$AB + BA = A \implies A$ and $B$ have a common eigenvector

Solution 1:

Observe that $$AB -\frac{1}{2}A +BA-\frac{1}{2}A=0$$ $$A(B -\frac{1}{2} I) =-(B -\frac{1}{2} I)A$$ Therefore matrixs $A$ and $C=B -\frac{1}{2} I$ satisfies identity $$AC=-CA.$$ Suposse that $\lambda$ is an eingevalue of $A$ and $x_{\lambda}$ is an suitable eingevector. Then $$CAx_{\lambda} =\lambda Cx_{\lambda}=-ACx_{\lambda}$$ hence $-Cx_{\lambda}$ is an eingevector of $A$ correspondent to eingenvalue $\lambda.$ If we assume that one of eingenspaces of $A$ is one dimensional then $-Cx_{\lambda}$ mus be a multiple of $x_{\lambda}$ hence $$-Cx_{\lambda}=\mu x_{\lambda}$$ $$(\frac{1}{2}I -B)x_{\lambda}=\mu x_{\lambda}$$ $$Bx_{\lambda}=\left(\frac{1}{2} -\mu \right) x_{\lambda}$$ Thus $A$ and $B$ have a common eingevector $x_{\lambda}.$