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The exercise is:

Given $V$ a vector space, $\dim V<\infty$ and $T, N\in \mathcal{L}(V)$, where $N$ is nilpotent and $TN=NT$. Prove that: $T$ is invertible iff $T+N$ is invertible. Furthermore, $\det(T)=\det(T+N)$ and $p_T(t)=p_{T+N}(t)$.

I need a hint about how can I start this kind of exercise, which involves a nilpotent matrix.

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    $\begingroup$ Try an example, $$ N = \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right) $$ along with $$ T = \left( \begin{array}{cc} a & b \\ c & d \end{array} \right). $$ In order to have $TN = NT,$ what are the conditions on $a,b,c,d?$ Then, what happens tothe requested determinants, and, I am guessing, the characteristic polynomials? $\endgroup$
    – Will Jagy
    Nov 30, 2020 at 19:16

3 Answers 3

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For a rather different approach one can directly prove $p_T(t)=p_{T+N}(t)$ which then gives you the equality of determinants and in turn the equivalence of invertibility. I assume $\dim V = m$ and assume the field has characterisitic zero (or if one wants to contemplate positive characteristic fields, then $\text{char}(\mathbb F)\gt m$)

lemma 1:
by nilpotence, $N^m = \mathbf 0$ and with commutativity this means you can apply the binomial theorem, so $\big(T+N\big)^j= T^j +\Big(\sum_{k=1}^{j-1} \binom{j}{k}T^kN^{j-k}\Big) + N^j$

lemma 2:
$T^k N^r$ for $r\geq 1$ is nilpotent, because
$(T^k N^r)^m = T^{km} N^{rm}= T^{km}\mathbf 0 = \mathbf 0$
and recall that since nilpotent matrices are similar to strictly upper triangular matrices, they necessarily have trace zero

main argument:
$\text{trace}\Big(T+N\Big)= \text{trace}\Big(T\Big)+\text{trace}\Big(N\Big)= \text{trace}\Big(T\big) + 0 $

and for $j\in\big\{1,2,...,m\big\}$
$\text{trace}\Big(\big(T+N\big)^j\Big)= \text{trace}\Big(T^j\Big) + \sum_{k=0}^{j-1} \binom{j}{k}\cdot\text{trace}\Big(T^kN^{j-k}\Big)= \text{trace}\Big(T^j\Big) + 0$

Thus by application of Newton's Identities $\Big(T+N\Big)$ and $\Big(T\Big)$ have the same characteristic polynomial.

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  • $\begingroup$ Thanks! I agreed! Can you help me with the other itemize? I proved the first implication: det(T+N)=det(T+T−1TN)=det(T)det(Id+T−1N)≠0, because 1 can not be a eigenvalue of T−1N who is nilpotent. How can i do the other implication? $\endgroup$ Dec 1, 2020 at 1:53
  • $\begingroup$ the characteristic polynomials agree, so evaluate them at zero and get $(-1)^m \det\Big(T\Big) = \det\Big(0I - \big(T\big)\Big) =p_T(0) = p_{T+N}(0) = \det\Big(0I - \big(T+N\big)\Big) = (-1)^m \det\Big(T+N\Big)$. Now multiply each side by $(-1)^m$. $\endgroup$ Dec 1, 2020 at 4:40
  • $\begingroup$ What are the Newton's identities and how are you using that to conclude that $\Big(T+N\Big)$ and $\Big(T\Big)$ have the same characteristic polynomial? Thanks! $\endgroup$
    – user402543
    Oct 3, 2022 at 20:35
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Hint: Note that $N$ and $T$ are simultaneously upper triangularizable.

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It suffices to prove that $T$ and $T+N$ have the same characteristic polynomial. The other two results follow immediately.

I shall give a proof that does not involve triangulation of $T$, because an attendee of a first course on linear algebra usually has not seen any proof of the existence of algebraic closure of an arbitrary field from any course.

Let $F$ be the underlying field and $x$ be an indeterminate. Denote by $F(x)$ be the field of fractions of $F[x]$. For ease of presentation we assume that $T$ and $N$ are matrices over $F$.

Since $\det(xI-T)$ is a monic polynomial in $x$, it is nonzero. Therefore $xI-T$ is invertible over $F(x)$. Then $(xI-T)^{-1}$ commutes with $N$. Hence $N_1=(xI-T)^{-1}N$ is nilpotent and it can be triangularised over $F(x)$ (we do not need $F(x)$ to be algebraically closed here because $0\in F(x)$). Therefore $\det(I-N_1)=1$ and in turn $\det(xI-T-N)=\det(xI-T)$.

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  • $\begingroup$ out of curiosity, can you adapt this technique to solve math.stackexchange.com/questions/4849941/… ? $\ker N$ being a $T$-invariant subspace yields an identical proof to each but I have a lingering curiosity about porting proofs from here to there $\endgroup$ Jan 24 at 1:11
  • $\begingroup$ @user8675309 Done, but I think your invariant subspace argument provides a better explanation. Before reading your answer to the linked question, I wasn’t aware that “$N$ is nilpotent and $NA=0$” implies that $N$ and $A$ are simultaneously triangularisable. $\endgroup$
    – user1551
    Jan 24 at 7:51
  • $\begingroup$ Thanks. I share your concern of people not knowing about algebraic closure, so I went for simultaneous block triangulation. It is interesting that we can port a variety of techniques from this problem to that one. $\endgroup$ Jan 24 at 19:06

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