# Determinant of $A=B-I$ where $B$ is nilpotent complex matrix

Let $B$ be a nilpotent $n\times n$ matrix with complex entries. Set $A=B-I$. I need to find the determinant of $A$.

Let $m$ be the least possible integer such that $B^m=0$. Then $$\det(A)=[\det(A^{-1})]^{-1}=[-\det(I+B+B^2+\cdots+B^{m-1})]^{-1}$$ But as $B$ is nilpotent I should be able to claim that $B$ is singular, because otherwise $B^m$ would have rank $n$, contrary to our assumption. But then $\det(B)=\det(B^k)=0$ for any positive integer $k$. Therefore $\det(A)=[-det(I)]^{-1}=-1$.

Is the proof correct?

• Hint: Every nilpotent matrix is conjugate to a strictly upper-triangular matrix. – darij grinberg Nov 18 '17 at 5:56

No. For instance $B=\pmatrix{0&0\\0&0}$ is nilpotent, but $B-I=\pmatrix{-1&0\\0&-1}$ has determinant $1$. In general all the eigenvalues of a nilpotent matrix $B$ are zero, so all eigenvalues of $B-I$ are $-1$. Now can you see how to find $\det(B-I)$?
• yeah, i get it. Then $\det(A)=\prod(\lambda_i-1)$ where $\lambda_i$ being the eigenvalue of a nilpotent matrix $B$, must be $0$. So $\det(A)=(-1)^n$, right? – Abishanka Saha Nov 18 '17 at 7:11