Find the vector of minimum norm... I'm working on this exercise:

Show that the subset $ M = ${$y= (\eta_j) ;  \sum \eta _j= 1$} of
complex space $\mathbb{C^{n}}$  is complete and convex. Find the
vector of minimum norm in M.

I've proved the completeness and convexity part, but I could not find the minimizing vector. I have the suspicion that it's $(\frac{1}{n},\frac{1}{n},...,\frac{1}{n})$, but I couldn't find an argument.
Could you help me? Thank you in advance.
 A: Your conjecture is correct. You can view this problem as a constraint optimization problem:
$$
\min\sum_j(a_j^2+b_j^2)\quad\text{s.t.}\;\sum_ja_j=1,\;\sum_jb_j=0.
$$
Set $f(a,b) = \sum_j(a_j^2+b_j^2)$, $g(a,b) = \sum_ja_j - 1$, and $h(a,b)=\sum_jb_j$. Then
$$
f'(a,b) = 2(a,b),\quad g'(a,b) = (\mathbf{1},0), \quad\text{and}\quad h'(a,b) = (0,\mathbf{1}).
$$
By Lagrange, if $(a,b)$ is an optimum, there exist numbers $\lambda,\mu$ such that
$$
f'(a,b) = \lambda g'(a,b) + \mu h'(a,b).
$$
Hence,
$$
2(a,b) = \lambda(\mathbf 1,0) + \mu(0,\mathbf 1) = (\lambda,\ldots,\lambda,\mu,\ldots,\mu).
$$
Therefore, we have $a_1=\ldots=a_n$ and $b_1 =\ldots = b_n$. The restrictions $g(a,b)=0$ and $h(a,b)=0$ now imply that $a_j = \frac 1n$ for all $j$ and $b_j = 0$ for all $j$, hence $\eta_j = \frac 1n$ for all $j$.
A: This also follows directly from the Cauchy-Schwarz inequality:
$$
\sum_{j=1}^n \eta_j=\sum_{j=1}^n \eta_j\cdot 1\leq \left(\sum_{j=1}^n |\eta_j|^2\right)^{1/2}\left(\sum_{j=1}^n 1\right)^{1/2}=n^{1/2}\left(\sum_{j=1}^n |\eta_j|^2\right)^{1/2}
$$
With equality if and only if there exists $\lambda\geq 0$ such that $\eta_j=\lambda$ for all $j\in\{1,\dots,n\}$. The constraint $\sum_j \eta_j=1$ is then satisfied if and only if $\lambda=1/n$.
