Nilpotent Matrix – Finding Determinant of A = B-I

Question

Let $B$ be a given nilpotent $n\times n$ matrix with complex entries. Let $A = B-I$ find out $\det(A)$. What if B is orthogonal or skew symmetric matrix? Then can we say anything about its trace and determinant?

Answer 1

If $\lambda$ is an eigenvalue of $B$, then $\lambda-1$ will be an eigenvalue of $B-I$. Hence $\det A = \prod (\lambda_i-1)$. Since $B$ is nilpotent, all eigenvalues are $0$. Hence $\det A = (-1)^n$.