Let $V$ be a finitely-generated inner product space and let $\alpha,\beta_1,\beta_2 \in \text{End}(V )$ satisfy $\alpha^* \alpha \beta_1 =\alpha^* \alpha \beta_2$ . Show that $\alpha \beta_1 = \alpha \beta_2$.
$\alpha^*$ is defined as the adjoint transformation of $\alpha$.
Any help/suggestions are greatly appreciated.
Suppose $x \in V$. Then our hypothesis implies that $\alpha^* \alpha (\beta_1 - \beta_2) x = 0$. Hence \begin{align*} 0 &= \langle \alpha^* \alpha (\beta_1 - \beta_2) x, \ (\beta_1 - \beta_2) x \rangle \\ &= \langle \alpha (\beta_1 - \beta_2) x, \ \alpha(\beta_1 - \beta_2) x \rangle. \end{align*} Thus $\alpha (\beta_1 - \beta_2) x = 0$. Because $x$ was an arbitrary element of $V$, this implies that $\alpha \beta_1 = \alpha \beta_2$, as desired.