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There are complex matrices $A$, $B$ such that $AB - BA = B$. Prove that $0$ is the only eigenvalue of $B$.

I know that $tr(B)=0$. I tried different transformations and I ran out of ideas. Can you please help me?

Maxim Nikitin
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2 Answers2

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Suppose $ \mathrm{sp}\left(B\right)\neq\left\lbrace 0\right\rbrace\cdot $ Since $ B\neq O_{n} $, then $ \left(\forall k\in\mathbb{N}\right),\ B^{k}\neq O_{n} \cdot $ (Or else $ B $ would be nilpotent and would have $ 0 $ as the only eigenvalue)

Define, an endomorphism $ \varphi_{A}:\mathscr{M}_{n}\left(\mathbb{C}\right)\rightarrow\mathscr{M}_{n}\left(\mathbb{C}\right),\ X\mapsto AX-XA $, then we have, for any $ k\in\mathbb{N}^{*} $, the following : \begin{aligned}\varphi_{A}\left(B^{k}\right)=AB^{k}-B^{k}A=\sum_{i=0}^{k-1}{\left(B^{i}A B^{k-i}-B^{i+1}AB^{k-1-i}\right)}&=\sum_{i=0}^{k-1}{B^{i}\left(AB-BA\right)B^{k-1-i}}\\&=\sum_{i=0}^{k-1}{B^{i}BB^{k-1-i}}\\ \varphi_{A}\left(B^{k}\right) &=kB^{k}\end{aligned}

Thus $ \left\lbrace 1,2,3,\cdots\right\rbrace\subset\mathrm{sp}\left(\varphi_{A}\right) $, which is impossible since $ \dim\left(\mathscr{M}_{n}\left(\mathbb{C}\right)\right)<+\infty $, and $ \left|\left\lbrace 1,2,3,\cdots\right\rbrace\right|=+\infty \cdot $

Thus, $$ \mathrm{sp}\left(B\right)=\left\lbrace 0\right\rbrace $$

CHAMSI
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Let's pretend that $B^{-1}$ exists. So we can have

$$A B B^{-1}-B A B^{-1}=B B^{-1}$$

$$A - B A B^{-1}= I$$

$$tr(A - B A B^{-1})= tr(I)$$

$$tr(A) - tr(B A B^{-1})= tr(I)$$ And $ tr(B A B^{-1}) = tr( A B^{-1} B) = tr(A)$

$$tr(A) - tr(A)= tr(I)$$

$$ 0 \neq tr(I) = n$$

It means that $B^{-1}$ cannot exist --> $B$ is a singular matrix($det(B)=0$) --> one eigenvalue is $0$

Capsn
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