Problems with understanding a proof of eigenspaces independence 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.


*

*Suppose that nonzero vectors ${x}_{i} \in \ker(\varphi - {\lambda}_{i}\varepsilon)$ satisfy ${x}_{1} + ... + {x}_{k}=0$. 

*Let us apply the operators $(\varphi - {\lambda}_{i}\varepsilon)$ to the identity above for $i = 2, 3,...,k$. 

*Then we get $({\lambda}_{1}-{\lambda}_{2})\cdot({\lambda}_{1}-{\lambda}_{3})\cdot...\cdot({\lambda}_{1}-{\lambda}_{k}){x}_{1}$

*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. 
 A: Yes the second step is what you think. Then to understand step 3) just keep in mind that $x_1$ was assumed to be in $\ker(\varphi - \lambda_1 \epsilon)$ i.e. $\varphi (x_1) = \lambda_1 x_1$.
A: There's something not really precise in point 1, which should be

Suppose that $x_1+x_2+\dots+x_k=0$ where at least one of the vectors $x_1,x_2,\dots,x_k$ is nonzero; without loss of generality we can assume $x_1\ne0$.

Now follow the hint; if we apply $\varphi-\lambda_2\varepsilon$ we get
$$
\varphi(x_1)-\lambda_2x_1+
\varphi(x_2)-\lambda_2x_2+
\varphi(x_3)-\lambda_2x_3+
\dots+
\varphi(x_k)-\lambda_2x_k
=0
$$
that, in view of the properties of the vectors, becomes
$$
\lambda_1x_1-\lambda_2x_1+
\lambda_3x_3-\lambda_2x_3+
\dots+
\lambda_kx_k-\lambda_2x_k
=0
$$
or
$$
(\lambda_1-\lambda_2)x_1+(\lambda_3-\lambda_2)x_3+\dots+
(\lambda_k-\lambda_2)x_k=0
$$
Upon applying $\varphi-\lambda_3\varepsilon$, the term with $x_3$ vanishes and the term with $x_1$ is multiplied by $\lambda_1-\lambda_3$.
Proceeding in the same way, we're indeed left with
$$
(\lambda_1-\lambda_k)\dots(\lambda_2-\lambda_1)x_1=0
$$
and this implies $x_1=0$, which is a contradiction.

I prefer a different kind of proof, by induction on $k$. The result is obvious for $k=1$, so assume we know it for $k-1$ eigenvalues. Suppose
$$
x_1+\dots+x_{k-1}+x_k=0
$$
with $x_i\in\ker(\varphi-\lambda_i\varepsilon)$ for $i=1,2,\dots,k$. Apply $\varphi$ to the relation, getting
$$
\lambda_1x_1+\dots+\lambda_{k-1}x_{k-1}+\lambda_kx_k=0
$$
and also multiply the relation by $\lambda_k$:
$$
\lambda_kx_1+\dots+\lambda_kx_{k-1}+\lambda_kx_k=0
$$
Now subtract getting
$$
(\lambda_1-\lambda_k)x_1+\dots+(\lambda_{k-1}-\lambda_k)x_{k-1}=0
$$
By the induction hypothesis,
$$
(\lambda_1-\lambda_k)x_1=0,\quad
\dots,\quad
(\lambda_{k-1}-\lambda_k)x_{k-1}=0
$$
and since $\lambda_i-\lambda_k\ne0$ for $i=1,2,\dots,k-1$, we get
$$
x_1=0,\quad \dots,\quad x_{k-1}=0
$$
which also implies $x_k=0$.
The proof is direct, as you see.
