Solution VS.T30 Robert Beezer

We have, \begin{align*} \zerovector &=\vect{x}-\vect{x}&&\knowl{./knowls/property.AI.knowl}{\text{Property AI}}\\ &=\alpha\vect{x}-\vect{x}&&\text{Hypothesis}\\ &=\alpha\vect{x}-1\vect{x}&&\knowl{./knowls/property.O.knowl}{\text{Property O}}\\ &=(\alpha-1)\vect{x}&&\knowl{./knowls/property.DSA.knowl}{\text{Property DSA}} \end{align*} So by Theorem SMEZV we conclude that $\alpha-1=0$ or $\vect{x}=\zerovector$. However, since our hypothesis was for every $\vect{x}\in V$, we are left with the first possibility and $\alpha=1$.

There is one flaw in the proof above, and as stated, the problem is not correct either. Can you spot the flaw and as a result correct the problem statement? (Hint: Example VSS).