Show that the sum of spectrum is direct

Question

Let $T\in End(V)$. For each $\lambda\in F$ let $V_\lambda=Ker(T-\lambda)$ be the corresponding eigenspace. Let $Spec_F(T)=\{\lambda\in F|V_\lambda\ne\{0\}\}$ be the set of eigenvalues of $T$. Show that the sum $\sum_{\lambda\in Spec_F(T)}V_\lambda$ is direct.

My answer:

To prove above sum is direct we need to show the representation is unique. Suppose it is not i.e. $\sum_{\lambda\in A}v_\lambda=\sum_{\lambda\in B}v_\lambda$ while $A\ne B$. It can be further reduced to $\sum_{\lambda\in C}v_\lambda=\sum_{\lambda\in D}v_\lambda$ where $C\cap D=\emptyset$. Then $\sum_{\lambda\in C\cup D}T(v_\lambda)=\sum_{\lambda\in C\cup D}\lambda v_\lambda=0$ where $\lambda$ not all zero, contradict to independence of eigenvectors.

Am I correct?

Answer 1

The main idea of your arguemnt is correct but I think that there are two issues with the arguemnt itself.

First, how do you get that $\sum_{\lambda\in C\cup D}\lambda v_\lambda = 0$? I see how to get $\sum_{\lambda C\cup D} v_\lambda = 0$ follows from the identity before but not how you get it with the $\lambda$ in the sum.

Second, if you already know that the eigenvectors are linearly independent why don't you use the identity $\sum_{\lambda C\cup D} v_\lambda = 0$ itself to conclude a contradiction?