Let $\{b_1,b_2,\ldots,b_n\}$ be a basis of the vector space $V$ over the field $\Bbb F$ and let
$$L=\left\{x=\sum_{i=1}^n\alpha_ib_i:\sum_{i=1}^n\alpha_i=0\right\}\leqslant V.$$ Find $\dim L$ and some direct complement of $L$ in $V$.
My work:
For arbitrary $\alpha_1,\alpha_2,\ldots,\alpha_{k-1},\alpha_{k+1},\ldots,\alpha_n\in\Bbb F$, $\begin{aligned}\sum_{i=1}^n\alpha_i&=0\\\implies\alpha_k&=-\sum_{i=1\\i\ne k}^n\alpha_i\\\implies x&=\underbrace{\sum_{i=1\\i\ne k}^n\alpha_ib_i}_{v_1}+\underbrace{\left(-\sum_{i=1\\i\ne k}^n\alpha_i\right)b_k}_{v_2}\\&=\sum_{i=1\\i\ne k}^n\alpha_i\left(b_i-b_k\right)\\\implies L&=\operatorname{span}\{b_1-b_k,b_2-b_k,\ldots,b_{n-1}-b_k,b_n-b_k\},\quad i\in\{1,\ldots,n\}\setminus\{k\}\\\implies\dim L&=n-1\end{aligned}$
Now, if we return to $v_1$ and $v_2$, we can see $x=0\iff v_1=v_2=0\iff \alpha_i=0\quad\forall i\in\{1,\ldots,n\}$ because $b_1,b_2,\ldots,b_k,b_{k+1},\ldots,b_n$ are linearly independent, so one direct complement of $L$ in $V$ is $M=\operatorname{span}\{b_k\}$.
May I ask if this is legitimate and if so, is there anything I should include in my solution?
Thank you in advance!
Note: Some data I copied in the first version of the post were inaccurate and I also replaced the notation $[\{\}]$ by $\operatorname{span}\{\}$.