Let $L$ be a semisimple Lie algebra. I am trying to understand root space decomposition of $L$ on my own. Since $L$ is semisimple, $L$ possesses an abelian maximal toral subalgebra i.e. an abelian subalgebra $H$ which is $\text {ad}$-semisimple, known as Cartan subalgebra. But then $\text {ad} (H)$ is a commuting family of semisimple operators on $L$ and hence they are simultaneously diagonalizable. So there exists a basis $\{e_1, \cdots, e_n \}$ of $L$ and $\lambda_i \in H^{\ast}$ corresponding to each basis element $e_i$ such that $$[h, e_i] = \lambda_i (h) e_i$$ for all $h \in H.$ Define $$L_{\lambda_i} : = \left \{x \in L\ |\ [h, x] = \lambda_i (h) x\ \text {for all}\ h \in H \right \}.$$
Then it's clear that $L = \sum\limits_{i = 1}^{n} L_{\lambda_i}.$ But I can't see why the sum is direct. First of all how can conclude that all the $\lambda_i$'s are distinct? Because if for some $i \neq j$ we have $\lambda_i = \lambda_j$ then clearly $L_{\lambda_i} = L_{\lambda_j}.$ But then the sum won't be direct. So at first we have to somehow show that all $\lambda_i$'s are distinct. If they are distinct they are all one dimensional.
In particular, if the sum is direct then $H$ is also one-dimensional. Is it always the case?
Could anyone please answer the questions? Also please let me know where I am going wrong if there is any.
Thanks for your time.
With your definitions / setup / notations, the $\lambda_i$ are not yet distinct in general: Because if we choose the basis $e_1, ..., e_n$ so that the first $r$ elements $e_1, ... e_r$ are a basis of $H$, which is its own $0$-space a.k.a. centralizer, then $\lambda_1 = ... = \lambda_r = 0 \in H^*$.
The correct way to define the root space decomposition starting from your notation is to define $R:= \{\lambda_i: 1\le i \le n\} \color{red}{\setminus \{0\}}$, and write
$$L = L_0 \oplus \bigoplus_{\alpha \in R} L_{\alpha}$$
Note that I'm kind of cheating here because I have now made the $\alpha$ mutually distinct by definition. From there, the standard way to proceed is to note $H=L_0$ and then show that each of the "proper" root spaces $L_\alpha$ ($\alpha \in R)$ has dimension $1$ -- which is more involved than one might think, cf. Why are root spaces of root decomposition of semisimple Lie algebra 1 dimensional? and When is the Lie bracket of two root spaces nonzero? including links from there, also discussion in https://math.stackexchange.com/a/4583911/96384.
Once one has that, of course it is clear in hindsight that the $\color{red}{\text{nonzero}}$ $\lambda_i$, i.e. in my above notation the $\lambda_i$ with $r+1 \le i \le n$, were mutually distinct.
After that comes the real fun in showing that $R$ is a root system in a natural way.
I also advertise my own answer here for better understanding of the root space decomposition by seeing it in a basic but telling matrix example. That should clear things up.