Determinant of a nilpotent matrix Let $A$ be a nilpotent matrix. Prove that $\det(I+A)=1$
Could someone at least give me a clue ?
 A: Since $A$ is nilpotent, we have
$A^m = 0 \tag{1}$
for some positive interger $m$.  This implies every eigenvalue of $A$ vanishes, since the equation
$Av = \lambda v \tag{2}$
for non-zero $v$ (recall eigenvectors are required to be non-zero) implies
$0 = A^mv = \lambda^m v, \tag{3}$
whence
$\lambda^m = 0, \tag{4}$
since $v \ne 0$.  (4) forces 
$\lambda = 0 \tag{5}$
Now use the fact that for any scalars $\lambda$ and $a$, $\lambda$ is an eigenvalue of $A$ if and only if $\lambda + a$ is an eigenvalue of $A + aI$; indeed we have, from (2),
$(A + aI)v = Av + av = (\lambda + a)v. \tag{6}$
(6) allows us to conclude that every eigenvalue of $A + I$ is $1$; hence $\det (A+I)$, being the product of its eigenvalues, satisfies
$\det(A+I) = 1. \tag{7}$
QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: If your matrices are over algebraically closed field, consider Jordan canonical form $J$ of $A$. It will have only zeros on the diagonal, since nilpotent matrix has only zeroes as eigenvalues. Thus, $A = S J S^{-1}$ where $J$ is upper triangular and has zeroes on the diagonal. Now, $I + A = S I S^{-1} + S J S^{-1} = S(I + J)S^{-1}$, and so $det(I+A) = det(S)det(I+J)det(S^{-1}) = det(I+J)$, but $I+J$ is upper triangular and has only $1$-s on the diagonal, so $det(I+J) = 1$.
If our base ring is an arbitrary domain (for instance a non algebraically closed field), we can embed this ring into an algebraically closed field, and repeat the argument above. 
If our ring is not necessarily domain, I don't know how to prove it, or whether it's true at all.
A: We have that:
 $A^m =0$.
 Then: 
\begin{equation}
(I+A)A^{m-1}=A^{m-1}+A^{m}\\
\therefore (I+A)A^{m-1}=A^{m-1}
\end{equation}
Now, comptuting:
\begin{equation}
det(I+A)det (A^{m-1})=det(A^{m-1}); \quad det(A^{m-1})\neq 0\\
\therefore det(I+A)=1
\end{equation}
