<p>If \(A\) is a non-diagonal involutory matrix, then</p>
Step-by-Step Solution
Key Concept: An involutory matrix satisfies A² = I, which means (A-I)(A+I) = 0. For non-diagonal involutory matrices, the characteristic polynomial has eigenvalues ±1, and the trace equals the difference between counts of +1 and -1 eigenvalues.
<p><strong>Step 1:</strong> Recall that an involutory matrix A satisfies A² = I, so A² - I = 0, giving (A-I)(A+I) = 0.</p><p><strong>Step 2:</strong> This means the minimal polynomial divides (x-1)(x+1), so eigenvalues of A are only ±1.</p><p><strong>Step 3:</strong> For a non-diagonal involutory matrix, the characteristic polynomial is det(A - λI) = 0, which has roots λ = ±1 only.</p><p><strong>Step 4:</strong> The trace of A equals the sum of eigenvalues = (number of +1 eigenvalues) - (number of -1 eigenvalues).</p><p><strong>Step 5:</strong> The determinant of A equals the product of eigenvalues = (-1)^(number of -1 eigenvalues), so det(A) = ±1.</p><p><strong>Step 6:</strong> Since A is non-diagonal and involutory with A ≠ ±I, it must have both +1 and -1 eigenvalues, making it non-singular with eigenvalues restricted to {-1, +1}.</p><p>∴ Answer: C</p>
Correct Answer: C