dimension of $M = \{ x \in \mathbb{C}^{n} \ | \ \sum_{i=1}^n x_i=0 \}$

What is the dimension of $M = \{ x \in \mathbb{C}^n \ | \ \sum_{i=1}^n x_i=0 \}$?

New Attempt: Let $f$ be the linear map $f: \mathbb{C}^n \to \mathbb{C}$ defined by $f(x) = \sum_{i=1}^n x_i$, then we notice that $\ker f = M$. By the rank theorem we have: $\text{rg}(f) + \ker{f}= \dim{\mathbb{C}^n} $, thus $\dim{M} = n - \dim{\mathbb{C}} = n -1 $. Is this right?

Solution 1:

Let $(x_{1},x_{2},...,x_{n})$ be an arbitrary vector in $M$. such that $x_{i}\in \mathbb{C}$

Then $\sum_{i=1}^{n}x_{i}=0$ so $x_{n}=-\sum_{i=1}^{n-1}x_{i}$

So it is $(x_{1},x_{2},....,-\sum_{i=1}^{n-1}x_{i}) = x_{1}(1,0,...0,-1)+x_{2}(0,1,...0,-1)+...+x_{n-1}(0,0,...1,-1)$

So $$M=span\left(\left\{(1,0,...0,-1),(0,1,...0,-1),...,(0,0,...1,-1)\right\}\right)$$

The above set $\left\{(1,0,...0,-1),(0,1,...0,-1),...,(0,0,...1,-1)\right\}$ is a basis for $M$ and clearly $M$ has dimension $n-1$.