Let $A,B$ $n\times n$ matrices such that $$AB-BA=A$$ and $A$ is a nonzero matrix. Prove that $B$ is not nilpotent.
I know why $A$ is nilpotent, but how can I prove $B$ is not nilpotent?
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2Why is $A$ nilpotent? – Ben Grossmann Dec 11 '16 at 22:34
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1@Omnomnomnom Because of this question. – Dietrich Burde Dec 11 '16 at 22:37
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Here's one way to see it: define the linear transformation $$ \Phi_B(X) = XB - BX $$ clearly, $A$ is an eigenvector of this transformation associated with $\lambda = 1$. Now, the eigenvalues of $\Phi_B$ are necessarily of the form $\lambda - \mu$ where $\lambda,\mu$ are eigenvalues of $B$ (can be seen via vectorization).
Conclude that since $\Phi_B$ has a non-zero eigenvalue, $B$ has a non-zero eigenvalue, which means it is not nilpotent.
Ben Grossmann
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Find the eigenvectors of $X\mapsto XB$ and $X\mapsto BX$, then note that the two operators commute. – Ben Grossmann Dec 12 '16 at 00:56
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If $B$ is nilpotent then there is $k$ such that $AB^k=0$.
So $-BAB^{k-1}=AB^k-BAB^{k-1}=(AB-BA)B^{k-1}=AB^{k-1}$.
Hence, $(Id+B)(AB^{k-1})=0$. Since B is nilpotent then $Id+B$ is invertible. Thus, $AB^{k-1}=0$.
We can repeat the argument to prove that $AB^{k-2}=\ldots=AB^1=AB^0=A=0$. Absurd!
Daniel
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