Suppose we have a cartan subalgebra (CSA) $\mathfrak{h} \subset \mathfrak{g}$. Assign a basis to this $\{H_{i}\}_{i=1...r}$ such that $ ad_{H_{i}}$ are all simultaneously diagonalized [ $\implies ad_{H}$ diagonal $\forall H \in \mathfrak{h}$] (which we know we can do from linear algebra). Want to find set of eigenvectors
Why is it true that the only eigenvectors with eigenvalue $0$ are the basis elements $H_{i}$ of $\mathfrak{h}$?Why is it true that the only eigenvectors with eigenvalue $0$ are the elements of $\mathfrak{h}$ only?
Attempt at Solution:
If $Y \in \mathfrak{g}$ is an eigenvector with eigenvalue $\lambda =0 $ then: $$ ad_{H_{i}}(Y) = [H_{i},Y] = 0 \; \forall i \implies ad_{Y}(H) =0 $$
If we can also show that $ad_{Y}$ is diagonalizable, then it follows that $Y \in \mathfrak{h}$, but why is this true? Note that we can't assume the Cartan-Weyl basis, since we need this property for the basis to be valid?
EDIT: Definition of CSA:
For $\mathfrak{h} \subset \mathfrak{g}$ to be a CSA, the following must be true: $$ H \in \mathfrak{h} \implies ad_{H} \; \text{is diagonalizable} $$ $$ H,H' \in \mathfrak{h} \implies [H,H'] = 0 $$ $$ \text{if} \; X \in \mathfrak{g} \; \text{is ad diagonalizable and } [X,H] =0 \; \forall H \in \mathfrak{h} \implies X \in \mathfrak{h} $$
Your statement is not true, because every element of $\mathfrak{h}$, i.e. not only the basis elements $H_i$ but every linear combination of those, is an eigenvector of eigenvalue $0$ (to all the $ad_{H_i}$ simultaneously).
Now, to show that these are indeed the only elements of $\mathfrak{g}$ which are eigenvectors to the eigenvalue $0$ to all $ad_{H_i}$, just notice that from your first attempt it follows that for any such eigenvector $Y$, we have $[\mathfrak{h}, Y] = 0$ or in other words, $Y$ is contained in the centraliser and hence in the normaliser of $\mathfrak{h}$. But by definition of CSA, a CSA $\mathfrak{h}$ equals its own normaliser. Hence $Y\in \mathfrak{h}$.