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?

  • $\begingroup$ Hint: Every nilpotent matrix is conjugate to a strictly upper-triangular matrix. $\endgroup$ – 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 $\det(B-I)$?

  • $\begingroup$ 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? $\endgroup$ – Abishanka Saha Nov 18 '17 at 7:11
  • $\begingroup$ @AbishankaSaha exactly! $\endgroup$ – Lord Shark the Unknown Nov 18 '17 at 7:12

If you have to ask whether your proof is correct, then your proof is not complete. What happens at the 2nd "equals sign" of your centered expression seems to require an explanation (to me). It is a good habit to create "simple concrete examples" to help double-check your work, like 2-by-2 matrices in this case (to corroborate your results).


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