Proofs in matrices

Question

Given that A² = I, prove that at least one of (A-I) and (A+I) is singular.

I don't know how to do proofs but i'd like to see how to approach these kinds of questions thank you!

Answer 1

Theorem. Let A be a real n×n matrix, and let I be the nxn identity matrix. If A² = I, then at least one of A + I, A −− I is singular.

Proof. We can write A² = I as A² − I = 0, the n×n zero matrix. Since A and I commute for all A, we can factor this as (A + I)(A − I) = 0. Taking the determinant of both sides, and recalling that the determinant of a product is the product of the determinants, we have det(A + I)det(A − I) = 0. Since both determinants are real numbers (and the real numbers are a field), they satisfy the property that if the product is zero, then at least one factor is zero. Thus, either det(A + I) = 0 or det(A − I) = 0. Now, a square matrix is invertible (nonsingular) if and only if its determinant is nonzero. Thus, at least one of A + I, A − I is singular, and the proof is complete.