Finding complement of linear subspace

Question

Consider $\mathbb{C}^n$ with its usual basis $\{e^1,\ldots,e^n\}$. Let $V$ be the line spanned by the vector $v=\sum_ie^i$. I want to know how to deduce that $W=\{(z_1,\ldots,z_n):z_1+\ldots+z_n=0\}$ is its complementary subspace, i.e. $\mathbb{C}^n=V+W$. So, I know that $$S=\{(1,-1,0,\ldots),(0,1,-1,\ldots),\ldots, (0,\ldots,-1,1)\}$$ is a set of $n-1$ linearly independent vectors whose span intersects $V$ only at $0$. Moreover, Span $S=W$. Thus $\mathbb{C}^n=V+W$.This proves the assertion, but I am looking for a more intuitive way of thinking about $W$. For example, consider any vector $v\in \mathbb{C}^n$, it is of the form $v=\sum_ia_ie^i$, we want to find the space of all vectors $w$ such that $v=b\sum_ie^i+w$, such that $b\in \mathbb{C}$. I am looking to work with this kind of approach.

Answer 1

Consider the standard inner product on $\Bbb C^n$:$$\bigl\langle(z_1,\ldots,z_n),(w_1,\ldots,w_n)\bigr\rangle=\sum_{k=1}^nz_k\overline{w_k}.$$If $U$ is a subspace of $\Bbb C$, then $\Bbb C^n=U\bigoplus U^\perp$. If you put $U=\Bbb Cv$, then $U^\perp=W$.