I've found this proof of the independence of eigenspaces in a textbook and I can't reproduce it myself.

**Theorem**. Let $\varphi: L \rightarrow L$ be a linear operator on a linear space L. If numbers ${\lambda}\_{1},...,{\lambda}\_{k}$ are different, then the subspaces $\ker(\varphi - {\lambda}\_{i}\varepsilon)$ are independent. 

**Proof**.

1. Suppose that nonzero vectors ${x}\_{i} \in \ker(\varphi - {\lambda}\_{i}\varepsilon)$ satisfy ${x}\_{1} + ... + {x}\_{k}=0$.
2. Let us apply the operators $(\varphi - {\lambda}\_{i}\varepsilon)$ to the identity above for $i = 2, 3,...,k$.
3. Then we get $({\lambda}\_{1}-{\lambda}\_{2})\cdot({\lambda}\_{1}-{\lambda}\_{3})\cdot...\cdot({\lambda}\_{1}-{\lambda}\_{k}){x}\_{1}$
4. Since the numbers ${\lambda}\_{i}$ are different from the identity above we get ${x}\_{1} = 0$. That is a contradiction, hence the subspaces are independent.

Well, first of all I don't quite get what the second step means. Does it mean a composition of all these operators is applied to the equation, i.e. $(\varphi - {\lambda}\_{k}\varepsilon)\circ...\circ(\varphi - {\lambda}\_{2}\varepsilon)({x}\_{1}+...+{x}\_{k})=0$?
If so, how do we get ${\lambda}\_{1}$ in the third step if we don't apply $(\varphi - {\lambda}\_{1}\varepsilon)$? Thank you in advance.