) = \sum \lambda_i = \operatorname{rank}(A)$ (C is true).
4. If $A$ is invertible, multiplying $A^2 = A$ by $A^{-1}$ yields $A = I$ (D is true).