Show that the span of eigenvalues of a compact operator is a closed Let $(X, ||\cdot||)$ be a real Banach space and $T: X \to X$ a compact operator (so $\{x_n\}_{n=1}^\infty$ bounded implies that $\{Tx_n\}_{n=1}^\infty$ has a convergent subsequence).
Let $x_1, \dots, x_n \in X\setminus \{0\}$ be eigenvectors of $T$, with corresponding nonzero, distinct eigenvalues $\lambda_1, \dots, \lambda_n \in \mathbb{R}$.

Prove that $\operatorname{span}\{x_1, \dots , x_n\}$ is a closed subspace of $X$. 

Some facts that I do know:


*

*For each $\lambda_i$, the map $T-\lambda_i$ has closed range.

*Because the $\lambda_i$ are all distinct, the $x_i$ are pairwise linearly independent. 


Do I have enough with these two facts to prove that $\operatorname{span}\{x_1, \dots , x_n\}$ is closed, or is there something else I am missing?
If we have a sequence $\{\sum_{i=1}^n a_{n,i}x_i\} \subset \operatorname{span}\{x_1, \dots , x_n\}$ converging to some $b \in X$, we ultimately want to show that $b \in \operatorname{span}\{x_1, \dots , x_n\}.$ But I'm unsure how to get the proof started.
Hints or solutions are greatly appreciated. 
 A: Let's prove that every finite-dimensional subspace of a topological vector space is closed. Suppose $M$ is the linear span of linearly independent vectors $x_1,\dots,x_n$, and we have a convergent sequence of the form $v_k = \sum_{i=1}^n c_{ik} x_i$. Consider two cases: 


*

*There is a constant $C$ such that $|c_{ik}|\le C$  for all $i,k$. Then we can choose a sequence of $k$-indices along which $c_{1k}\to c_1$; from it choose a subsequence in which $c_{2k}\to c_2$, and so forth. Conclude that $\lim v_k = \sum_{i=1}^n c_i x_i$, which is an element of $M$.  

*No constant $C$ as above exists. This means we can choose a sequence of $k$-indices along which $\mu_k:=\sum_i |c_{ik}|$ tends to $\infty$. Work with this sequence from now on. Let $b_{ik}=c_{ik}/\mu_k$, which makes $b_{ik}$ bounded. As in part 1, find a subsequence in which $b_{ik}\to b_i$ for $i=1,\dots,n$. Note that $\sum_i |b_i|=1$. Since
$$0 = \lim \frac{v_k}{\mu_k} = \lim \sum_{i=1}^n b_{ik} x_i = \sum_{i=1}^n b_{i} x_i$$ we have a contradiction: the vectors $x_i$ were assumed to be linearly independent.



If you'd like  a shorter proof and have the Hahn-Banach theorem, then proceed differently: find   linear functionals $\phi_i$ such that $\phi_i(x_j)=\delta_{ij}$, and define   $Tx =x-\sum \phi_i (x)x_i$. Observe that $M=\ker T$.
