# [高等代数] 一道高代题

 证明：如果数域$K$上的$2$级矩阵$A$满足$AB-BA=A$,那么$A^2=0.$

 Without loss, we may consider the problem in the complex field $\mathbb{C}$. Since the trace of $A$ is zero, we need only to check that $A$ has $0$ as an eigenvalue (and so all eigenvalues of $A$ are zero). Or equivalently, we have to find some nonzero vector $x$ such that $Ax=0$. Suppose that $x\ne0$ and $Bx=kx$. Then by $AB-BA=A$, we deduce that $(B-(k-1)E)Ax=0$. If $Ax=0$, then we are done. Suppose that $Ax\ne0$, then $k-1$ is also an eigenvalue of $B$ with $Ax=0$ as its eigenvector. Since $B$ is $2\times 2$, $k-2$ is not its eigenvalue. We must have $A(Ax)=0$, as required.

 It is easy to see that $A^2B-BA^2=2A^2$. This along with $AB-BA=A$ give that ${\rm tr}A={\rm tr}A^2=0$. Thus if $a,b$ are eigenvalues of $A$, then $a+b=0$ and $a^2+b^2=0$. It follows that $a=b=0$, which shows that $A^2=0$.

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