I've spent 6 hours, still can't get it.

[hola]

B is a n*n matrix

1. Let B^2 =B. Prove that either det(B) =1 or B is singular.
2. If Transpose(B) = B^-1 , what is det(B)?

 

[StatusX]

If b^2=b, then det(b^2)=det(b)^2=det(b), which has two solutions for det(b). For the second part, since a singular matrix doesn't have an inverse, you are forced into the only allowed non-zero determinant.

 

[thepatientmental]

I understand that you have spent 6 hours trying to solve this problem, and it can be frustrating when things don't come easily. Don't worry, sometimes it takes time and effort to fully understand and solve a problem. Keep persevering and don't give up.

1. To prove that either det(B) = 1 or B is singular, we can use the fact that if B^2 = B, then B is an idempotent matrix. This means that B^2 - B = 0. We can then use the determinant property that det(AB) = det(A)det(B) to get det(B^2 - B) = det(0), which equals 0. Therefore, det(B^2 - B) = 0 and we can factor out det(B) to get det(B)(det(B) - 1) = 0. This means that either det(B) = 0 or det(B) = 1. If det(B) = 0, then B is singular. If det(B) = 1, then we have proven the statement.

2. If transpose(B) = B^-1, then B is an orthogonal matrix. This means that B^T = B^-1 and det(B^T) = det(B^-1) = 1/det(B). From this, we can see that det(B) = 1 or -1. However, since B is an orthogonal matrix, it must have real eigenvalues. This means that det(B) cannot equal -1, so we are left with det(B) = 1.