Vrftsjtryk

Let $U$ be an orthonormal matrix. The dimension of $U$ is equal to $n times n$. The first row of $U$ has the following form:
$$bigg( frac{1}{sqrt{n}}, ldots , frac{1}{sqrt{n}} bigg). tag{1}$$
How can I prove that $U$ with property (1) exists?

Take your row vector (call it $v_1$) and extend this set to a basis of $mathbb{R}^n$. Then orthonormalize this basis, (using Gram-Schmidt), obtaining {$v_1, dots, v_n$}. Then, the matrix U such that its jth row is $v_j$ is an orthonormal matrix.

$u:=(frac{1}{sqrt{n}},ldots,frac{1}{sqrt{n}})$ is a unit vector in your vector space. You can extend it to a basis ${u,v_{2},ldots,v_{n}}$ of $V$ and then apply the Gram-Schmidt algorithm on this basis to obtain an orthonormal basis: ${e_{1},e_{2},ldots,e_{n}}$ of $V$, where $e_{1}=u$. The matrix with rows $e_{1},ldots e_{n}$ will then be a unitary matrix.

Consider the vectors $v_1=e_1+dots+e_n$, $v_i=e_i$ for $i=2,dots,n$ and apply Gram-Schmidt to them. Then you obtain an orthonormal basis whose first vector is $(e_1+dots+e_n)/sqrt n$, which is precisely the vector you want.