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Let $A$ be a $n \times n$ complex matrix. Show (1) is equivalent to (2)

(1) There exists $B$ such that $AB-BA = A$

(2) $A^n=0$

Furthermore, prove $B^n\neq 0$ in (1) if $A \neq 0$.

Attmept for (1) $\implies$ (2)

$A^2 = A^2B - ABA = ABA-BA^2$. Therefore $A^2 = \frac12 (A^2B - BA^2)$. Inductively $A^n = \frac 1 n (A^nB- BA^n)$. Take submultiplicative norms, we obtain $\| A^n \|\leq \frac 2n \|A^n \| \|B\| $. But here I don't know how to preceed.

Rodrigo de Azevedo
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Kim
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    Hint for (1) implies (2): Notice that $\text{tr}(A^k) = 0$ for all $k$ and conclude that $A$ is nilpotent – Mike Daas Jul 13 '20 at 13:25
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    Almost there. What if $|A^n|\ne0$, divide in the inequality, and take $n\to\infty$? – Chrystomath Jul 13 '20 at 13:44
  • @Chrystomath oh now I get it – Kim Jul 13 '20 at 13:52
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    A rather generic hint for $(2)\implies (1)$: Consider the case $n=2$ when we also find vectors $x_2,x_1$ such that $Ax_2=x_1$, $Ax_1=0$. What if $B$ is the linear transformation defined by $x_1\mapsto x_1$, $x_2\mapsto 2 x_2$? To generalize you may need longer "chains" of vectors $\cdots x_3\mapsto x_2\mapsto x_1\mapsto 0$, and possibly several chains so that together they form a basis. If you hace ever studied quantum mechanics, you may have heard of ladder operators. $A$ can be viewed as such. – Jyrki Lahtonen Jul 13 '20 at 14:02
  • Cf this older post which confirms your attempt & offers Lie algebra background: https://math.stackexchange.com/a/3062183/316749 – Hanno Jul 13 '20 at 16:23

2 Answers2

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Let $ k $ be a positive integer, Given $ A,B\in\mathscr{M}_{n}\left(\mathbb{C}\right) $ such that $ AB-BA=A $, we have for any $ i \in\mathbb{N} :$

$$ A^{k}=A^{k-1-i}\left(AB-BA\right)A^{i}=A^{k-i}BA^{i}-A^{k-1-i}BA^{i+1} $$

Thus : $$ \sum_{i=0}^{k-1}{A^{k}}=\sum_{i=0}^{k-1}{\left(A^{k-i}BA^{i}-A^{k-1-i}BA^{i+1}\right)} $$

Hence, for any $ k\in\mathbb{N} $ : $$ kA^{k}=A^{k}B-BA^{k} $$

Define the following endomorphism : $ \varphi_{B}: \mathscr{M}_{n}\left(\mathbb{C}\right)\rightarrow\mathscr{M}_{n}\left(\mathbb{C}\right),\ X\mapsto XB-BX \cdot $

Suppose $ A $ isn't nilpotent, then $ \mathbb{N}\subset\mathrm{sp}\left(\varphi_{B}\right) $, because every $ k\in\mathbb{N} $ is an eigenvalue of $ \varphi_{B} $ and $ A^{k}\neq O_{n} $ is the associated eigenvector. But that can't be because we're working on $ \mathscr{M}_{n}\left(\mathbb{C}\right) $ and $ \mathrm{dim}\left(\mathscr{M}_{n}\left(\mathbb{C}\right)\right)<+\infty $ means $ \mathrm{sp}\left(\varphi_{B}\right) $ must be finite.

Thus $ A $ is nilpotent.

CHAMSI
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[(1) implies (2)]

As Mike Daas said in comment, $\operatorname{tr}(A^k)=0$ for all $k \geq 1$. Let $\lambda_ 1, \dots, \lambda_n$ be eigenvalues of $A$. After triangulization, you can see that $\operatorname{tr}(A^k)=\lambda_1^k + \dots + \lambda_n^k=0$, whence Newton's identity yields $\lambda_i = 0$ for all $i$.

Alternatively, we may assume that $A$ and $B$ are linearly independent and apply Lie's theorem for solvable Lie algebra $\mathbb C A \oplus \mathbb C B \subset gl_n(\mathbb C)$. As a result, $A$ and $B$ are simultaneously triangulizable. But then $A=AB-BA=[A, B]$ can be represented as a strictly (upper) triangular matrix. It follows that $A$ is nilpotent.

[(2) implies (1)]

Write $A$ in a Jordan canonical form and use the following identities to construct $B$. Here $e_{ij}$ is a matrix with $(i, j)$-entry $1$ and otherwise $0$.$$[e_{ij}, e_{kl}]=\delta_{jk}e_{il} - \delta_{il}e_{kj}$$

In particular, if $i<j$ we have $[e_{ij}, e_{jj}]=e_{ij}$, and $[e_{ij}, e_{kk}]=0$ holds if $i<j<k$.

[(1) and $A \neq 0$ implies that $B$ is not nilpotent]

Let $\varphi_{B}: Mat_n({\mathbb C}) \rightarrow Mat_n({\mathbb C}) $ as in the CHAMSI's answer. That is, $\varphi_{B}(X)=XB-BX$. Put $\varphi_{B}^1 = \varphi_{B}$ and $\varphi_{B}^{n+1}=\varphi_{B} \circ \varphi_{B}^{n}$. Then you can verify that $$A=\varphi_{B}(A)=\varphi_{B}^{k}(A)=\sum_{i=1}^{k} (-1)^{i} \binom{k}{i}B^i A B^{k-i} $$ for all $k$. If $B$ is nilpotent, letting $k$ large yields $A=0$; contradiction.

luxerhia
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