# commutes with a nilpotent matrix and invertible

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.

• 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? Nov 30, 2020 at 19:16

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.

• 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? Dec 1, 2020 at 1:53
• 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 $(-1)^m$. Dec 1, 2020 at 4:40
• 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! Oct 3, 2022 at 20:35

Hint: Note that $$N$$ and $$T$$ are simultaneously upper triangularizable.

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 $$\det(xI-T-N)=\det(xI-T)$$.

• 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 Jan 24 at 1:11
• @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. Jan 24 at 7:51
• 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. Jan 24 at 19:06