# Let $B$ be a nilpotent $n\times n$ matrix with complex entries let $A = B-I$ then find $\det(A)$

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?

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Since $B$ is nilpotent, its trace is 0, so the trace of $A$ is $-n$... – SL2 May 10 '12 at 22:03
@rschweb i apologize for that. – srijan May 10 '12 at 22:07
@rschwieb, Since the question seems to have been fixed, you can remove that comment. – Mariano Suárez-Alvarez May 11 '12 at 0:10

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 $\det A = (-1)^n$.

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$A=B-I$ implies $B=A+I$ and $B$ is nilpotent implies $B^n=0$

i.e., $(A+I)^n=0$ i.e., $A^n+\binom{n}{1}A^{n-1}+\cdots+\binom{n}{n-1}A+I=0$

This is characteristic polynomial for $A$

Sum of eigen values is same as sum of roots of corresponding polynomial which is $(-1)^n$ times that of constant term in this case it is $(-1)^n.1=(-1)^n$

I have used $(x-a_1)(x-a_2)(x-a_3)\cdots(x-a_n)=x^n+\cdots+(-1)^n(a_1a_2a_3\cdots a_n)$

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