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.
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.
$(1) \implies (2)$ which holds for field $\mathbb F$ with the mild restriction that $\text{char }\mathbb F =0$ or $\text{char }\mathbb F \gt n$.
This really is a duplicate and has been addressed many times on this site. E.g. see here for more ideas:
If $A=AB-BA$, is $A$ nilpotent?
re: "Furthermore prove $B^n\neq \mathbf0$ for $A\neq \mathbf 0$"
define $T: M_n\big(\mathbb F\big)\longrightarrow M_n\big(\mathbb F\big)$ given by $T(X) = XB-BX$
Then $T\big(A\big) = A\neq \mathbf 0\implies A$ is an eigenvector of $T$ with eigenvalue one. If $B$ is nilpotent then all eigenvalues of $T$ are zero so this is impossible. (This may be verified most easily by using the Kronecker sum and showing it is similar to a strictly upper triangular matrix, or if preferred we can observe that in general $T$ has eigenvalues of $\lambda_i-\lambda_j$ for $i, j \in \big\{1,2, \dots, n\big\}$, taking eigenvalues from a splitting field for $B$.)
$(2) \implies (1)$, which holds over any field without limitation. No knowledge of Jordan Forms is needed.
The underlying Lemma to be used is $A\mathbf x = \mathbf b$ has no solution iff there is some $\mathbf y^T A = \mathbf 0^T$ but $\mathbf y^T\mathbf b=1$. What this says is the only obstacle to a solution is the obvious one, i.e. if there was both a solution and such a $\mathbf y$ we'd conclude $0=1$. Proof for this lemma has been asked elsewhere on the site; I suggest proving it via use of Rank Normal Form.
We can translate (2) into standard linear equations by recognizing we are trying to solve a Sylvester Equation $AB + B(-A) = AB - BA = A $ which is equivalent to solving $\big(A\oplus (-A)^T\big)\operatorname{vec}\big(B\big) = \operatorname{vec}\big(A\big)$, where $\oplus$ denotes Kronecker Sum and the vec operator is given by $
\operatorname{vec}\big(B\big) =\text{vec}\left(\bigg[\begin{array}{c|c|c|c}
\mathbf b_1 & \mathbf b_2 &\cdots & \mathbf b_{n}
\end{array}\bigg]\right) = \begin{bmatrix}
\mathbf b_1 \\
\mathbf b_2\\
\vdots \\
\mathbf b_n
\end{bmatrix}$
Suppose for contradiction that $\text{vec}\big(A\big)\not\in \text{image}\Big(A\oplus (-A)^T\Big)$. The Lemma tell us there is some $\text{vec}\big(Y\big)$ such that
$1=\text{vec}\big(Y\big)^T\text{vec}\big(A\big)$ but $\mathbf 0^T=\text{vec}\big(Y\big)^T\big(A\oplus (-A)^T\big) \implies 0=\text{vec}\big(Y\big)^T\big(A\oplus (-A)^T\big)\text{vec}\big(X\big)$ for all $X\in M_n\big(\mathbb F\big)$ or equivalently
$1=\text{trace}\big(Y^TA\big ) $ and $0=\text{trace}\Big(Y^T\big(AX -XA\big)\Big)=\text{trace}\Big(\big(Y^TA - AY^T\big)X\Big)$ for all $X\in M_n\big(\mathbb F\big)$
The bilinear form $M_n\big(\mathbb F\big)\times M_n\big(\mathbb F\big)\longrightarrow \mathbb F$ given by the trace of the product of two matrices is non-degenerate and $\text{trace}\Big(\big(Y^TA - AY^T\big)X\Big)=0$ for all $X$ means $\big(Y^TA - AY^T\big)$ is a null vector $\implies Y^TA = AY^T$ by non-degeneracy. But that implies $\big(Y^TA\big)^n = (Y^T)^n A^n = (Y^T)^n \mathbf 0 =\mathbf 0\implies 0=\text{trace}\big(Y^TA\big )=1$ which is a contradiction.
[(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.
Many Answers use trace $tr(A^k)$ and characteristic polynomial to solve this problem, but I think it is not necessary. I hope this post is another view.
First we note on a complex field $B$ must have an eigenvector $u_1$ such that $Bu_1 = \beta u_1$. Then we act $A$ on $u_1$ to get $\beta Au_1-BAu_1 = Au_1$ which is $$B(Au_1) = (\beta-1)Au_1$$ So $B$ has another eigenvalue $\beta-1$ if $Au_1 \neq 0$. Then let $Au_1 = u_2$, we act $A$ another time on $u_2$ and get $ABu_2-BAu_2 = Au_2$ which is $(\beta-1)Au_2-B(Au_2) = Au_2$ which is $$B(Au_2) = (\beta-2)Au_2$$ It means $B$ has another eigenvalue $\beta-2$. But because $\beta-i \neq \beta-k$ if $i \neq k$, then the eigenvectors $\{Au_i\}$ are linearly independent and so makes $B$ infinitely dimensional. It is a contradiction. So all the eigenvectors $u$ of $B$ has the property $A^ku = 0$ for some $k$.
Then if we can prove all the generalized eigenvectors of $B$ are vanished by $A^k$ for some $k$, we can probe all the vectors in $V$ are vanished by $A^k$ for some $k$ and so $A$ is nilpotent because the genralized eigenvectors consist of a basis. First we decomposed $V$ by $B$'s generalized eigenspace, then consider $Bu_2 = au_1+\lambda u_2$ where $u_2$ is in the generalized eigenspace but not in eigenspace. We have known $A^ku_1 = 0$, then act $AB-BA = A$ on $u_2$ we get $$B(Au_2) = (\lambda-1)Au_2+aAu_1$$ Then act $AB-BA = A$ on $Au_2$ we get $$B(A^2u_2) = (\lambda-2)A^2u_2+aA^2u_1$$ Proceeding this process we can prove $A^ku_2$ is the eigenvector of $B$ because the remainder term $aA^ku_1 = 0$, and so it again can be reproduced infinitely if $A^mu_2 \neq 0$. So for all the basis of generalized eigenvectors we can say $A^mu = 0$ for some $m$. And so $A$ is nilpotent.
I do not know if it can be proved without using $B$'s generalized eigenspace, and if we use $A$'s eigenvector we cannot reproduce infinite dimensions to induce contradiction.
I shall give another interesting example which uses the same method of contradiction. Suppose $A$ is invertible and finite, and $AB-BA = B^2A$, we can prove $B$ cannot have a non-zero real eigenvalue. If $\lambda$ is the real eigenvalue of $B$, then $Bu = \lambda u$ for some $u$. But we note that $A$ is invertible and so $Ak = u$ for some $k$, then act $AB-BA = B^2A$ on $k$ we can see $ABk = (\lambda^2+\lambda)Ak$, then because $A$ is invertible we get $$Bk = (\lambda^2+\lambda)k$$ which means there is an eigenvalue $\lambda^2+\lambda$ strictly larger than $\lambda$, implying a distinct, eigenvalue for $B$. And by the same process we can reproduce $(\lambda^2+\lambda)^2+(\lambda^2+\lambda)$ as a distinct eigenvalue for $B$. Then this implies $B$ is infinite dimension and so contradiction. So all the eigenvalues can only be $0$ or complex. Done.
