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Solving for Invertibility of (B+I) Given B=B^2
If det(B) = 1, then B-1B = B-1B2, B = I so det(B+I) = det(2I) != 0, so B+I is invertible. I'm still stuck on if det(B) = 0.. I'm quite sure that if B = B2, then B must be a diagonal matrix with entries being either 1 or 0, but I don't know how to prove it.- transcendency
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- Forum: Calculus and Beyond Homework Help
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Solving for Invertibility of (B+I) Given B=B^2
Homework Statement Given a matrix B, if B = B2, is (B+I) invertible? 2. The attempt at a solution det(B) = 0 or 1 rref(rref(B) + I) is I, so (rref(B) + I) is invertible if det(B) = 1: let E1E2...En = B then E1E2...En(rref(B) + I) = B + E1E2...En I'm not sure if what I did is...- transcendency
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- Forum: Calculus and Beyond Homework Help