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Let $A$ be a nilpotent matrix. Prove that $\det(I+A)=1$

Could someone at least give me a clue ?

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  • $\begingroup$ A hint is that $1+x$ divides $1+x^m$ if $m$ is odd and positive, and divides $1-x^m$ if $m$ is even and positive. $\endgroup$ – rogerl Oct 29 '13 at 16:33
  • $\begingroup$ @rogerl: I don't quite see how you want to conclude it from this hint. $\endgroup$ – xyzzyz Oct 29 '13 at 16:37
  • $\begingroup$ @rogerl But your hint enables me only to conclude that $\det(I+A)|\det(I+A^n)=1$ .. $\endgroup$ – user43418 Oct 29 '13 at 16:38
  • $\begingroup$ @xyzzyz True enough. $\endgroup$ – rogerl Oct 29 '13 at 16:55
  • $\begingroup$ The characteristic polynomial must be $X^n$. So $A$ is triangularizable with null diagonal. So $I+A$ is trigularizable with diagonal... $\endgroup$ – Julien Oct 29 '13 at 18:02
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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!!!

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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.

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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}

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  • $\begingroup$ Why is that determinant nonzero? (It is not.) $\endgroup$ – darij grinberg Mar 13 '17 at 3:38

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