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Suppose that $A,B\in M_n(\mathbb{C})$ such that $AB-BA=A $. Prove that $A$ is not invertible.

My work:

Suppose $A$ is invertible. Then $ABA^{-1}=I+B$ . So $B$ is similar to $I+B$ .Let $B$ have eigenvalues $c_1,c_2,\ldots,c_n \in \mathbb{C}$. So $B$ has basis such that $B$ is upper triangular with respect to it and has $c_1,\ldots,c_n$ as diagonal entries . It is easy to see that $I+B$ is upper-triangular with respect to this basis and has entries $1+c_1,\ldots, 1+c_n$ .

Hence $$c_1+c_2+\ldots +c_n=\operatorname{trace}(B)$$ $$=\operatorname{trace}(I+B)=(1+c_1)+\ldots+(1+c_n)=n+c_1+\ldots +c_n .$$

So $n=0$, contradiction. I'm not sure if my solution is correct one. It seems alright. I will be very thankful if you can confirm that the proof indeed is a correct one.

Any other possible solutions are welcomed.

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    $\begingroup$ It is correct, but too complicated. After proving that $B$ and $I+B$ are similar, you can simply say that this is no possible, since the trace of $I+B$ is $n$ plus the trace of $B$. No need to mention upper triangular matrices. $\endgroup$ – José Carlos Santos May 28 '17 at 16:38
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Let $A$ be invertible.

We have $(AB-BA)A^{-1}=I$, or $ABA^{-1}-B=I.$

But $\operatorname{tr} (ABA^{-1})=\operatorname{tr}{B},$ which is a contradiction.

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It is probably worth showing that $A$ is in fact nilpotent. Indeed, by induction we get $$A^m B - B A^m = m A^m$$ for all $m\ge 0$. Taking the traces on both sides we get $$0=m \operatorname{Trace}A^m$$ so (assuming char $0$) $$\operatorname{Trace}A^m = 0$$ for all $m\ge 1$. This implies $A$ nilponent.

$\bf{Added:}$ Based on an idea of @Hans: , we can generalize this . For $X$, $Y$ matrices, let $[X,Y]=X Y - Y X$.

Assume that $[C,B]=A$, and $[C,A]=0 $ ( $C$, $A$ commute). Then $A$ nilponent.

Indeed, for all $m\ge 1$ we have $[C,A^{m-1} B] = A^{m-1} [C,B]= A^m$, and so $\operatorname{Trace} A^m=0$ . Conclude $A$ nilpotent (assume char $0$).

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    $\begingroup$ Nice conclusion. +1. You only need to show $A^mB-A^{m-1}BA=A^m\implies \text{tr}A^m=0, \forall m\in\mathbf N$. $\endgroup$ – Hans Oct 8 '17 at 7:34
  • $\begingroup$ @Hans; yes, your approach is simpler and better! I should add this idea. $\endgroup$ – orangeskid Oct 8 '17 at 17:46
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$\def\m{\mathfrak }$This result (and, in fact, orangeskid's observation) has the following far reaching generalization:

if $\m g$ is a finite dimensional (complex) Lie algebra and $\mathfrak r$ is its radical, then $[\m g,\m r]$ acts on any finite dimensional representation of $\m g$ nilpotently.

Indeed, in the question the matrices $A$ and $B$ span a Lie algebra $\m g$ of dimension (at most, really, but let us suppose) equal to $2$ which is solvable, so that the radical of $\m g$ is simply $\m g$ itself, and the theorem above tells us that $[\m g,\m g]$ acts nilpotently on finite dimensinal modules: since $[\m g,\m g]$ is spanned by $A$, the desired result follows.

Of course, this is immensely more general.

As an example:

if $A$, $B$ and $C$ are square matrices such that $CA-AC = B$, $CB-BC = aA + B$, $AB-BA = 0$, then $B$ is nilpotent and, more generally, al linear combinations of $A$ and $B$.

This comes from looking at a random 3-dimensional solvable Lie algebra.

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  • $\begingroup$ One can find the proof of this, which depends on Levi's theorem, in the book on Lie algebra by Hossein Abbaspour and Martin A. Moskowitz; it is Theorem 3.5.6 there. $\endgroup$ – Mariano Suárez-Álvarez Oct 12 '17 at 17:41
  • $\begingroup$ Excellent answer! I will have to review my rusty rep theory... $\endgroup$ – orangeskid Oct 12 '17 at 19:16

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